An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery in series. Calculate the potential difference across the electric lamp.

Applying Ohm’s law,

The electric lamp and conductor will have separate potential differences as the connection is series.

V = 6 V

R₁ = 20 $\Omega$; R₂ = 4 $\Omega$ 

Equivalent Resistance, R_s = R₁ + R₂ = 20 + 4 = 24 $\Omega$

The current through the circuit is given by:
I = V/R_s = 6V / 24 = 0.25 A

 
Potential across the lamp, V₁ = 20 Ω × 0.25 A = 5 V
Potential across the conductor, V₂ = 4 Ω × 0.25 A = 1V