Question

(a) What is meant by "electric power"? Write the formula for electric power in terms of potential difference and current.
(b) The diagram below shows a circuit containing a lamp

L

, a voltmeter and an ammeter. The voltmeter reading is $3 V$ , and the ammeter reading is

0.5 A

.

(i) What is the resistance of the lamp?
(ii) What is the power of the lamp?
(c) Define

kilowatt - hour

. How many

joules

are there in one

kilowatt - hour

?
(d) Calculate the cost of operating a heater of $500 W$ for

20 hours

at the rate of

₹ 3.90 p e r unit

.

Accepted Answer

(a) Electrical power is defined as the rate of energy consumption in an electrical circuit.
The expression for electric power in terms of potential difference and current is:
$P = V \times I$ .

(b) Given,
Potential difference = reading of the voltmeter, $V = 3 V$
Current in the circuit = reading of the ammeter, $I = 0.5 A$
(i) The resistance of the lamp, $R = \frac{V}{I} = \frac{3}{0.5} = 6 Ω$ .
(ii) Power, $P = V \times I = 3 \times 0.5 = 1.5 W$ .

(c) One $kilowatt - hour$ is the amount of electrical energy consumed when an electric instrument having a power rating of one $kilowatt$ is used for one $hour$ .
$1 kilowatt - hour = 3.6 \times 10^{6} joules$ .

(d) Power rating of the heater, $P = 500 W$
Duration of the operation, $t = 20 hours$
The electrical energy consumed,
$E = P \times t = 500 \times 20 = 10000 Wh = 10 kWh$
Rate for $1 k W h = ₹ 3.90$
Total cost of operating, $\cos t = u n i t s c o n s u m e d \times r a t e = 10 \times ₹ 3.90 = ₹ 39$ .

Lakhmir Singh and Manjit Kaur Solutions for Chapter: Electricity, Exercise 33: Long Answer Type Questions

Lakhmir Singh Physics Solutions for Exercise - Lakhmir Singh and Manjit Kaur Solutions for Chapter: Electricity, Exercise 33: Long Answer Type Questions

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