Question

As shown in the fig. a simple potentiometer circuit for measuring a small emf produced by a thermocouple. The meter wire PQ has a resistance 5 $\Omega$ and the driver cell has an emf of 20011 (f a balance point is obtained 0.600 m along PO when measuring an emf of 6.00 mV, what is the value of resistance R ?

• A. $995\Omega$
• B. $1995\Omega$
• C. $2995\Omega$
• D. none of these

Answer 1

The correct option is A $995\Omega$

The voltage per unit light of the metre wire $PQ$ is $\left(\frac{6.00mV}{0.600m}\right)$ i.e. $10mV/m$.

Hence potential difference across the metre wire is $10mV\times 1mV$

The current drawn from the driver cell is $i=\frac{10mV}{5\Omega }=2mA$.

The resistance $R=\frac{(2V-10mV)}{2mA}=\frac{1990mV}{2mA}=995\Omega$.