NCERT Solutions of Electricity Class 10
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Electricity

Solution: NCERT Book - Exercise: Part -1

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 the ratio R/R€ is

(a) 1/25 (b) 1/5 (c) 5 (d) 25

Answer:- (d) 25

Solution:-

The piece of wire having resistance equal to R is cut into five equal parts.

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Question: 2:- Which of the following terms does not represent electrical power in a circuit?

(a) I2R (b) IR2 (c) VI (d) V2/R

Answer:- (b) IR2

Solution:-

We know that Power (P) = VI

After substituting the value of V = IR in this we get

P = (IR) I = I x R x I = I2R, Thus P = I2R

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Thus, P cannot be expressed as IR2

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) 75 W (c) 50 W (d) 25 W

Answer:- (d) 25 W

Solution:- Given,

Potential difference, V = 220V, Power, P = 100W

Therefore, power consumption at 100V =?

To solve this problem, first of all resistance of the bulb is to be calculated.

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Thus, bulb will consume power of 25W at 110V

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

Solution:

Let the potential difference = V,

Resistance of the wire = R

Resistance when the given wires connected in series = Rs

Resistance when the given wires connected in parallel = Rp

Heat produced when the given wires connected in series = Hs

Heat produced when the given wires connected in parallel = Hp

Thus, resistance Rs when the given two wires connected in series = R + R = 2R

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Question: 5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer: Voltmeter is connected into parallel to measure the potential difference between two points in a circuit.

ncert questions electricity voltmeter

Question: 6. A copper wire has diameter 0.5 mm and resistivity of 1.6 — 10 8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Answer:

Given, Diameter of wire = 0.5 mm

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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

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

Answer:

electricity graph voltage vs current

Since slope of the graph will give the value of resistance, thus

Let consider two points A and B on the slope.

Draw two lines from B along X-axis and from A along Y-axis, which meets at point C

Now, BC = 10.2 V 3.4 V = 6.8 V

AC = 3 1 = 2 ampere

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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:

Given, Potential difference, V = 12V

Current ( I ) across the resistor = 2.5mA = 2.5 x 10 -3 = 0.0025 A

Resistance, R =?

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Question: 9. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

Answer:

Given, potential difference, V = 9V

Resistance of resistors which are connected in series = 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω respectively

Current through resistor having resistance equal to 12Ω =?

Total effective resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω

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Since, there is no division of electric current, in the circuit if resistors are connected in series, thus, resistance through the resistor having resistance equal to 12 Ω = 0.671 A


NCERT Solutions of Electricity Class 10
<< Solution of Intext Questions (NCERT Book)- Part-2Solution: NCERT Book - Exercise: Part :-2 >>

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