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

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Aicharya · Accepted Answer

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 $I_{F C}=\frac{R}{2 R+2 R}=\frac{V}{4 R}$

Five equal resistances each ' 2 R ' are connected as shown in figure. A battery of ' $V$ ' volts connected…

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