Five equal resistances each ' 2 R ' are connected as shown in figure. A battery of ' $V$ ' volts connected between $A$ and B. Then current through FC is

• Correct answer
  A

  $\frac{V}{4 R}$
• B

  $\frac{V}{8 R}$
• C

  $\frac{V}{R}$
• D

  $\frac{V}{2 R}$

Solution

The circuit diagram can be redrawn as
$\frac{\mathrm{R}_{\mathrm{FC}}}{\mathrm{R}_{\mathrm{FD}}}=\frac{2 \mathrm{R}}{2 \mathrm{R}}=1, \frac{\mathrm{R}_{\mathrm{CE}}}{\mathrm{R}_{\mathrm{DE}}}=\frac{2 \mathrm{R}}{2 \mathrm{R}}=1 \quad \therefore \quad \frac{\mathrm{R}_{\mathrm{FC}}}{\mathrm{R}_{\mathrm{FD}}}=\frac{\mathrm{R}_{\mathrm{CE}}}{\mathrm{R}_{\mathrm{DE}}}$ $\Rightarrow$ The circuit is a balanced Wheatstone bridge. $\therefore$ The current through FC is $I_{F C}=\frac{R}{2 R+2 R}=\frac{V}{4 R}$

Hence, the correct answer is option: 'A'.