Question

In the circuit shown, a four - wire potentiometer is made of a $400\text{cm}$ long wire, which extends between $\text{A}$ and $\text{B}$. The resistance per unit length of the potentiometer wire is $x=0.01\Omega /\text{cm}$. If an ideal voltmeter is connected as shown with jockey $J$ at $50\text{cm}$ from end $\text{A}$, the expected reading of the voltmeter will be

• A. $0.20\text{V}$
• B. $0.75\text{V}$
• C. $0.25\text{V}$
• D. $0.50\text{V}$

Answer 1

The correct option is C $0.25\text{V}$

In the given potentiometer, resistance per unit length is $x=0.01\Omega /\text{cm}$.


Length of potentiometer wire is $L=400\text{cm}$

Net resistance of the wire $\text{AB}$ will be

$R_{AB}$ $=$ resistance per unit length $\times$ length of $\text{AB}$

$R_{AB}=0.01\times 400=4\Omega$

Net internal resistance of the cells connected in series,

$r=0.5+0.5=1\Omega$

Current in the given potentiometer circuit is,

$I=\frac{\text{Net emf}}{\text{Total resitance}}=\frac{\text{Net Emf}}{r+R+R_{AB}}$

$I=\frac{1.5+1.5}{1+1+4}=0.5\text{A}$

Reading of the voltmeter when the jockey is at $l^{\prime }=50\text{cm}$ from one end $\text{A}$,

$V=IR=I\left(x{l}^{\prime }\right)=0.5\times (0.01\times 50)$

$\therefore V=\frac{1}{4}=0.25\text{V}$

Hence, option $\left(c\right)$ is correct.