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

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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 \(\Omega\)
\(R_2\) = 6 \(\Omega\)
\(R_3\) = 4 \(\Omega\)
\(I_1\)=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, \[\begin{align*} I_1\times R_1&=I_2\times R_2\\ 0.8 \times 3&=I_2 \times 6\\ \therefore I_2&=0.4 \hspace{0.1cm} A \end{align*}\] The total current is then, \[I=I_1+I_2=1.2 \hspace{0.1cm} A\] So, the potential difference across the 4 Ω resistor is, \[\begin{align*} &=I\times 4\\ &=1.2 \times 4\\ &=4.8 \hspace{0.1cm} V\\ \end{align*}\]

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