Five equal resistances, each of resistance R, are connected as shown in the figure. A battery of V volts is connected between A and B. The current flowing in AFCEB will be

Solution:

The given circuit can be redrawn as shown. From circuit,

$\frac{\mathrm{FC}}{\mathrm{CE}}=\frac{\mathrm{FD}}{\mathrm{DE}}=1$

Thus, it is balanced Wheatstone's bridge, so resistance in arm CD is ineffective and so, no current flows in this arm.

Net resistance of the circuit is

$\begin{array}{l}\frac{1}{{\mathrm{R}}^{\text{'}}}=\frac{1}{\left(\mathrm{R}+\mathrm{R}\right)}+\frac{1}{\left(\mathrm{R}+\mathrm{R}\right)}\\ =\frac{1}{2\mathrm{R}}+\frac{1}{2\mathrm{R}}=\frac{2}{2\mathrm{R}}=\frac{1}{\mathrm{R}}\\ {\mathrm{R}}^{\text{'}}=\mathrm{R}\end{array}$

So, net current drawn from the battery

$i^{\text{'}}=\frac{V}{R^{\text{'}}}=\frac{V}{R}$

As from symmetry, upper circuit AFCEB is half of the whole circuit and is equal to AFDEB. So, in both the halves, half of the total current will flow.
Hence, in AFCEB, the current flowing is

$i=\frac{i\text{'}}{2}=\frac{V}{2R}$