Question

Figure shows a potentiometer. Length of the potentiometer wire $AB$ is $100 \,cm$ and its resistance is $100\, \Omega$ . $EMF$ of the battery $E$ is $2 V$ . 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. $2/3 \,V$
• B. $1\, v$
• C. $4/3 \,V$
• D. $3/2 \,V$

Answer 1

Answer: (A)

$2/3 \,V$

Answer 2

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

$AC=CB=\frac{l_{AB}}{2}=50\,cm$
$R_{AB}=100\,\Omega$
$R_{AC}=R_{CB}=\frac{R_{AB}}{2}=\frac{100}{2}$
$\therefore R_{AC}=R_{CB}=50\,\Omega$
$\therefore$ Net resistance of the circuit, $R_{net}=R_{CB}+\frac{R\cdot R_{AC}}{R+R_{AC}}$
$=50+\frac{50\times50}{50+50}=50+25$
$R_{net}=75\,\Omega$
$\therefore$ Current from the battery, $I=\frac{E}{R_{net}}=\frac{2}{75}A$
Using $KVL$ in mesh $1$,
$E-V_{AC}-V_{CB}=0$
$\Rightarrow V_{AC}+V_{CB}=2$
$\Rightarrow V_{AC}+IR_{CB}=2$
$\Rightarrow V_{AC}=2-\frac{2}{75}\times50$
$\Rightarrow V_{AC}=2-\frac{4}{3}=\frac{6-4}{3}=\frac{2}{3}V$