An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected to a 6 V battery as shown in the circuit. Calculate:
(a) the total resistance of the circuit.
(b) the current through the circuit.
(c) the potential difference across the
(i) electric lamp and
(ii) conductor, and
(d) power of the lamp.
- Resistance of electric lamp, R 1 = 20 ohm
- Resistance of conductor, R 2 = 4 ohm
- Potential difference, V = 6 V
Since R 1 and R 2 are connected in series,
Total resistance, R = R 1 + R 2
= 20 + 4
= 24 ohm
Therefore, total resistance of the circuit is 24 ohm.
Let the current through the circuit be I.
According to ohm’s law,
V = I*R
I = V / R
= 6 / 24
= 0.25 A
Therefore, the current through the circuit is 0.25 A
(i) For electric lamp,
V = I*R 1
= 0.25 * 20
= 5 V
Therefore, the potential difference across the electric lamp is 5 V.
(ii) For conductor,
V = I*R 2
= 0.25 * 4
= 1 V
Therefore, the potential difference across the conductor is 1 V.
Let the power of the lamp be P.
We know that,
P = V*I
= 5 * 0.25
= 1.25 W
Therefore, power of the lamp is 1 . 25 W