A resistance of $R \Omega$ draws current from a potentiometer as shown in the figure. The potentiometer has a total resistance $R_o\Omega$ . A voltage $V$ is supplied to the potentiometer. Derive an expression for the voltage across $R$ when the sliding contact is in the middle of the potentiometer.
CBSE OUTSIDE DELHI - SET 3 2014
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$\text{R}_{total} = \frac{\text{R}_{o}}{2} + \frac{\frac{\text{R}_{o}}{2}.\text{R}}{\frac{\text{R}_{o}}{2} + \text{R}}$
$ = \frac{\text{R}(\text{R}_{o} + 4\text{R})}{2(\text{R}_{o} + 2 \text{R})}$
$\text{I}_{(total)} = \frac{\text{V}}{\text{R}_{total}}$
Current through $R = I_{2 } = \text{I}_{total}\text{ x}\frac{\frac{\text{R}_{o}}{2}}{\frac{\text{R}_{o}}{2}+\text{R}}$
$ = \text{I}_{total}\text{ x}\frac{\text{R}_{o}}{\text{R}_{o} + 2 \text{R}}$
$ = \frac{\text{V}.2(\text{R}_{o} + 2 \text{R})}{\text{R}(\text{R}_{o} + 4\text{R})}\text{x}\frac{\text{R}_{o}}{\text{R}_{o} + 2 \text{R}}$
$ = \frac{2\text{VR}_{o}}{\text{R}(\text{R}_{o} + 4 \text{R})}$
Voltage across $R = I_2 R = \bigg(\frac{2\text{VR}_{o}}{\text{R}_{o} + 4 \text{R}}\bigg).$
$ = \frac{\text{R}(\text{R}_{o} + 4\text{R})}{2(\text{R}_{o} + 2 \text{R})}$
$\text{I}_{(total)} = \frac{\text{V}}{\text{R}_{total}}$
Current through $R = I_{2 } = \text{I}_{total}\text{ x}\frac{\frac{\text{R}_{o}}{2}}{\frac{\text{R}_{o}}{2}+\text{R}}$
$ = \text{I}_{total}\text{ x}\frac{\text{R}_{o}}{\text{R}_{o} + 2 \text{R}}$
$ = \frac{\text{V}.2(\text{R}_{o} + 2 \text{R})}{\text{R}(\text{R}_{o} + 4\text{R})}\text{x}\frac{\text{R}_{o}}{\text{R}_{o} + 2 \text{R}}$
$ = \frac{2\text{VR}_{o}}{\text{R}(\text{R}_{o} + 4 \text{R})}$
Voltage across $R = I_2 R = \bigg(\frac{2\text{VR}_{o}}{\text{R}_{o} + 4 \text{R}}\bigg).$
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