Question

An electrc bulb of resistance $20\Omega$ and a a resistance wire of $4\Omega$ are connected in series with a 6 V battery.
(a) total resistance of the circuit.
(b) current through the circuit.
(c) potential difference across the electric bulb.
(d) potential difference across the resistance wire.

Answer 1

A. The total resistance of the circuit = 20Ω + 4Ω =24 Ω

B. The current through the circuit = current through the bulb = current through the conductor
The current through the circuit = Voltage applied/ the voltage of the battery
The current through the circuit = (6/24) amp
The current through the circuit =0.25 amp

C. The potential difference across the electric lamp = 0.25 ×20 = 5volt

D. The potential difference across the resistance wire = 0.25 × 4 = 1volt