Question

Figure shows a simple potentiometer circuit for measuring a small emf produced by a thermocouple.

The meter wire $PQ$ has a resistance of $5\Omega$, and the driver cell has an emf of $2.00V$. If a balance point is obtained $0.600m$ along $PQ$ when measuring an emf of $6.00mV$, what is the value of resistance $R$ ?

• A. $95\Omega$
• B. $995\Omega$
• C. $195\Omega$
• D. $1995\Omega$

Answer 1

Answer: (B)

The voltage per unit length on the meter wire $PQ$ is
$\frac{6.00mV}{0.60m}$
or $10mV{m}^{-1}$
Hence, potential across the meter wire $PQ$ is $10mV{m}^{-1}$ $(1m)=10mV$.
Current drawn from the driver cell is
$I=\frac{10mV}{5\Omega }=2mA$
Resistance of the resistor $R$ is
$R=\frac{2V-10mV}{2mA}=\frac{1990mV}{2mV}=995\Omega$