An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery. Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor.

Fig. An electric lamp connected in series with a resistor of 4 Ω to a 6 V battery

Given,
Voltage of the battery, V =6V

Resistance of electric lamp, R₁ = 20 Ω 

Resistance of series conductor, R₂ = 4 Ω 

a) Total resistance in the circuit, R_s = R₁ + R₂ 
                                                     = 20 Ω + 4 Ω
                                                     = 24 Ω. 

(b) Using Ohm’s law, 

Current through the circuit is, $\mathrm{I} = \frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{s}}} = \frac{6 \mathrm{V}}{24 \mathrm{\Omega }} = 0.25 \mathrm{A}.$ 

(c) Potential difference across the electric lamp,
V₁ = IR₁ = 0.25 A x 20 Ω = 5 V.  

Potential difference across the conductor is,
V₂ = IR₂ = 0.25 A x 4 Ω = 1 V.