## 2nd PUC Political Science Model Question Papers with Answers 2019-20 Karnataka

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## 2nd PUC Electronics Previous Year Question Paper June 2017

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## Karnataka 2nd PUC Electronics Previous Year Question Paper June 2017

### 2nd PUC Electronics Previous Year Question Paper June 2017

Time : 3 hrs. 15 min.
Max Marks: 70

Instructions:

• The question has five Parts A, B, C and D.
• Part – A has no choice
• Part – D has two parts. Part -1 is from problems, Part – II is of essay type questions.
• Circuit diagram/timing diagram/truth tables are drawn wherever necessary.
• Problems without necessary formula/formulae carry no marks.

Part – A

Answer all questions: 10 x 1 = 10

Question 1.
What is amplification factor?
Answer:
It is the ratio of change in drain to source voltage VDS to corresponding change in gate to source voltage VGS for constant drain current ID.

Question 2.
Define CMRR.
Answer:
CMRR is defined as the ratio of differential mode .gain to common mode gain.

Question 3.
Mention the speed of electromagnetic waves in free space.
Answer:
3 x 108 ms-8

Question 4.
Mention the value of IF for AM reception.
Answer:
455 KHz

Question 5.
An FM signal has a freuency deviation of 12 KHz and modulating frequency of 6KHz. Calculate the modulation index.
Answer:
Δ= 12KHz fm = 6KHz mf = ?
$\mathrm{m}_{t}=\frac{\Delta \mathrm{f}}{\mathrm{f}_{\mathrm{m}}}=\frac{12 \mathrm{KHz}}{6 \mathrm{KHz}}$

Question 6.
Name the lightly doped layer in the power diode.
Answer:
Drift layer or epitaxial layer.

Question 7.
What is an octet?
Answer:
An octet is a group of eight adjacent is either in two horizontal rows or in two vertical columns.

Question 8.
Write excess-3 code for (69)10.
Answer:

 Decimal 6 9 BCD 0110 1001 Add0011 0011 0011 Excess 3 code 1001 1100

Question 9.
Why 8051 microcontroller is known as 8 bit processor?
Answer:
8051 microcontroller is called 8 bit processor because it processes 8 bits of data.

Question 10.
What is the size of an integer in C-programming?
Answer:
2 bytes.

Part – B

Answer any five questions. 5 x 2 = 10

Question 11.
Write the steps involved in drawing DC equivalent circuit of a CE amplifier.
Answer:

• Reduce all the AC sources to zero.
• Open all the capacitors.

Question 12.
A amplifier of gain 500 reduces to 100 after feedback. Calculate the feedback factor. Answer:
A=500,Af= 100, β = ?

Question 13.
Distinguish between sinusoidal and non-sinusoidal oscillator.
Answer:

• Sinusoidal oscillator: Oscillator output is a sinusoidal waveform
• Non-sinusoidal oscillator: Oscillator output is a non-sinusoidal waveform.

Question 14.
Mention any two applications of power devices.
Answer:

• AC to DC rectifier.
• AC to AC voltage controller.
• DC to DC chopper
• DC to AC inverter.

Question 15.
A silicon power diode has Vj of 0.4V, R0N in drift region of 0.0020 and Ir = 75A
Answer:

Question 16.
Briefly explain the function of accumulator and program counter in microcontroller 8051.
Answer:

• Accumulator : It stores a number and on receipt of another number, adds the two and stores the sum. It can also sense the signal, clear and complement etc.
• Program counter: It holds the address of a byte in memory. It also specifies the address at the next instruction to be fetched and executed.

Question 17.
Write the meanings of the following operators in C programming,
(i) = = (ii) &&
Answer:
(i) Equal to
(ii) LoginalAND

Question 18.
Mention any two applications of RADAR.
Answer:

• Used as altimeters to measure height
• Used in defence weapons system
• Used in landing of planes in bad weather conditions.

Part – C

Answer any five questions.

Question 19.
Explain the construction of n-channel JFET.
Answer:

Consider n channel JFET with gate to source voltage VGG and drain to source voltage VGG. When VGG is zero, drain current is maximum and flows from drain to source called as Lcc. When V0G is increased, reverse bias across junction increases and width of depletion region across the junctions increases. Due to this channel width decreases and hence the drain current decreases. Thus the reverse voltage applied across gate to source terminal controls the drain current. When the gate to source voltage is further increased, a stage will be reached at which two depletion regions touch each other and the drain current becomes zero. Pinch off voltage is the gate voltage at which drain current becomes zero.

Question 20.
Explain the terms thermal runaway, leakage current and heat sink.
Answer:
The self destruction of an unbiased transistor due to increase in temperature and hense increase in leakage current.

Heat sink is a device which absorbs unwanted heat generated in the transistor and radiates to the surroundings. A copper conductor with different shapes are connected to the device to take out heat from the device. With heat sink, power rating of the device increases.

Leakage current is current through a device due to the motion of minority charge carriers under reverse bias conditions.

Different leakage currents are:

• Collector to base leakage current with emitter open (ICB0).
• Collector to emitter leakage current with base open (1CE0).

Question 21.
Derive an expression for input impedance of negative feedback amplifier.
Answer:
An amplifier should have high input impedance so that it will not load the preceding stage or input voltage source. High input impedance for an amplifier can be achieved with Voltage series negative feedback.

Let Ii and Vi be input current and input voltage to the basic amplifier. Vs is input to feedback amplifier, Vf is feedback voltage and vo is the output voltage.

Input impedance of basic amplifier is
$z_{i}=\frac{V_{i}}{I_{i}}-(1)$

Input impedance of feedback amplifier is
$z_{v}=\frac{V_{s}}{I_{i}} \rightarrow(2)$

Input voltage to basic amplifier with negative feedback is

$=\left(\frac{V_{i}}{I_{i}}\right)(1+A \beta)$
Zif = Zi (1+Aβ)from eq ………… (1)
As (1 + Aβ) > 1 for negative feedback, Zif > Zi
Hence with negative feedback, input impedance of an amplifier increases.

Question 22.
Define critical angel, critical frequency and skip distance
Answer:
Skip distance is the minimum distance measured between transmitting antenna and the first receiving antenna measured along the surface of the earth after propagation through the ionosphere.

Critical angle for a given layer is the maximum angle of incidence of a radio wave under ionospheric propagation that gets returned to the earth.

Critical frequency for a given layer is the highest frequency of a beamed radio wave that will be returned to the earth.

Question 23.
What is demodulation? Draw the circuit diagram of linear diode detector.
Answer:
Demodulation is the process of recovering a modulating signal from the modulated wave.

LC tank circuit is used as a parallel resonant circuit. The resonant frequency of the circuit can be varied by varying the value of capacitance C and hence RF signal of any desired frequency can be tuned in.

When a selected modulated signal is applied to the diode, the diode conducts only during the positive half of the modulated wave. Thus the diode removes the entire negative half cycles. Across R1C1 only positive half cycles of carrier modulated wave appears. Thus the low pass filter made up of R1 C1 removes the RF.

During positive half cycle, diode conducts the capacitor C1 gets charged. During negative half cycle of the carrier and diode doesnot conduct but C1 discharges through R1. Hence output is obtained. The spikes can be reduced by proper choice of R1 C1 and depth of modulation. The capacitor CB removes Dc component produced by the detector.

Question 24.
Explain non-punch through type power diode.
Answer:
In non-punch through diodes, the depletion region boundary does not reach the end of drift layer. In punch through diode, the depletion layer spans the entire drift region and is in contact with n+ cathode.

Question 25.
Draw the logic circuit of a full-adder using 3 input XOR gate. Write the Boolean expression for outputs SUM and CARRY.
Answer:
Full adder is a combinational logic circuit that performs arithmetic sum of three bits and gives their sum and carry .

Question 26.
List the additional features of 3G and 4G cell-phone systems.
Answer:
Additional features of 3G and 4G phone systems are:

1. Colour LCD screens
2. Digital cameras
3. E-mail, Games
4. GPS, Bluetooth
5. Internet access
6. Video conferencing

Part – D

Answer any Five questions. 3 x 5 = 16

Question 27.
For CE amplifier circuit using Germanium transistor given below. Find
i) Voltage across 10KQ
ii) IE
iii) re
iv) Av [Given VBE = 0.32V]
Answer:

Question 28.
Calculate output voltage if V1 = 300mV and V2 = 700mV.

Answer:

Question 29.
A colpitts oscillator oscillates at 1.13MHz. If inductor in the feedback network has a value of 20μH and one of the capacitor value is 0.1 μF. Calculate the value of the other capacitor.
Answer:

Question 30.
A sinusoidal carries voltage Vc = 80 sin 2π x 105 t is amplitude modulated by a sinusoidal voltage
Vm = 32 sin 2π x 103t . Write the equation of the AM wave and draw its output frequency spectrum.
Answer:

Question 31.
Simplify the given equation using K-map. Y = Σm (1, 3, 4, 6, 9) + Σd (11,12,14) and identify the logic gate for the simplified output expression.
Answer:

Part – E

Answer any Four questions.

Question 32.
What is direct-coupled amplifier? With neat circuit diagram explain the working of direct-coupled amplifier and draw the frequency response curve.
Answer:

The direct coupled amplifier using two npn transistors is as shown in the diagram. In this method, output of first stage is directly coupled to the input of second stage. Resistors R1 and R2 form voltage divider circuit. Any signal at the base of the first transistor is amplified A, times and appears at the collector of first stage.

It is then applied to the base of transistor in the second stage. Hence it is amplified A2 times. The gain of the direct coupled amplifier is A = A x A. Direct-coupled amplifiers are used to amplify extremely low frequency signals.

Frequency Response of Direct coupled transistor amplifier.

Question 33.
What is op-amp subtractor? Derive an expression for the output of op-amp subtractor with a circuit diagram.
Answer:

Voltage follower is Obtained by shorting Rf. The output is directly fed to inverting input terminal. Output voltage V of non-inverting

It is used as buffer amplifier for impedance matching.

Question 34.
With a neat block diagram explain the working of FM transmitter.
Answer:

The modulating signal is applied to the pre-emphasis circuit, which improves signal to noise ratio, AF amplifier amplifies output of pre-emphasis. The processed signal is fed to reactance modulator. The reactance modulator uses a transistor or FET connected across tank circuit of carrier oscillator.

The oscillator frequency depends on the tank reactance which in turn depends on the instantaneous amplitude of modulating signal. Thus, output of reactance modulator will be a frequency modulated wave.

The oscillator is followed by a buffer amplifier which isolates oscillator from subsequent stages. The limiter maintains the amplitude constant. Class C power amplifier amplifies modulated wave to required power levels. The FM signal is then fed to the transmitting antenna.

Question 35.
Explain the working of JK flip-flop with logic circuit and truth-table. Draw its timing diagram.
Answer:

1. When J = 0, K= 0, the S and R inputs are both at 0 and hence the output is in the HOLD state.
2. When J = 0, K = 1, S input is at 0 while R can be either at 0 or 1, but output is always in the RESET state.
3. When J = 1, K = 0, S input can be 0 or 1 but R input is always at logic 0 and hence the output remains at a stable SET state.
4. When J = 1, K = 1 the Flip flop goes to complementary’ state of previous output i.e, flip flop toggles

Question 36.
Write an assembly language program to add two numbers 1FH and B4H and store the result in R7. Verify the result by binary addition.
Answer:

Question 37.
What is debugging? Explain the different types of error that occur in C programming language?
Answer:
The process of detecting and correcting errors in the program is debugging.
C is a powerful, flexible, portable and structured programming language.

## 2nd PUC Electronics Previous Year Question Paper March 2017

Students can Download 2nd PUC Electronics Previous Year Question Paper March 2017, Karnataka 2nd PUC Electronics Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka 2nd PUC Electronics Previous Year Question Paper March 2017

### 2nd PUC Electronics Previous Year Question Paper March 2017

Time : 3 hrs. 15 min.
Max Marks: 70

Instructions:

• The question has five Parts A, B, C and D.
• Part – A has no choice
• Part – D has two parts. Part -1 is from problems, Part – II is of essay type questions.
• Circuit diagram/timing diagram/truth tables are drawn wherever necessary.
• Problems without necessary formula/formulae carry no marks.

Part – A

Answer all questions: 10 x 1 = 10

Question 1.
Write the function of source in JFET.
Answer:
Source supplies charge carriers for current conduction.

Question 2.
Name the power amplifier in which conduction angle is 360°.
Answer:
Class A.

Question 3.
Define slew rate.
Answer:
Slew rate is the rate of change of output.

Question 4.
What is the frequency of modulating signal if the bandwidth of AM wave is 10 KHz?
Answer:
$f_{\mathrm{m}}=\frac{\mathrm{BW}}{2}=\frac{10 \mathrm{KHz}}{2}=5 \mathrm{KHz}$

Question 5.
Define demodulation.
Answer:
Demodulation is the process of recovering original modulating signal from the modulated wave.

Question 6.
Draw the circuit symbol of IGBT.
Answer:

Question 7.
What is a counter?
Answer:
Counter is a logic circuit for counting the pulses.

Question 8.
Convert (1101)2 into gray code.
Answer:
(1011)G

Question 9.
How many interrupt sources are there in 8051 microcontroller?
Answer:
6 including RESET or 5 excluding RESET.

Question 10.
Write C equivalent expression for the mathematical expression $\sqrt{a^{2}+b^{2}}$
Answer:
sqrt (a*a+ b*b)

Part – B

Answer any five questions.

Question 11.
Write any two differences between FET and BJT.
Answer:

 S.No. BJT JFET 1. It is a bipolar device. It is a unipolar device. 2. It is a current controlled device. It is a voltage controlled device. 3. Current conduction is by both holes and electrons. Current conduction is by holes or electrons. 4. Input resistance is low. Input resistance is very high. (MCI) 5. BJT is more noisy. JFET is less noisy. 6. Switching speed is less. Switching speed is high. 7. Fabrication in IC is difficult. Fabrication in IC is simpler.

Question 12.
Mention the steps involved to obtain DC equivalent circuit of an amplifier.
Answer:

• Reduce all the AC source to zero.
• open all the capacitors.

Question 13.
Distinguish between positive feedback and negative feedback.
Answer:

• Positive feedback is the process of applying a portion of output in phase with the input.
• Negative feedback is the process of applying a portion of output out of phase with the input.

Question 14.
Write any two advantages of RC oscillators over LC oscillators.
Answer:

• RC oscillators provide constant output and good stability.
• RC oscillators do not require any inductor and hence size and cost get reduced.

Question 15.
Write Shockley’s diode equation for current through the power diode and explain its terminology.
Answer:
$I=I_{ s }\left( e^{ \frac { qv }{ KT } }-1 \right)$

• Is = Reverse saturation current
• T = Absolute temperature
• K = Boltzmann constant
• q = charge of electron
• V = voltage across diode.

Question 16.
Determine average value of DC from chopper. Given T= 2 ms, TQN= 0.5 ms and supply voltage is 24V.
Answer:
Duty ratio = $\frac{T_{o N}}{T}=\frac{0.5}{2}=0.25$
Vdc = Duty ratio x Vs = 0.25 x 24 = 6V.

Question 17.
Realise NOT and OR-gate using NOR-gate
Answer:

Question 18.
Distinguish between uplink and downlink signals
Answer:

 Uplink Downlink 1)  Signal transmitted from ground station to satellite 2) Frequency range is 6 GHz 1) Signal transmitted from satellite to ground station. 2) Frequency range is 4 GHz.

Part – C

Answer any five questions.  5 x 3 = 15

Question 19.
What is a DC load line? Mention any two advantages of voltage divider bias.
Answer:
Operating point is the zero signal values of collector Ic and collector to emitter voltage VCE of a transistor. DC load line is a straight line drawn on the output characteristics representing output DC voltage and corresponding direct current under no signal conditions.

The operating point does not depend upon the value of β of the transistor. Hence the operating point does not change its position due to rise in temperature or replacing a transistor of different p value. Hence voltage divider bias provides excellent stabilisation and is preferred to other biasing methods.

Question 20.
Calculate the input impedance of a negative feedback amplifier if input impedance without feedback is 10 KQ. Given A = 100 and β = 0.01.
Answer:
Zif = Zi(1 +Aβ)
= 10 x 103 (1 + 100 x 0.01)
= 20 x 103 Ω
= 20kΩ

Question 21.
With block diagram, explain basic communication system.
Answer:

Question 22.
Draw the equivalent circuit of transmission lines for low frequency. Mention any two types of antennas.
Answer:

Types of Antenna:

• Helical antenna
• Yagi antenna
• Loop antenna

Question 23.
Write the circuit diagram, input and output waveforms of SCR halfwave rectifier with RC triggering circuit.
Answer:

The SCR is connected with a diode bridge rectifier. The rectifier circuit converts AC to a unidirectional current through the thyristor. The voltage on the capacitor follows the diode rectifier waveform because RC circuit is connected to diode rectifier. During each half cycle when the voltage across the capacitor reaches trigger level which is just enough to supply gate trigger current to thyristor, then thyristor starts conducting. Firing angle $\alpha=\left(\frac{x}{\mathrm{y}}\right)_{180}$

Question 24.
Draw the labelled block diagram of microcontroller.
Answer:

Question 25.
How do you represent.
i) logical AND
ii) logical OR
iii) logical NOT operators in C programming?
Answer:
Internet is worldwide interconnection of millions of computers by means of complex network which serves as an inexpensive communication medium.

1. Long-distance telephone and cable TV systems.
2. To provide LAN connections.
3. Closed-circuit TV systems used for security.
4. Aircraft communication and controls.

Question 26.
What is internet? Write few applications of fibre optic communication system.
Answer:
Internet

1. Using internet, access to required information is easy, economical, reliable and fast.
2. With E-mail, messages reach destination with in minutes.
3. E-mail can be sent and received from any place.
4. Using E-mail, message can be sent to several persons simultaneously.

Applications

1. Long-distance telephone and cable TV systems.
2. To provide LAN connections.
3. Closed-circuit TV systems used for security.
4. Aircraft communication and controls.

Part – D

Answer any Three questions:   3 x 5 = 15

Question 27.
For the given CE amplifier circuit using silicon transistor. Calculate
i) Voltage across R2
ii) IE
iii) Zin(base)
iv) Zin (and)
v) Zo
Given $\text { Given } r^{\prime} e=\frac{26 m V}{I_{E}}$ and β = 100

Answer:

Zin = R1 || R2|| Zin(base) =1.38kΩ = 2kΩ
Zb =Rc || RL =(10kΩ) || (10kΩ) = 5kΩ

Question 28.
Calculate the output voltage if V1= 300mV and V2 = 700 mV in the following circuit.

Answer:

Question 29.
A Hortley oscillator oscillates at 15 KHz. If the capacitor in the tank circuit has a value of 0.01 pF and one of the inductors value is 1 mH, calculate the value of the other inductor.
Answer:

L1 + L2 = LT= 11.25 mH
L,1+ 1mH= 11.25mH, L1 = 10.25mH

Question 30.
A frequency modulated signal is given by 10sin[6 x 108t + 5 sin 1250t].
Determine
i) Carrier frequency
ii) Modulating frequency
iii) Modulation index
iv) Maximum deviation
v) Carrier swing
Answer:

Maximum frequency deviation = δ= mf fm. = 5 x 199.04= 995.2 Hz
Carrier swing = 28 = 2(995.2) = 1990.4 Hz.

Question 31.
Simplify the Boolean expression Y = sm(l,2,3,8,9,11,13) + sd(0,10,15) using K-map. Draw the logic diagram using NAND-gates to realise the simplified expression.
Answer:

Part – E

Answer any Four questions.  4 x 5 = 20

Question 32.
Give the comparison between CB, CC and CE amplifiers on the performance characteristics.
Answer:

 Parameter Types of Amplifiers. CE CB CC Current gain High (P) Less than 1 Highest (1+ p) Voltage gain High Moderate Low Power gain Highest Moderate Moderate Phase shift 180° 0° 0- Input impedance Moderate Low High Output impedance Moderate High Low Bandwidth Narrow Wide Wide Applications Used as AF amplifies Used as RF amplifies impedance matching

Question 33.
What is a logarithmic amplifier? With circuit diagram, obtain an expression for output voltage of logarithmic amplifier using OP-amp.
Answer:
Answer:

∴ The output voltage is proportional to natural logarithm of input voltage.

Question 34.
With a labelled block diagram, explain the working of AM transmitter.
Answer:
The transmitter performs modulation and raises the power level of a modulated wave to required level for effective radiation.

The microphone converts the audio signal into an equivalent audio electrical signal. The output is filtered to have a bandwidth of 10kHz.

The pre amplifier provides the required voltage amplification. The driver amplifier increases the power level of the signal as required by the high power modulation amplifier.

The high frequency carrier wave is generated by a crystal oscillator which has highest frequency stability. The Carrier signal is fed to a buffer amplifer which is a low gain, high input impedance amplifer which isolates crystal oscillator and the power level of the carrier signal to drive the modulated class C amplifier.

The modulating audio signal and carrier signal are applied to the modulator. The collector modulation is used for modulation in high power transmitter. The transmitting antenna transmits the signal.

Question 35.
Explain with circuit diagram and truth table, the working of clocked SR flip flop using NAND gates.
Answer:

It is constructed using inverters inserted into inputs of cross coupled NAND gates.
Working:

1. When S=0, R=0, it does not respond and hence outputs Q and Q will remain in their previous state. This is called HOLD condition.
2. When S=1 and R=0, the output Q and Q change to 1 and 0. This condition is called SET state.
3. When S=0, R=1, the output Q and Q change to 0 and 1. This condition is called RESET state.
4. When S=1 and R=l, it drives output Q and Q both to HIGH which is FORBIDDEN or INVALID condition.

Question 36.
Write an assembly language instructions to move 45H into register A and 5 EH into register R0. Then add them together and save the result in Rt. What is the content of R, after the execution of the program.
Answer:

Question 37.
Write a C program to accept three integer and find their sum and average. Find the sum and average of 5, 6 and 7.
Answer:

## 2nd PUC Electronics Previous Year Question Paper June 2018

Students can Download 2nd PUC Electronics Previous Year Question Paper June 2018, Karnataka 2nd PUC Electronics Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka 2nd PUC Electronics Previous Year Question Paper June 2018

### 2nd PUC Electronics Previous Year Question Paper June 2018

Time : 3 hrs. 15 min.
Max Marks: 70

Instructions:

• The question has five Parts A, B, C and D.
• Part – A has no choice
• Part – D has two parts. Part -1 is from problems, Part – II is of essay type questions.
• Circuit diagram/timing diagram/truth tables are drawn wherever necessary.
• Problems without necessary formula/formulae carry no marks.

Part – A

Answer all questions: 10 x 1 = 10

Question 1.
Expand JFET.
Answer:
Junction Field Effect Transistor.

Question 2.
Define CMRR.
Answer:
CMRR is the ratio of differential mode gain to common mode gain.

Question 3.
How many frequency components are present in amplitude modulated wave?
Answer:
3

Question 4.
Mention any one application of digital communication system.
Answer:
TV transmission or mobile signal transmission.

Question 5.
Draw the circuit symbol of TRIAC.
Answer:

Question 6.
Write Excess-3 code of (26)10.
Answer:
(26)10 = (010110O1)2

Question 7.
What is a Full-Adder?
Answer:
It is circuit that performs arithmetic sum of 3 binary bits.

Question 8.
Give the meaning of CLR A.
Answer:
Clear the content of Accumulator.

Question 9.
Write the C-equivalent expression for the mathematical expression
Answer:
(X * Y) / Z

Question 10.
What is Transponder?
Answer:
Transponder is transmitter receiver combination in a satellite.

Part – B

Answer any five questions.  5 x 2 = 10

Question 11.
Define amplification factor. Write its relation with rd and gm.
Answer:
Amplification factor is the ratio of small change in drain to source voltage to the corresponding change in Gate to source voltage for a constant drain current.
ICBO and ICEO

Question 12.
Write the steps involved in drawing AC equivalent circuit of an amplifier.
Answer:

• Reduce all dc sources to zero.
• Short all capacitors.

Question 13.
The open loop gain and closed loop gain of an amplifier are 100 and 50 respectively. Calculate the feedback fraction.
Answer:
Aβ = 100
Af = 50
$\mathrm{Af}=\frac{\mathrm{A}}{1+\mathrm{A} \beta}$
β = 0.01

Question 14.
Write any two advantages of negative feedback.
Answer:

• Stabilisation of Gain
• Increase in the bandwidth

Question 15.
Draw the neat circuit diagram of Wein-bridge oscillator.
Answer:

Voltage gain A = $1+\frac{\mathrm{R}_{3}}{\mathrm{R}_{4}}=3$

Question 16.
Explain double-injection meahanism in power diode.
Answer:
At higher forward bias, the holes from p+ region reach n~ n+junction and attract electrons from n+ cathode. This leads to electron injection into n~ drift region from n+ cathode layer. This mechanism is called double injection method.

Question 17.
Realize XOR gate using only NAND gates.
Answer:

Question 18.
Mention any two applications of bluetooth.
Answer:

• Wireless communication between PC or laptops or cell phones.
• Used in wireless audio head sets.

Part – C

Answer any five questions.  5 x 3 = 15

Question 19.
What is meant by leakage current? Name two types of leakage currents.
Answer:
The flow of current through the device due to the motion of minority charge carriers under reverse bias condition is called leakage current.
ICBO and ICEO

Question 20.
What is Ionosphere? Explain F-layer.
Answer:
Ultraviolet radiation from the sun cause the upper atmosphere to ionize i.e. to become electrically charged. Hence a thick layer of ions formed at heights of 50 km to 400km. The ionosphere supports MF and HF wave propagation.
Its three layers are:

D layer: It is at a height of 70 km and has an average thickness of 10 km. It disappears in the night time. It reflects VLF and LF waves and absorbs MF and HF waves.

E layer: It is at a height of 100 km with a thickness of 25 km. This layer disappears at night. It helps in MF and HF wave propagation.

F layer: It is at a height of 150 km and extends upto 400 km. This layer exists during both day and night time. During day time, it splits into two layers F, and F2, and combines to form a single F layer during night time.

 F2 region. 250-400 km. F1 region. 160-250 km. E region. 95-130 km D region. 50-95 km. Troposphere Earth.

Question 21.
With a neat circuit diagram explain linear diode detector.
Answer:
LC tank circuit is used as a parallel resonant circuit. The resonant frequency of the circuit can be varied by varying the value of capacitance C and hence RF signal of any desired frequency can be tuned in.

When a selected modulated signal is applied to the diode, the diode conducts only during the positive half of the modulated wave. Thus the diode removes the entire negative half cycles. Across R1C1 only positive half cycles of carrier modulated wave appears. Thus the low pass filter made up of R1 C1 removes the RF.

During positive half cycle, diode conducts the capacitor C1 gets charged. During negative half cycle of the carrier and diode doesnot conduct but C1 discharges through R1. Hence output is obtained. The spikes can be reduced by proper choice of R1 C1 and depth of modulation. The capacitor CB removes Dc component produced by the detector.

Question 22.
Draw the circuit diagram, gate pluses and load voltage waveform of power inverter.
Answer:

Question 23.
Determine Vdc and Idc of SCR full wave frectifier. Given firing angle is 60°. The rms voltage of ac input to the rectifier is 150V, when load resistance of 15Ω is connected.
Answer:

Question 24.
Name the addressing modes of the following instructions:
(i) MOV R2, A
(ii) MOV A, 25H
(iii) MOV A, @ R0.
Answer:
(i) Register Addressing mode
(ii) Direct Addressing mode
(iii) Indirect Addressing mode

Question 25.
What is Debugging in C-program? Write the syntax of the if else statement.
Answer:
Debugging is detecting and correcting errors.
if (condition)
{
statement – 1 ;
}
else
{
statement -2;
}

Question 26.
Draw the block diagram of RADAR system and mention the function of duplexer.
Answer:

• Transmitter: The radar transmitter produces high-frequency RF pulses and is transmitted into space by the antenna.
• Duplexer: It switches the antenna between transmitter and receiver alternately. This is required because high power pulses of the transmitter will destroy the receiver if energy were allowed to enter the receiver.
• Receiver: It amplifies and demodulates the received RF signals. It provides video signals on the output.
• Radar antenna: The antenna transfers the transmitting signal to space with the required distribution and efficiency.

Part – D

Answer any Three questions:  3 x 5 = 15

Question 27.
For the CE amplifier circuit shown below, find:
(i) Voltage drop across 8.2 Ω
(ii) IZ
(iii) re
(iv) Zin(base)
(v) Z out

Answer:

Question 28.
Find the output voltage ‘V0‘ for the following circuit: Given: V1, = IV, V2 = 2V, V3 = -1 V

Answer:

Question 29.
Calculate the frequency and feedback fraction of the circuit shown below:

Answer:

Question 30.
A 100 KW carrier power is amplitude modulated at 50% depth of modulation by a sinusoidal modulating signal. Calculate:
(i) Total power
(ii) Sideband power
(iii) Power in each sideband.
Answer:

Question 31.
Simplify the following Boolean expression by using K-MAP. If(A, B, C, D) = Σm(0, 1, 2, 4, 5, 8, 9, 10, 12, 13) Draw the logic circuit for the simplified expression using only NAND gates.
Answer:
.

Part – E

Answer any Four questions.   4 x 5 = 20

Question 32.
Explain with circuit diagram the working of class-B Push-Pull power amplifier.
Answer:

Class B push-pull amplifier is as shown in the diagram. T1 and T2 are two centre tapped transformers and Q1, Q2 are two identical transistors. Transformer T1 produces signal voltages V1 and V2 which are 180° out of phase with each other.

These two signals are applied to the Polarities reverse during negative half cycle of input voltage. Q2then is ON and Q1 is OFF. Q2amplifies the signal and the alternate half cycle appears across the loud speaker.

two transistors. Transformer T2 couples AC output signal from collector to loud speaker. The two emitters are connected to centre tap of transformer T1 secondary and Vcc to the centre tap of T2 secondary.

During positive half cycle of input voltage, secondary winding of T1 has voltages V, and V2. Transistor Q1, conducts and Q2 is cut off. The collector current through Q1 produces an amplified and inverted voltage which applied to loud speaker through a transformer.

Question 33.
What is a differentiator? With the help of circuit diagram obtain an expression for
output voltage of OP-AMP differentiator.
Answer:
Answer:

Question 34.
Along with waveforms, derive an expression for instantaneous voltage of frequency modulated wave.
Answer:
Frequency modulation is the process of varying the frequency of high frequency carrier wave in accordance with instantaneous amplitude of the modulating signal.
Let the modulating signal be given by vm=vm coswt.
Where w is the angular frequency and vm is the peak amplitude. Let the carrier voltage be given by the equation vc= vcsin (Wct +Ø) where wc is the angular frequency of the carrier and Vc is the peak amplitude of the carrier phase angle θ = Wct + θ.
This is instantaneous phase angle of carrier voltage.
Then carrier voltage Vc = Vc sin Ø
The angular frequency $\omega _{ c }=\frac { d\emptyset }{ dt }$
or Ø = ∫ ωdt
The angular frequency of carrier after modulation is
ω = ω + kvm
ω = ω+ kVmcos ωt.
Where K is a constant of proportionality called as the modulation deviation constant. On integrating,

Question 35.
Explain the working of four bit synchronous up counter with the help of logic diagram. Write its truth table.
Answer:
Counter is a logic circuit used for counting the pulses. It is a set of flip flops whose state changes in response to pulses applied to their inputs.

Counters are of two types:

• Synchronous counters and
• Asynchronous counters.

In a synchronous circuit, the clock signal is applied to all flip flops simultaneously.

Question 36.
Write an assembly language program to multiply 08 H and OB H. What are the contents of register A and register B after the execution of the program?
Ans.
MOV A, # 08H → Load the number 08H into the accumulator.
MOV B, # OBH → Load the number OBH into register B.
MOV AB → 08H  OBH
08H= 1000(2)              0BH=1011(2)
1000 x 1011 =01011000(2) = 58H
∴ A = 58H    B = OOH

Question 32.
Write a C-program to find the sum of first n
Ans.
# include <stdio.h>
# include < conico.h>
main ( )
{
int n, i, sum = 0;
printf (” Enter the value of n”);
scanf(” % d”, & n);
i= i;
while (i< = n)
{
sum = sum + i; i + +;
}
printf(“The sum = %d”, sum); getch  ( );
}

## 2nd PUC Electronics Previous Year Question Paper March 2018

Students can Download 2nd PUC Electronics Previous Year Question Paper March 2018, Karnataka 2nd PUC Electronics Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka 2nd PUC Electronics Previous Year Question Paper March 2018

### 2nd PUC Electronics Previous Year Question Paper March 2018

Time : 3 hrs. 15 min.
Max Marks: 70

Instructions:

• The question has five Parts A, B, C and D.
• Part – A has no choice
• Part – D has two parts. Part -1 is from problems, Part – II is of essay type questions.
• Circuit diagram/timing diagram/truth tables are drawn wherever necessary.
• Problems without necessary formula/formulae carry no marks.

Part – A

Answer all questions: 10 x 1 = 10

Question 1.
Define Pinch off voltage.
Answer:
Pinch off voltage is the drain to source voltage in JFET at which the drain current reaches constant satenation level.

Question 2.
What is the input impedance of an ideal op-amp?
Answer:
Input impedance, Zin = ∞

Question 3.
Mention the number of sidebands present in an AM wave.
Answer:
2 sidebands.

Question 4.
Define frequency modulation.
Answer:
Frequency modulation is the process in which the frequency of the carrier is varied in accordance with the modulating signal, keeping amplitude and phase of carrier wave constant.

Question 5.
Expand MOSFET.
Answer:
Metal Oxide Semiconductor Field Effect Transistor.

Question 6.
Convert (1001)G to binary code.
Answer:
(1001)G = (1110)2

Question 7.
Draw the symbol of XOR-gate.
Answer:

Question 8.
Write the meaning of MOV Rs, A.
Answer:
Copy the content of accumulator (register A) to R5

Question 9.
If a = 6 and b = 9, what is the content of a after the execution of a+ = b in C programming?
Answer:
a = 6 + 9 = 15

Question 10.
What is cell splitting?
Answer:
Cell splitting is the process of dividing the geographical area into many {smaller areas.

Part – B

Answer any Five questions.  5 x 2 = 10

Question 11.
Derive the relation μ = gm x rd
Answer:

• Ac drain resistance (rd) is the ratio of small change in drain to source voltage VDS to the corresponding change in drain current (ID) for a constant VGS.
• Transconductance (gm) is the ratio of change in drain current ID to the corresponding change in gate to source voltage VGS for a constant VDS.

Question 12.
Mention any two characteristics of CC amplifier.
Answer:

• Voltage gain is nearly equal to 1
• Input and output signals are in phase

Question 13.
Draw the block diagrams of voltage shunt and current series negative feedback.
Answer:
1.Voltage series negative feedback.

2. Voltage shunt negative feedback:

3.Current series negative feedback:

4. Current shunt negative feedback:

Question 14.
An amplifier has Z0 = 10KQ, voltage gain A = 150 and P = 0.02. Find the output impedance of feedback amplifier.
Answer:

Question 15.
Write the pin diagram of IC-555 timer.
Answer:

Question 16.
Mention the four modes of operation of TRIAC.
Answer:
A triac can conduct in both the directions. The conduction of a triac is initiated by injecting a current pulse into the gate terminal. The gate loses control over the conduction once TRIAC is turned ON. The TRIAC turns OFF only when the current through the main terminals becomes zero.

The three terminals of TRIAC are marked MT, (Main terminal 1), MT2 (Main terminal 2) and the gate G. As TRIAC is a bidirectional device and can have its terminals at various combinations of positive and negative voltages, there are four possible electrons potential combinations, as given below:

1. MT2 positive with respect to MT1, G positive with respect to MT1.
2. MT2 positive with respect to MT1, G negative with respect to MT1.
3. MT2 negative with respect to MT1, G negative with respect to MT1.
4. MT2 negative with respect to MT1, G positive with respect to MT1.

In trigger mode-1, the gate current IG, flows mainly through P2N2 junction like an ordinary thyristor. When the gate current has injected sufficient charge into P2 layer, the TRIAC starts conducting through the P1, N1, P2, N2, layers like an ordinary thyristor.

In trigger mode-3, the gate current IG forward biases the P2 P3 junction and large number of electrons are introduced in the P2 region by Nr Finally the structure P2 N1, P1, N4 turns on completely.

Question 17.
What is a half subtractor? Write its truth table.
Answer:
Half subtractor.

 Inputs Outputs A B Difference Borrow 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 0

Question 18.
Give any two applications if Wi-Fi.
Answer:

1. Wi-Fi is used to access the internet via hotspots with a laptop computer.
2. It is one of the most reliable, flexible and affordable wireless networking technology.
3. Wi-Fi allows cheaper deployment of local area network in outdoor areas like historical buildings.
4. The price of chipsets for Wi-Fi continues to drop, making it an economical networking operation.
5. The standard Wi-Fi device will work anywhere in the world.

Part – C

Answer Any FiveQuestions

Question 19.
Define the terms:
(i) Thermal runaway
(ii) Leakage current
(iii) Heat sink.
Answer:
(i) The self destruction of an unbiased transistor due to increase in temparature and hence increase in leakage current is called thermal runaway.

(ii) Leakage current is the current through a device due to the motion of minority charge carriers under reverse biased condition.

(iii) Heat sink is a device which absorbs the unnecessary heat generated in the transistor and radiates it to surroundings and protects it.

Question 20.
Explain the three modes of radio wave propagation.
Answer:

• Ground Waves
• Space Waves
• Skywaves

Question 21.
Draw the equivalent circuit of transmission lines for high frequency. Mention any two uses of microstrip antenna.
Answer:
For low frequency

Primary constants:

• The series resistance per unit length R is proportional to the square root of frequency.
• The inductance L per unit length for a 2 wire T-line is $\mathrm{L}=\frac{\mu}{2} \log \left(\frac{2 \mathrm{D}}{\mathrm{d}}\right)$ where pt is the permeability.D is distance between two wires. And d is diameter of each wire.
• The capacitance C per unit length,
$C=\cfrac{\mu}{\log \frac{2 D}{d}}$
• Where e is the permittivity of free space, D is distance between 2 wires and d is the distance between the plates of capacitor.

Secondary constants of T-line:

• The series impedance of T-line is Z=R+jwL.
• The shunt admittance of T-line is y=G+JWC are the two secondary constants.

Question 22.
Explain the operation of power diode under forward biased condition.
Answer:
Power diodes of largest power rating are required to conduct several kilo amps of current in the forward direction with very little power loss while blocking several kilovolt in the reverse direction. Large blocking voltage requires wide depletion region. Charge density in the depletion layer should also below in order to get a wide depletion region. For these two requirements, drift layer is introduced between two heavily doped p and n layers.
$\mathrm{V}_{\mathrm{m}}=\sqrt{2} \mathrm{V}_{\mathrm{rms}}=\sqrt{2} \times 30=42.42 \mathrm{V}$

Question 23.
Determine Vdc and Idc of SCR half wave rectifier. Given firing angle is 30° and rms voltage of ac input to the rectifier is 30V and load is 10Ω.
Answer:

Question 24.
Mention any three features of PIC microcontroller.
Answer:
24/35 I/O pins with individual direction control:

• High current source/sink for direct LED drive
• Interrupt-on-Change pin
• Individually programmable weak pull-ups
• Ultra Low-Power Wake-up (ULPWU)

Analog Comparator module with:

• Two analog comparators
• Programmable on-chip voltage reference (CVREF) module (% of VDD)
• Fixed voltage reference (0.6V)
• Comparator inputs and outputs externally accessible
• SR Latch mode
• External Timer 1 Gate (count enable)

A/D Converter:

• 10-bit resolution and 11/14 channels
• Timer O: 8-bit timer/counter with 8-bit programmable prescaler

Enhanced Timer 1:

• 16-bit timer/counter with prescaler
• External Gate Input mode
• Dedicated low-power 32 kHz oscillator

Timer2: 8-bit timer/counter with 8-bit period register, prescaler and postscaler

Enhanced Capture, Compare, PWM+ module:

• 16-bit Capture, max. resolution 12.5 ns
• Compare, max. resolution 200 ns
• 10-bit PWM with 1, 2 or 4 output channels, programmable “dead time”, max. frequency 20 kHz
• PWM output steering control

Capture, Compare, PWM module:

• 16-bit Capture, max. resolution 12.5 ns
• 16-bit Compare, max. resolution 200 ns
• 10-bit PWM, max. frequency 20 kHz

Enhanced USART module:

• Supports RS-485, RS-232, and LIN 2.0
• Auto-Baud Detect
• Auto-Wake-Up on Start bit
• In-Circuit Serial Programming TM (ICSPTM) via two pins
• Master Synchronous Serial Port (MSSP) module – supporting 3-wire SPI (all 4 modes) and 12C™ Master and Slave Modes with I2C address mask

Question 8.
Briefly explain different addressing modes in the 8051.
Answer:
There are four addressing modes in 8051.

• Immediate addressing mode: the data source is available immediately as a part of instruction itself. Example: MOV A, #8bit data
• Register addressing mode: the source and destination register names are parts of the opcode in instruction. Example: MOV A, R0
• Direct addressing mode: the source and/or destination may be the internal RAM locations. Example: MOV 90h, #50h
• Indirect addressing mode: the source or destination address may be indicated the content of index register. Example: MOV @Rl,#40h

Question 25.
For a 8-bit computer, if A = 0000 0101 and B = 0011 0100. Compute the following bitwise operators.
(i) A & B
(ii) A | B
(iii) A ∧ B
Answer:
A&B = 0000 0100 A | B = 0011 0101 A∧B = 0011 0001

Question 26.
Explain with diagram, the working of a satellite transponder system.
Answer:

Part – D

Answer any Three question  3 x 5 = 15

Question 27.
For the CE amplifier using silicon transistor given below, calculate
(i) Voltage across R2
(ii) Emitter current IE
(iii) Voltage gain Av
(iv) Power gain Ap Given that $r_{e}^{\prime}=\frac{26 \mathrm{mV}}{\mathrm{I}_{\mathrm{E}}} \text { and } \beta=100$

Answer:

Question 28.
Calculate output voltage V0for the circuit given below.

Answer:

Question 29.
A Colpitts oscillator circuit generates a frequency of 24 KHz. The capacitors used in the tank circuit are Ct = 0.2pF and C2 = 0.22pF. Calculate the value of an inductor L and feedback factor β.
Answer:

Question 30.
The output of an AM transmitter is given by 400[1 + 0.4 sin(6280) t] sin(3.14 x 107f). This voltage is fed to an antenna of resistance 500Ω. Determine

• Carrier frequency
• Modulating frequency
• Carrier power
• Mean power output.

Answer:

Question 31.
Simplify the Boolean expression Y =∑m (0,2,6,8,10,12,13,14) + ∑d (4,9) using K-map. Draw the NAND-gate equivalent circuit to realize the simplified expression.
Answer:

Part – E

Answer any Four questions.       4 x 5 = 20

Question 32.
Draw the circuit diagram of two stage RC-coupled amplifier. Explain its frequency response.
Answer:

RC coupling is most widely used method of coupling signal from one stage to next stage in multistage amplifiers. The two stage RC coupled amplifier is as shown in the diagram. The signal developed across the collector resistor Rc of the first stage is capacitively coupled to the input of second stage through the coupling capacitor Cc.

RC coupled transistor amplifier is widely used as it provides excellent audio fidelity over wide range of frequency.

R1R2,Rc,RE provide biasing resistors. RE provides stabilisation and CE is used for bypassing ac across Rg.

Working:

When an AC signal is applied to the input of first stage, it gets amplified and appears at its output with a phase reversal of 180°. Through the coupling capacitor Cc, the amplified signal is fed to the input of second stage.

Cc blocks DC and allows AC to pass through. The second stage further amplifies the signal, further providing a phase reversal of 180° with respect to its input. The input and output signal are in phase as the signal is reversed twice. Overall gain of RC coupled amplifier is the product of gain of individual stages.

At low frequencies, reactance of coupling capacitor is high. Hence it allows only a small part of the signal to pass from one stage to next stage. Hence gain is low at low frequencies. At high frequencies, the coupling capacitor Cc offers low reactance and acts like a short. Hence loading effect of next stage increases and the gain decreases.
In mid frequency range, as frequency increases, the capacitive reactance of coupling capacitor decreases, which tries to increase gain. As capacitive reactance decreases, due to loading effect of next stage, the gain reduces. Hence, voltage gain is constant.

Question 33.
What is an integrator? With circuit diagram, obtain an expression for output voltage of an op-amp integrator.
Answer.

Integrator is a circuit whose output is proportional to integral of its input.

Question 34.
With a labelled block diagram explain the working of FM superhetrodyne radio receiver.
Answer:

when turned to the desired frequency. The RF amplifier is designed to handle large bandwidth of 150 kHz.

Mixer: the incoming RF signal of frequency fm is applied to a mixer which also receives the output from the local oscillator. A new frequency called intermediate frequency IF is produced whose value is difference of local oscillator signal f and signal frequency f.

Local oscillator: the receiver converts incoming carrier frequency to the IF by using local oscillator frequency higher than incoming tuned frequency. Colpitts oscillator is used as the local oscillator.

IF amplifier: IF signal is amplified by one or more number of amplifiers, which raises the strength of IF signal. It has multistage class A amplifier providing better selectivity and gain.

Limiter: It removes all the amplitude variation in FM signal caused by noise. Differential amplifiers are preferred for limiter.

Discriminator: It recovers the modulating signal from the IF signal. It converts frequency variation into corresponding voltage variation and produces the modulating signal. De-emphasis network: It reduces the relative amplitude of high frequency signals that are boosted in the transmitter and brings them back to their original level.

AF amplifier: It amplifier the modulating signal recovered by the FM detector. The speaker converts the electrical signal into sound signal.

Question 35.
(a) What is a register? Mention the different types of shift registers.
Answer:

• Serial-in, serial-out.
• Serial-in, parallel-out
• Parallel-in, serial-out.
• Parallel-in, parallel-out.

(b) Draw the logic diagram of clocked SR flip-flop.
Answer:

It is constructed using inverters inserted into inputs of cross coupled NAND gates.
Working:

1. When S=0, R=0, it does not respond and hence outputs Q and Q will remain in their previous state. This is called HOLD condition.
2. When S=1 and R=0, the output Q and Q change to 1 and 0. This condition is called SET state.
3. When S=0, R=1, the output Q and Q change to 0 and 1. This condition is called RESET state.
4. When S=1 and R=l, it drives output Q and Q both to HIGH which is FORBIDDEN or INVALID condition.

Question 36.
Write an assembly language program to subtract 23 H from 3 F H when CY = 0 and save the result in Rs.
Answer:

Question 37.
Write a C program to accept the radius of a circle and compute its area and perimeter.
Answer:
# include < stdio.h>
# include <conio.h>
void main ( )
{
float radius, area, perimeter, PI = 3.142; clrscr ( );
printf (“Enter the radius\n”); scanf (“%f’, & radius);
Area = PI * radius * radius;
Perimeter = 2.0 * PI * radius;
printf (“Area of circle is % f \n”, Area);
printf (“perimeter of the circle is % f \n”, perimeter); getch ( );
}

## 2nd PUC Electronics Previous Year Question Paper March 2019

Students can Download 2nd PUC Electronics Previous Year Question Paper March 2019, Karnataka 2nd PUC Electronics Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka 2nd PUC Electronics Previous Year Question Paper March 2019

### 2nd PUC Electronics Previous Year Question Paper March 2019

Time : 3 hrs. 15 min.
Max Marks: 70

Instructions:

• The question has five Parts A, B, C and D.
• Part – A has no choice
• Part – D has two parts. Part -1 is from problems, Part – II is of essay type questions.
• Circuit diagram/timing diagram/truth tables are drawn wherever necessary.
• Problems without necessary formula/formulae carry no marks.

Part – A

Answer all questions: 10 x 1 = 10

Question 1.
Name the charge carriers in N-channel JFET.
Answer:
Electrons.

Question 2.
Write the value of bandwidth of an ideal OPAMP.
Answer:
Infinity.

Question 3.
Draw the waveform of damped oscillations.
Answer:

Question 4.
What is the function of transmitter in a radio communication system?
Answer:
Transmitter processes the information and transmits it.

Question 5.
Draw the waveform of AM wave when modulation index ma=1
Answer:

Question 6.
Define fidelity in a radio receiver.
Answer:
Fidelity is the ability of the receiver to reproduce the input faithfully.

Question 7.
Mention one application of TWAC.
Answer:
TRIAC can be used in fan motor control.

Question 8.
Write the decimal equivalent of BCD code (00110101)2.
Answer:
(00110101)2 = 3510.

Question 9.
Name anyone arithmetic operator in C-language.
Answer:
Addition or Subtraction.

Question 10.
Expand RADAR. or = r v                                                                       ‘
Answer:
RADAR – Radio Detection And Ranging.

Part – B

Answer any Five question.

Question 11.
Mention the steps to obtain DC-equivalent circuit of a CE amplifier.
Answer:

• Reduce all ac sources to ground.
• Open all the capacitors.

Question 12.
Calculate the gain of a negative feedback amplifier with an internal gain, A = 100 and feedback factor β= 1/10.
Answer:
A= 100, β = 1/10.

Question 13.
Write any two applications of crystal oscillator.
Answer:

• Digital watches.
• Clock, Calculators.
• Transmitters and Receiver.

Question 14.
Realise XOR gate using basic logic gates.
Answer:

Question 15.
Write the difference between combinational and sequential logic circuits.
Answer:

 Combinational Logic Circuit Sequential Logic Circuit (i)  It does not have memory (ii)  Its output depends only on present input. (i) It has memory. (ii) Its output depends on present and past input

Question 16.
Write any two instruction set of 8051 microcontroller.
Answer:
These instructions move the content of one register to another one. Data can also be transferred to stack with the help of PUSH and POP instructions.

Question 17.
Mention any two features of C-language.
Answer:
C is a powerful, flexible language that has gained worldwide acceptance in recent years. C is concise, yet powerful in its scope. It is a high-level language that also incorporates features of low-level languages. It is becoming the standard for program development on small machines and microcomputers.

It is possible to apply C to nearly all application areas. Modularity {process of dividing problems into sub-problems) makes C ideal for projects involving several programmers. C permits nearly every programming technique. C is widely available for most brands and type of computers. It is portable. With all the features C is not a difficult language, it is one of the easiest languages to learn and implement.

C programming can be used to write any complex program because of its rich set of built-in functions and operations. C compiler combines the capabilities of an assembly language with the feature of high-level language and therefore it is very much suited for uniting both system software and business packages. Because of its variety of data types, powerful operator programs in C are efficient and fast. C has the ability to extend, it means user-defined functions can also be added to C library.

Question 18.
Briefly explain the internet.
Answer:

1. Using internet, access to required information is easy, economical, reliable and fast.
2. With E-mail, messages reach destination with in minutes.
3. E-mail can be sent and received from any place.
4. Using E-mail, message can be sent to several persons simultaneously.

Part – C

Answer any five questions.   5 x 3 = 15

Question 19.
Compare the FET and BJT.
Answer:

 FET BJT. 1. It is unipolar device. 1. It is a bipolar device. 2. It is a voltage controlled device. 2. It is a current controlled devices. 3. Its input resistance is very high. 3. Input resistance is very low. 4. Current conduction is only by majority charge carriers either hole or electron. 4. Current conduction is both by electrons and holes. 5. It has high switching speed. 5. It has low switching speed. 6. Less noisy 6. More noisy. 7. Fabrication is easier in IC. 7. Difficult to fabricate in IC.

Question 20.
What is transistor biasing? Briefly explain the need for biasing a transistor
Answer:
The application of suitable DC voltage across the transistor terminals is called biasing of the transistor.

The reasons for biasing of a transistor are:

1. To make the operating point to be at the centre of the load line.
2. To make the collector current independent of temperature variations.
3. To maintain the operating point constant even though transistor is replaced by another of same type.

Question 21.
With a block diagram derive an expression for output impedance of aN amplifer with negative feedback.
Answer:

An amplifier should have a low output impedance so that it can deliver power to load without much loss. With negative feedback, the output impedance of the amplifier decreases which is desirable.

Let Zo be output impedance of basic amplifier and Zof be the output impedance with negative feedback. A hypothetical source of voltage Vo is applied at the output. By applying KVL to the output loop,

Vo = IoZo +AVi.
With negative feedback, input to basic amplifier with negative feedback is
Vi = Vs – Vf
V= o – Vf
V– Vf
{As input is short circuited, V = o}

As 1+Aβ>1, output impedance of amplifier decreases by a factor(1+Aβ)

Question 22.
Explain the importance of ionosphere in the radio communication.
Answer:
Ultraviolet radiation from the sun cause the upper atmosphere to ionize i.e. to become electrically charged. Hence a thick layer of ions formed at heights of 50 km to 400km. The ionosphere supports MF and HF wave propagation.
Its three layers are:

D layer: It is at a height of 70 km and has an average thickness of 10 km. It disappears in the night time. It reflects VLF and LF waves and absorbs MF and HF waves.

E layer: It is at a height of 100 km with a thickness of 25 km. This layer disappears at night. It helps in MF and HF wave propagation.

F layer: It is at a height of 150 km and extends upto 400 km. This layer exists during both day and night time. During day time, it splits into two layers F, and F2, and combines to form a single F layer during night time.

 F2 region. 250-400 km. F1 region. 160-250 km. E region. 95-130 km D region. 50-95 km. Troposphere Earth.

Question 23.
Derive an expression for the total power of AM wave.
Answer:
The total power in AM wave is equal to the sum of powers of the carrier, the upper sideband and the lower sideband.

Question 24.
In a SCR full wave rectifier the firing angle is 90°, rms voltage of an input to the rectifier is 230 V and the load resistance is 25Ω. Calculate Vdc and Idc
Answer:

Question 25.
With a neat circuit diagram explain the working of power diode under forward bias condition.
Answer:

Question 26.
With a neat block diagram explain the satellite Transponder.
Answer:

Part – D

Answer any Three question  3 x 5 = 15

Question 27.
Calculate voltage gain (Ay) and input impedence (Zin) of a given CE amplifer with values
$r_{ e }=\frac { 26{ mV } }{ { I }_{ { E } } } ,{ V }_{ { BE } }=0.7V$ β = 100.

Answer:

Question 28.
Calculate the output voltage (V0) for a given circuit below

Answer:

Question 29.
A transistor Colpitts oscillator has L = 4 mH, C1 = 10 nF and C2 = 10 nF. Calculate the frequency of oscillations.
Answer:

Question 30.
The current of a AM transmitter is 8A when the carrier is sent, it increases to 8.65 A when the carrier is modulated. Find the percentage of modulation. Also calculate the antenna current when the depth of modulation is 0.75.
Answer:

Question 31.
Simplify the Boolean expression using K-map. Y(ABCD) = Σm(0, 2, 4, 5, 8, 10, 12, 14) + Σd(6, 13)  Write the logic circuit for the simplified expression.
Answer:

Part – E

Answer any Four questions.

Question 32.
Compare CE, CB and CC amplifiers.
Answer:

 Parameter Types of Amplifiers. CE CB CC Current gain High (P) Less than 1 Highest (1+ p) Voltage gain High Moderate Low Power gain Highest Moderate Moderate Phase shift 180° 0° 0- Input impedance Moderate Low High Output impedance Moderate High Low Bandwidth Narrow Wide Wide Applications Used as AF amplifies Used as RF amplifies impedance matching

Question 33.
(a) What are active filters? (1 + 4)
Answer:
An active filter is a type of analog circuit implementing an electronic filter using active components, typically an amplifier. Amplifiers included in a filter design can be used to improve the cost, performance and predictability of a filter.

(b) With a neat circuit diagram explain the working of first order low pass active filter and draw its frequency response curve.
Answer:
Filter are circuits used to select a band of frequencies. The two types of filters are

• low pass filter and
• high pass filter.

Cut off frequency $\mathrm{f}_{C}=\frac{1}{2 \Pi \mathrm{R}_{1} \mathrm{C}_{1}}$
Frequency response of low pass filter.

Question 34.
Draw a block diagram and explain the different stages of FM transmitter.
Answer:
The modulating signal is applied to the pre-emphasis circuit, which improves signal to noise ratio, AF amplifier amplifies output of pre-emphasis. The processed signal is fed to reactance modulator. The reactance modulator uses a transistor or FET connected across tank circuit of carrier oscillator.

The oscillator frequency depends on the tank reactance which in turn depends on the instantaneous amplitude of modulating signal. Thus, output of reactance modulator will be a frequency modulated wave.

The oscillator is followed by a buffer amplifier which isolates oscillator from subsequent stages. The limiter maintains the amplitude constant. Class C power amplifier amplifies modulated wave to required power levels. The FM signal is then fed to the transmitting antenna.

Question 35.
(a)What is flip-flop?
Answer:
A flip flop is an electronic circuit with two stable states that can be used to store binary data. Flip-flops and latches are fundamental building blocks of digital electronics systems used in computers, communications, and many other types of systems. Flip-flops and latches are used as data storage elements

(b) With a logic diagram and truth table, explain the working of SR flip-flop, (1+4)
Answer:

Question 36.
Write a flow chart and explain the steps used in creating assembly language program in 8051 microcontroller
Answer:

Question 37.
Write a C-program to find largest among three numbers.
Answer:

Output:
Enter the three numbers 15
24
48
48 is largest number among 15,24 and 48