The sliding contact C is at one-fourth of the | vedclass

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(A) The sliding contact $C$ divides the wire $AB$ into two parts: $AC$ and $CB$. Since $C$ is at one-fourth of the length,the resistance of $AC$ is $R

Vedclass · Accepted Answer

$\frac{4 V_0 R}{3 R_0 + 16 R}$

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(A) The sliding contact $C$ divides the wire $AB$ into two parts: $AC$ and $CB$. Since $C$ is at one-fourth of the length,the resistance of $AC$ is $R_{AC} = R_0 / 4$ and the resistance of $CB$ is $R_{CB} = 3 R_0 / 4$.The resistor $R$ is connected in parallel with the part $AC$. The equivalent resistance of the parallel combination of $R$ and $R_{AC}$ is $R_p = \frac{R \cdot (R_0 / 4)}{R + (R_0 / 4)} = \frac{R R_0}{4 R + R_0}$.Now,the circuit consists of $R_p$ and $R_{CB}$ in series across the voltage source $V_0$. The potential drop $V$ across the resistor $R$ (which is the same as the potential drop across $R_p$) is given by the voltage divider rule:$V = V_0 \cdot \frac{R_p}{R_p + R_{CB}}$Substituting the values:$V = V_0 \cdot \frac{\frac{R R_0}{4 R + R_0}}{\frac{R R_0}{4 R + R_0} + \frac{3 R_0}{4}}$Multiplying the numerator and denominator by $4(4 R + R_0)$:$V = V_0 \cdot \frac{4 R R_0}{4 R R_0 + 3 R_0(4 R + R_0)} = V_0 \cdot \frac{4 R R_0}{4 R R_0 + 12 R R_0 + 3 R_0^2} = V_0 \cdot \frac{4 R R_0}{16 R R_0 + 3 R_0^2}$Dividing by $R_0$:$V = \frac{4 V_0 R}{16 R + 3 R_0}$

The sliding contact $C$ is at one-fourth of the length of the potentiometer wire $(AB)$ from $A$ as shown in the circuit diagram. If the resistance of the wire $AB$ is $R_0$,then the potential drop $(V)$ across the resistor $R$ is

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