Question

An electric 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

Given:
Resistance of the electric bulb = 20 Ω
Resistance of the wire = 4 Ω
Potential difference of the battery = 6 V

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

(B.)
Everything is connected in series, Hence
The current through the circuit = current through the bulb = current through the conductor = I
The current through the circuit, I = $\frac{V}{R}$
I = $\frac{6}{24}$
The current through the circuit, I = $0.25$ amp

(C.)
We know V = I $\times$ R
The potential difference across the electric lamp = 0.25 ×20
= 5 volt

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