Electricity Worksheet11

An electric heater draws a current of 5 A and its element has a resistance of 50 Ω. If the heater is switched on for 5 minutes, calculate the energy released in kilojoules.

An electric room heater has a resistance of 25 Ω and operates at 220 V for 12 minutes. Calculate heat energy dissipated by it in kilojoules.

Calculate the total power of 5 fans if each of them draws a current of 0.8 A at a potential difference of 220 V.

An electric bulb of resistance 400 Ω, draws a current of 0.5 A. Calculate the power of the bulb and the potential difference at its ends.

A soldering rod iron draws an energy of 45000 J in 4 minutes when the current flowing through its element is 6 A. Calculate the resistance of its heating element.

An electric bulb is marked 250 W200 V. what information does it convey? How many joules of energy is consumed by this bulb in one hour? How long will it take for the bulb to consume 1 k Wh?

100 J of heat are produced each second in a 4 Ω resistance. Find the potential difference across the resistor.

An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?

An electric refrigerator rated 400 W operates 8 hour/day. What is the cost of the energy to operate it for 30 days at Rs. 3.00 per kWh?

Two bulbs 0f 100 W each and two coolers of 250 W each, work on an average 6 hours a day. If the energy costs Rs. 1.75 per kWh, calculate the monthly bill and the minimum fuse rating when power is supplied at 250 V.
Answer:

W = I^{2}Rt = (5)^{2} × (50) × (5 × 60) = 375000 J = 375 kJ


P = VI = 220 × 0.8 = 176 W. Power of 5 fans = 5 × 176 = 880 W

P = I^{2}R = (0.5)^{2} × 400 = 100 W
As


A 250 W – 200 V bulb consumes an electric power of 250 W when working on 200 V and the resistance of such a bulb is (200)^{2}/250 = 160 Ω (as P = V^{2}/R or R = V^{2}/P)
W = Pt = 250 × (60 × 60) = 900000 J
As

V = IR = 5 A × 4 Ω = 20 V
= 20 V

P = VI = 220 V × 0.50 = 110 W

Total energy consumed in 30 days, i.e., W = Pt = 400 W × 8 hour/day × 3 days = 96000 Wh = 96 kWh.
Cost of energy = 96 kWh × Rs.3.00 kWh = Rs. 2880.00

Total power, P = 2 × 100 + 2 × 250 = 700 W
Electric energy consumed per month
= P × t = (700 W) (6 h) (30)
= 126000 Wh = 126 kWh (1 month = 30 days)
Monthly bill = 126 × 1.75 = Rs. 220.50
As P = VI, I (minimum fuse rating)