❋ 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:
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:
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