Consider the figure below. The current through 6 Ω resistor is 4 A in the direction shown. What are the currents through the 25 Ω and 20 Ω resistors?

Consider the figure below. The current through 6 Ω resistor is 4 A in the direction shown. What are the currents through the 25 Ω and 20 Ω resistors?

Solution:
current through 6 \(\Omega\) resistor, \(I_1\) = 4 A
current through 25 \(\Omega\) resistor, \(I\) = ?
current through 20 \(\Omega\) resistor, \(I_4\) = ?

Since \(R_1\) and \(R_2\) are parallel, potential difference across them is equal.
\[\begin{align*} \text{p.d. across}\hspace{0.1cm} R_1&=\text{p.d. across}\hspace{0.1cm} R_2\\ I_1 R_1&=I_2 R_2\\ 4 \times 6 &=I_2 \times 8\\ \therefore I_2&=3 \hspace{0.1cm} A\\ \end{align*}\]
Current through upper branch is, \[\begin{align*} I&=I_1+I_2\\ &=7 \hspace{0.1cm} A\\ \end{align*}\] So the current through 25 Ω resistor is 7 A.
Equivalent resistance of R₁ and R₂ is, \[\begin{align*} \frac{1}{R_{12}}&=\frac{1}{R_1}+\frac{R_2}\\ &=\frac{1}{6}+\frac{1}{8}\\ &=\frac{24}{7} \hspace{0.1cm} \Omega\\ \end{align*}\] Then, total resistance in the upper branch is, \[\begin{align*} R&=R_{12}+R_3\\ &=\frac{24}{7}+25\\ &=\frac{199}{7} \hspace{0.1cm} \Omega\\ \end{align*}\] Since the two branches i.e., upper and lower branch are parallel to each other, their potential difference is equal.
\[\begin{align*} \text{p.d. across upper branch}&=\text{p.d. across lower branch}\\ I \times R&=I_4\times R_4\\ 7 \times \frac{199}{7}&=I_4 \times 20\\ \therefore I_4&= 9.95 \hspace{0.1cm} A\\ \end{align*}\] Hence, the current through 25 Ω resistor is 7 A and through 20 Ω resistor is 9.95 A.