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.
Answer:
Given,
- Resistance of electric lamp, R _{ 1 } = 20 ohm
- Resistance of conductor, R _{ 2 } = 4 ohm
- Potential difference, V = 6 V
(a)
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.
(b)
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
(c)
(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.
(d)
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