Question

❋ An electric lamp of resistance 20Ω and a conductor of resistance 4Ω re connected to a 6V battery as shown in the circuit Calculate:

(a) The current through the circuit
(b) The potential difference across the
(i) Electric lamp and
(ii) Conductor, and
(c) Power of the lamp. ​

Answer 1

Answer:

Answer

a). the total resistance of the circuit =20Ω+4Ω=24Ω

b). The current through the circuit= current through the bulb= current through the conductor.

The current through the circuit=

the voltage of the battery

voltage applied

.

the current through the circuit =

24

6

amp

The current through the circuit = 0.25amp

c). The potential difference across the lamp and the conductor.

(i) The potential difference across the electrical lamp =0.25×20=5volt

(ii) The potential difference conductor =0.25×4=1 volt

d). Power can be calculated as

current through the lamp

voltage across the lamp

power=

0.25

5

→20W.

Answer 2

Answer:

current I= 6÷(20+4)=1/4A

Explanation:

potential across lamp= 1/4×20=5V

potential drop across conductor=1/4×4=1V

power of lamp P= I²R= 1/16×20=5/4 Watt