Question

The figure shows a simple potentiometer circuit for measuring a small $\text{EMF}$ produced by a thermocouple.


The metre wire $PQ$ has a resistance of $5\Omega$ and the driver cell has an $\text{EMF}$ of $2.00\text{V}$. If a balance point is obtained $0.6\text{m}$ along $PQ$, when measuring an $\text{EMF}$ of $6.00\text{mV}$, what is the value of resistance $R$?

• A. $1005\Omega$
• B. $995\Omega$
• C. $1000\Omega$
• D. $2000\Omega$

Answer 1

The correct option is B $995\Omega$

The voltage per unit length on the metre wire $PQ$ is,

$\frac{6.00\text{mV}}{0.60\text{m}}=10\text{mV/m}$

Hence, potential across the metre wire $PQ$ $\left(1\text{m}\right)$ is,

$V_{\text{PQ}}=10\text{mV/m}\times \text{1 m}=10\text{mV}$,

Current drawn from the driver cell is,

$I=\frac{V_{\text{PQ}}}{R_{\text{PQ}}}=\frac{10\text{mV}}{5\Omega }=2\text{mA}$

The resistance of the resistor $R$ is,

$R=\frac{2\text{V}-10\text{mV}}{2\text{mA}}=\frac{1990\text{mV}}{2\text{mA}}=995\Omega$

Hence, option $\left(b\right)$ is the correct answer.