Question

Find the current through the middle branch of the unbalanced Wheatstone's bridge circuit, as shown in figure.

• A. $0.460\text{A}$
• B. $0.023\text{A}$
• C. $0.1\text{A}$
• D. $0.046\text{A}$

Answer 1

The correct option is B $0.023\text{A}$

We can solve this question by distributing the potentials at different nodes as shown in figure.


Writing, $\text{KCL}$ equation for the node $x$, we get

$\frac{x}{2}+\frac{x-y}{4}+\frac{x-2}{1}=0$

$\Rightarrow 7x-y=8.....\left(1\right)$

Writing, $\text{KCL}$ equation for the node $y$, we get

$\frac{y}{3}+\frac{y-x}{4}+\frac{y-2}{2}=0$

$\Rightarrow 13y-3x=12.....\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$, we get

$88y=108$

$\Rightarrow y=\frac{108}{88}=\frac{27}{22}\text{V}$

Using equation $\left(1\right)$,

$x=\frac{y+8}{7}=\frac{\frac{27}{22}+8}{7}=\frac{203}{154}\text{V}$

Thus, the current in middle branch is,

$I_{4\Omega }=\frac{x-y}{4}=\frac{\left(\frac{203}{154}\right)-\left(\frac{27}{22}\right)}{4}$

$\Rightarrow I_{4\Omega }=0.023\text{A}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.