Question

Five equal resistors each of resistance $R$ are connected as shown in the Figure. A battery of voltage $V$ is connected between $\text{A}$ and $\text{B}$. The current flowing in $\text{AFCEB}$ will be

• A. $\frac{V}{R}$
• B. $\frac{V}{2R}$
• C. $\frac{2V}{R}$
• D. $\frac{3V}{R}$

Answer 1

The correct option is B $\frac{V}{2R}$


Rearranging the given circuit we get,


Since it is a balanced Wheatstone bridge, we can remove the middle resistance $R$. So network becomes

Here, $R_{eq}$ for the branch $\text{AFCEB}$ will be $2R$.

$\therefore i_{AFCEB}=i_1=\frac{V}{2R}$

Hence, option $\left(b\right)$ is correct.

Why this question ? Key concept: Balancing condition of Wheatstone's bridge is, $\frac{R_1}{R_2}=\frac{R_3}{R_4}$