Question

A potentiometer wire of length $10\text{m}$ and resistance $30\text{ohm}$ is connected in series with a battery of emf $2.5\text{V}$, internal resistance $5\text{ohm}$ and external resistance $R$. If the fall of potential along the potentiometer wire is $50\text{mV/m}$ the value of $R$ in $\Omega$ is

• A. 115
• B. 115.0
• C. 115.00

Answer 1

Answer: (A), (B), (C)

Potential gradient $\frac{V}{l}=50\times {10}^{-3}\text{V/m}$

The P.D across the wire is,

$V_{\text{AB}}=\frac{v}{l}\times 10=50\times {10}^{-3}\times 10=0.5\text{V}$

$I=\frac{V_{\text{AB}}}{R_{\text{w}}}=\frac{0.5}{30}=\frac{1}{60}\text{A}$

$E=I[R_{\text{w}}+R+r]$

$2.5=\frac{1}{60}[30+R+5]$

$2.5\times 60=R+35$

$R=150-35$

$R=115\Omega$