**KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity** are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 12 Electricity.

## Karnataka SSLC Class 10 Science Solutions Chapter 12 Electricity

### KSEEB SSLC Class 10 Science Chapter 12 Intext Questions

Text Book Part I Page No. 94

Question 1.

What does an electric circuit mean?

Answer:

A continuous and closed path of an electric current is called an electric circuit. It consists of electric devices like switch, sources of electricity etc.

Question 2.

Define the unit of current.

Answer:

The unit of electric current is ampere (A). 1A is defined as the flow of 1C of charge through a wire in Is of time.

Question 3.

Calculate the number of electrons constituting one coulomb of charge.

Answer:

Given: Q = IC, e = 1.6 × 10^{-19} C, n = ?

We know, Q = ne

Substitute the values we get,

IC = n × 1.6 × 10^{-19} \(=\frac{1}{1.6 \times 10^{-19}}\) = n

n = 6 × 10^{18} electrons.

Text Book Part I Page No. 96

Question 1.

Name a device that helps to maintain a potential difference across a conductor.

Answer:

Voltmeter.

Question 2.

What is meant by saying that the potential difference between two points is 1V?

Answer:

The potential difference between the two points is called to be 1 V, if 1 joule (J) of work is done to move a charge of amount 1 C from one point to another.

Question 3.

How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer:

According to question,

Potential difference (V) = 6V

Amount of charge (q) = 1 coulomb

Work done or energy = ?

Potential difference = Work done/Charge

or Workdone = V × q

∴ Work done = 6 × 1 = 6 J.

Text Book Part I Page No. 103

Question 1.

On what factors does the resistance of a conductor depend?

Answer:

The resistance of the conductor depends

- on its length
- on its area of cross section and
- on the nature of its material.

Question 2.

Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Answer:

Resistance, \(\mathrm{R} \alpha \frac{1}{\mathrm{A}}\). The resistance of a conductor is inversely proportional to its area of cross-section. A thick wire has a greater area of cross-section whereas a thin wire has a smaller area of cross-section. Thus, thick wire has less resistance and a thin wire has more resistance therefore current will flow more easily through a thick wire.

Question 3.

Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Answer:

Ohm’s law:

V = IR

According to question, Potential difference (V) is decreased to half.

Hence, new potential difference, V’ = V/2.

Resistance remains constant.

So, the new current I’ = V’/R

= (V/2)/R

= (1/2) (V/R)

= (1/2) 1 = 1/2

Therefore, the amount of current flowing through the electrical component is also reduced by half.

Question 4.

Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer:

Due to two main following reasons the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal:

- Resistivity of an alloy is higher than the pure metal.
- At high temperatures, the alloys do not melt easily.

Text Book Part I Page No. 107

Question 1.

Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8Ω resistor, and a 12Ω resistor, and a plug key, all connected in series.

Answer:

R_{s} = R_{1} + R_{2} + R_{3}

= 2V + 2V + 2V

= 6W

Question 2.

Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12Ω resistor. What would be the readings in the ammeter and the voltmeter?

Answer:

Resistance, R = 5 + 8 + 12 = 25 Ω

\(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6}{25}=0.24 \mathrm{A}\)

As per ohm’s law =0.24A

V_{1} = IR

Potential difference

V_{1} = IR = 0.24 × 12 = 2.88 V.

reading in ammeter = 0.24A

reading in voltmeter = 2.88 v.

Text Book Part I Page No. 110

Question 1.

Judge the equivalent resistance when the following are connected in parallel –

(a) 1 Ω and 10^{6} Ω,

(b) 1 Ω and 10^{3} Ω, and 10^{6} Ω.

Answer:

(a) If two resistances are connected in parallel

equivalent resistance = 1Ω.

(b) If three resistances 1Ω, 10^{3}Ω and 10^{6}Ω are connected in parallel,

equivalent resistance = 0.999Ω

Question 2.

An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all the three appliances, and what is the current through it?

Answer:

According to question,

Resistance of electric lamp = 100 Ω

Resistance of a toaster = 50 Ω

Resistance of a water filter = 500 Ω

Potential difference = 220 volt

So, total resistance of the three appliances in parallel

= 1/R = 1R_{1} + l/R_{2} + 1/R_{3}

= 1/100 + 1/50 + 1/500

= 1/R = (5 + 10 + 1) 500

1/R = 16/500 = 4/125

= 125/4 = 31.25 ohm

V = I × R

220 = 1 × 31.25

∴ I = 7.04 amp

So, current through the electric iron = 7.04 A.

Question 3.

What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Answer:

Advantages of parallel connection are:

(a) In parallel circuit, all other appliances keep working normally even though one electric appliance stop working.

(b) In parallel circuit, all the electrical appliance has its own switch, they can be turned on or off without affecting other appliances.

(c) Each electrical appliance receive the same voltage (220 V) as that of the power supply line.

(d) In parallel connection of electrical appliances, power supply is high because the overall resistance of the circuit is reduced.

Question 4.

How can three resistors of resistances 2Ω, 3Ω, and 6Ω be connected to give a total resistance of (a) 4Ω, (b) 1Ω?

Answer:

In this diagram 2 resistors of resistance 3W and 3 W are connected in parallel.

If 2Ω and 2Ω are connected in series 2Ω + 2Ω = 4Ω

∴ Total resistance = 4Ω

If resistors are connected in series

∴ Total resistance = 1Ω

Question 5.

What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4Ω, 8Ω, 12Ω, 24Ω?

Answer:

(a) If four resistors are connected in series then resistance

= 4 + 8 + 12 + 24

= 48Ω

(b) If these are connected in parallel

∴ We can get lowest resistance.

**Text Book Part I Page No. 112**

Question 1.

Why does the cord of an electric heater not glow while the heating element does?

Answer:

The resistance of cord is extremely smaller than heating element, so the heat produced in cord is less than the heating element. So, the heating element begins to glow but cord does not glow.

Question 2.

Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Answer:

As per Joule’s law H = VIt.

Here V = 50 V.

Question 3.

An electric iron of resistance 20 takes a current of 5 A. Calculate the heat developed in 30 s.

Ans. According to question,

Resistance of the iron = 20 Ω

Current = 5 amp

Time = 30 seconds

As we know, Heat produced = I^{2} RT

= 5^{2} × 20 × 30

= 15000 Joules

or = 1.5 × 10^{4} J.

Text Book Part I Page No. 114

Question 1.

What determines the rate at which energy is delivered by a current?

Answer:

Electrical power determines the rate at which the energy is delivered by a current.

Question 2.

An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Answer:

Given, Current in the motor, I = 5 Amp

Potential difference, V = 220 V

Time = 2 hours

So, Power of the motor = V × I

= 220 × 5

= 1100 watt or 1.1 kW

Now, Energy = power × time

= 1.1 kW × 2 h.

= 2.2 kWh

### KSEEB SSLC Class 10 Science Chapter 12 Textbook Exercises

Question 1.

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then ratio R/R’?

(a) 1/25

(b) 1/5

(c) 5

(d) 25

Answer:

(d) 25

Question 2.

Which of the following terms does not represent electrical power in a circuit?

(a) I^{2}R

(b) IR^{2}

(c) VI

(d) V^{2}/R

Answer:

(b) IR^{2}

Question 3.

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:

(a) 100 W

(b) 15 W

(c) 50 W

(d) 25 W

Answer:

(d) 25 W

Question 4.

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be:

(a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4 : 1

Answer:

(c) 1 : 4

Question 5.

How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer:

It is connected in parallel.

Question 6.

A copper wire has diameter 0.5 mm and resistivity 1.6 10^{-8}Ω m. What will be the length of this wire to make its resistance 10Ω? How much does the resistance change if the diameter is doubled?

Answer:

Area of the copper wire A \(=\pi\left(\frac{\mathrm{D}}{2}\right)^{2}\)

diameter = 0.5 mm = 0.0005 m

Resistance, R = 10W.

Diameter of copper wire is doubled, then diameter = 2 × 0.5 = 1 mm = 0.001 m

∴ Length of wire = 122.7 m

Resistance of wire = 2.5 W

Question 7.

The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –

I (amperes) | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |

V (volts) | 1.6 | 3.4 | 6.7 | 10.2 | 13.2 |

Plot a graph between V and I and calculate the resistance of that resistor.

Answer:

In the following graph, voltage is taken along the x-axis and the current is taken along the y-axis. Different values are as follows:

V (volts) | 1.6 | 3.4 | 6.7 | 10.2 | 13.2 |

I (amperes) | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |

Slope indicates Resistance

∴ Resistance of the resistor = 3.4Ω.

Question 8.

When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Answer:

Here I = 2.5 mA = 2.5 × 10^{-3} A.

∴ \(R=\frac{12}{2.5 \times 10^{-3}}\)

= 4.8 × 10^{3}Ω

= 4.8 K cal.

Question 9.

A battery of 9 V is connected in series with resistors of 0.2Ω, 0.3Ω, 0. 4Ω, 0.5Ω and 12Ω, respectively. How much current would flow through the 12Ω resistor?

Answer:

A battery of 9V is connected in series with resistors 0.2Ω, 0.3Ω, 0.4Ω,

R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4Ω.

V = 9V

\(I=\frac{9}{13.4}=0.671 \mathrm{A}\)

∴ 12 V battery, 0.671 A current flows.

Question 10.

How many 176Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer:

Let the resistors be x’

Resistance = 176Ω

As per ohm’s law,

V = IR

∴ Four resistors (in parallel) are required to carry 5A on a 220 V line.

Question 11.

Show how you would connect three resistors each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω (ii) 4 Ω

Answer:

(i) To get 9 Ω, connect two resistors of 6 Ω in parallel and third resistor in series.

\(\frac{1}{\mathrm{R}_{p}}\) = \(\frac { 1 }{ 6 } \) + \(\frac { 1 }{ 6 } \) + \(\frac { 2 }{ 6 } \) = \(\frac { 1 }{ 3 } \)

(ii) To get 4 Ω, connect 2 resistors in series and the third in parallel to both of them

R_{s} = 6 + 6 = 12 Ω

Question 12.

Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Answer:

Given, P = 10 W

V = 220 V

I = ?

Using formula I = \(\frac { p }{ v } \)

= \(\frac { 10 }{ 220 } \) = \(\frac { 1 }{ 22 } \) A

If number of bulbs connected to current of 5 is n.

So, \(\frac { 1 }{ 22 } \) n = 5

= n = 5 × 22 = 110 bulbs

Question 13.

A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer:

(i) When used separately

Then, I = \(\frac { V }{ R } \) = \(\frac { 220 }{ 24 } \) = 9.2 A.

(ii) When connected in series

Given R_{1} = 24 Ω, R_{2} = 24 Ω

Then R_{s} = R_{1} + R_{2} = 24 + 24 = 48 Ω

So, I = \(=\frac{V}{R_{s}}\) = \(\frac { 220 }{ 48 } \) = 4.6 A

(iii) When connected in parallel

Given, R_{1} = 24 Ω, R_{2} = 24 Ω

Question 14.

Compare the power used in the 2 Ω resistor in each of the following circuits:

(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and

(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors

Answer:

(i) Given, V = 6 V

R_{1} = 1 Ω

When, R_{1} and R_{2} are in series.

Then, R_{s} = R_{1} R_{2} = 1 + 2 = 3 Ω

Now current flowing through the circuit, I = \(\frac { V }{ R } \),

So, I = \(\frac { 6 }{ 3 } \) = 2 A

Using, Power = I^{2}R_{s} = (2)^{2} × 3 = 12 W

Hence, 12 W power is used in the 2Ω resistor in this circuit.

(ii) Given, R_{1} = 12 Ω

R_{2} = 2 Ω voltage = 4V p_{2} = ?

∴ I_{1} through 2 Ω = \(\frac{4 \mathrm{V}}{2 \Omega}\) = 2 A

So, P_{2} = I_{1} ^{2}R_{2} = 2^{2} × 2 = 4 × 2 = 8 W

Question 15.

Two lamps, one rated 100W at 220V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Solution:

Given, P_{1} = 100 W

P_{2} = 60 W

So, P = P_{1} + P_{2} = 160 W

V = 220

I = ?

As we know, P = IV or I = \(\frac { p }{ v } \) = \(\frac { 160 }{ 220 } \) = 0.73 A

Question 16.

Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer:

Energy consumed, H = Pt.

(i) By TV set = 250 × 3600 = 9 × 10^{5} J

(ii) By toaster = 1200 × 600 = 7.2 × 10^{5} J

It is consumed more by TV set as compared to toaster.

Question 17.

An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Answer:

Given, Resistance (R) = 8 Ω

Current (I) = 15 A

Time (t) = 2 hours = 2 × 3600s

= 7200s

Using, H = I^{2} Rt

= (15)^{2} × 8 × 7200s

= 1.296 × 10^{7} J.

Question 18.

Explain the following.

(a) Why is the tungsten used almost exclusively for filament of electric lamps?

Answer:

Tungstan is a strong and metal having high melting point. This will not melt in high temperature. Because of this tungsten is used almost exclusively for filament of electric lamps.

(b) Why are the conductors of electric heating devices, such as bread- toasters and electric irons, made of an alloy rather than a pure metal?

Answer:

The resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (burn) readily at high temperatures. Because of this reason the conductors of electric heating devices, such as bread-toasters and electric irons made of an alloy rather than a pure metal.

(c) Why is the series arrangement not used-for domestic circuits?

Answer:

In case of series connection, when one component fails, the circuit it broken and none of the components work. But in case of parallel connection, circuit divides the current throughout the electrical gadgets. Because of this reason series arrangement is not used for domestic purposes.

(d) How does the resistance of a wire vary with its area of cross-section?

Answer:

Resistance is inversely proportional to its cross section. As area is increasing resistance is less. Thus resistance of wire is changing with area of cross section.

(e) Why are copper and aluminium wires usually employed for electricity transmission?

Answer:

Copper and Aluminium are good conductors of electricity and they have less resistance. Hence they are used in electricity transmission.

### KSEEB SSLC Class 10 Science Chapter 12 Additional Questions and Answers

Question 1.

Draw a Neat diagram of Electric circuit for studying ohm’s law.

Answer:

Question 2.

What is the S.I. Unit of resistivity?

Answer:

ΩM

Question 3.

How is fuse wire made of?

Answer:

It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, iron, lead etc.

Question 4.

Write the formula of Electric power.

Answer:

\(P\quad =\quad \frac { { V }^{ 2 } }{ R } \)

Question 5.

Give examples where we find heating of electric current.

Answer:

Electric heater, electric iron etc.

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