Explanation:
The Wheatstone bridge in given circuit is balanced, we can remove its middle branch. After this remaining resistance of \(16\,\, \Omega, 8\,\, \Omega\) and \(16\,\, \Omega\) will be in parallel for which we can find the equivalent resistance,\(R_{e q}=\left(\dfrac{1}{16}+\dfrac{1}{8}+\dfrac{1}{16}\right)^{-1}+1=5 \Omega\)
Current supplied by the battery, \(I=\dfrac{V}{R_{e q}}=\dfrac{60}{5}=12 {~A}\)
In the circuit 12 A will be distributed in the three branches in parallel. The potential drop across the three branches in parallel,\(X Y=I R_{X Y}=12 \times 4=48 {~V}\)
Thus, current in top \(16 \Omega\) resistance branch, \(I_{6 \Omega}=\dfrac{V_{K Y}}{16}=\dfrac{48}{16}=3 {~A}\)
