Question

An electric bulb of resistance 200ω draws a current of 1 ampere. calculate the power of the bulb the potential difference at its ends and the energy in kwh consumed burning it for 5h.

Answer 1

We need to find the voltage .

V= IR
V = 200 volts.

Power = V * I = 200 * 1 = 200 w.
Totally in 5 hours the power consumed Is 200 * 5 = 1000 watt hour.
= 1 k w h.


#Be Brainly❤️

Answer 2

An electric bulb of resistance 200Ω draws a current of 1 Ampere.

We have to calculate the power of the bulb , the potential difference at its ends and the energy in kWh consumed burning it for 5h.

the power of bulb :

power is given by, P = i²R

where i is current through wire, and R is resistant of wire.

∵ i = 1 A , R = 200 Ω

∴ P = 1² × 200 = 200W

Therefore the power of bulb is 200 W.

potential difference at its ends :

using Ohm's law, V = iR

= 1 A × 200 Ω

= 200 Volts

Therefore the potential difference at its ends is 200v.

energy consumed :

Energy consumed , E = power of bulb × time

= 200W × 5h

= 1000Wh

= 1 kWh

Therefore the energy in kWh consumed burning it for 5h is 1 kWh.