Question

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

The meter wire $P Q$ has a resistance of $5\, \Omega$, and the driver cell has an emf of $2.00 \, V$. If a balance point is obtained $0.600\, m$ along $P Q$ when measuring an emf of $6.00 \, mV$, 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 $P Q$ is
$\frac{6.00 \,mV }{0.60 \,m }$
or $ 10 \,m\,Vm ^{-1}$
Hence, potential across the meter wire $P Q$ is $10\, m\,Vm ^{-1}$ $(1 m )=10\, mV$.
Current drawn from the driver cell is
$I=\frac{10 \,mV }{5 \,\Omega}=2 \,mA$
Resistance of the resistor $R$ is
$R=\frac{2 V -10\, mV }{2 \,mA }=\frac{1990\, mV }{2 mV }=995\, \Omega$