In the figure, the current through the 3 Ω resistor is 0.8 Ω A. Find the potential difference across the 4 Ω resistor.

Solution:
Let,

R_{1}

= 3

Ω

R_{2}

= 6

Ω

R_{3}

= 4

Ω

I_{1}

=0.8 A

Since R₁ and R₂ are parallel to each other, the potential difference across each of these are equal.
p.d across R₁ = p.d. across R₂ Then,

\begin{aligned} I_{1} \times R_{1} & = I_{2} \times R_{2} \\ 0.8 \times 3 & = I_{2} \times 6 \\ ∴ I_{2} & = 0.4 A \end{aligned}

The total current is then,

I = I_{1} + I_{2} = 1.2 A

So, the potential difference across the 4 Ω resistor is,

\begin{aligned} = I \times 4 \\ = 1.2 \times 4 \\ = 4.8 V \end{aligned}

Physics in Depth