Question

A potentiometer wire of length $100 cm$ has a resistance of $10 \Omega$. It is connected in series with a resistance and a cell of emf $2 V$ and of negligible internal resistance. A source of $emf 10 mV$ is balanced against a length of $40 cm$ of the potentiometer wire. What is the value of external resistance ?

• A. $790 \Omega$
• B. $890 \Omega$
• C. $990 \Omega$
• D. $1090 \Omega$

Answer 1

Answer: (A)

The current in the potentiometer wire $AC$ is
$I=\frac{2}{10+R}$

the potential difference across the potentiometer wire is
$V =$ current $\times$ resistance $=\frac{2}{10+ R } \times \frac{10}{100} \ldots(i)$
The length of the wire is $l=100 cm$
So, the potential gradient along the wire is
$k =\frac{ V }{l}=\left(\frac{2}{10+ R }\right) \times \frac{10}{100} \cdots$(ii)
The source of emf $10 mV$ is balanced against a length of $40 cm$ of the potentiometer wire
i.e. $10 \times 10^{-3}= k \times 40$
or $10 \times 10^{-3}=\frac{2}{(10+ R )} \times \frac{40}{10} \quad$ (Usingii)
or $R =790 \Omega$