Question

Figure shows a potentiometer. Length of the potentiometer wire $AB$ is $100cm$ and its resistance is $100\Omega$ . $EMF$ of the battery $E$ is $2V$ . A resistance $R$ of $50\Omega$ draws current from the potentiometer. What is the voltage across $R$ when the sliding contact $C$ is at the mid-point of $AB$ ?

• A. $2/3V$
• B. $1v$
• C. $4/3V$
• D. $3/2V$

Answer 1

Answer: (A)

Given, $E=2V$, $R_{AB}=100\Omega$
$l_{AB}=100cm$ and $R=50\Omega$
Circuit according to the question,

$AC=CB=\frac{l_{AB}}{2}=50cm$
$R_{AB}=100\Omega$
$R_{AC}=R_{CB}=\frac{R_{AB}}{2}=\frac{100}{2}$
$\therefore R_{AC}=R_{CB}=50\Omega$
$\therefore$ Net resistance of the circuit, $R_{net}=R_{CB}+\frac{R\cdot R_{AC}}{R+R_{AC}}$
$=50+\frac{50\times 50}{50+50}=50+25$
$R_{net}=75\Omega$
$\therefore$ Current from the battery, $I=\frac{E}{R_{net}}=\frac{2}{75}A$
Using $KVL$ in mesh $1$,
$E-V_{AC}-V_{CB}=0$
$\Rightarrow V_{AC}+V_{CB}=2$
$\Rightarrow V_{AC}+I{R}_{CB}=2$
$\Rightarrow V_{AC}=2-\frac{2}{75}\times 50$
$\Rightarrow V_{AC}=2-\frac{4}{3}=\frac{6-4}{3}=\frac{2}{3}V$