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 $\mathrm{\Omega }$ resistor, $I_1$ = 4 A
current through 25 $\mathrm{\Omega }$ resistor, $I$ = ?
current through 20 $\mathrm{\Omega }$ resistor, $I_4$ = ?

Since $R_1$ and $R_2$ are parallel, potential difference across them is equal.
$$\begin{array}{rl}\text{p.d. across}{R}_1& =\text{p.d. across}{R}_2\\ I_1{R}_1& =I_2{R}_2\\ 4\times 6& =I_2\times 8\\ \therefore I_2& =3A\end{array}$$
Current through upper branch is, $$\begin{array}{rl}I& =I_1+I_2\\ & =7A\end{array}$$ So the current through 25 Ω resistor is 7 A.
Equivalent resistance of R₁ and R₂ is, $$\begin{gathered} \begin{array}{rlr}\frac{1}{R_{12}}& =\frac{1}{R_1}+\frac{R_2}{ \\ }& =\frac{1}{6}+\frac{1}{8}\\ & =\frac{24}{7}\mathrm{\Omega }\end{array} \end{gathered}$$ Then, total resistance in the upper branch is, $$\begin{array}{rl}R& =R_{12}+R_3\\ & =\frac{24}{7}+25\\ & =\frac{199}{7}\mathrm{\Omega }\end{array}$$ Since the two branches i.e., upper and lower branch are parallel to each other, their potential difference is equal.
$$\begin{array}{rl}\text{p.d. across upper branch}& =\text{p.d. across lower branch}\\ I\times R& =I_4\times R_4\\ 7\times \frac{199}{7}& =I_4\times 20\\ \therefore I_4& =9.95A\end{array}$$ Hence, the current through 25 Ω resistor is 7 A and through 20 Ω resistor is 9.95 A.