Question

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

• A. $890\Omega$
• B. $600\Omega$
• C. $650\Omega$
• D. $790\Omega$

Answer 1

Answer: (D)

$790\Omega$

Let, R be the resistance external resistance. The current through the circuit is
$i=\frac{2}{10+R}$

We know, $x=\frac{V}{l}=\frac{2\times 10}{(R+10)}\frac{1}{100}$
where, $x$ is the resistance of the wire.

Also, A source of emf of 10 mV is balanced against a length of 40 cm of the potentiometer wire.

Hence, $V_1=xl$
$10\times {10}^{-3}=\frac{2\times 10}{(R+10)}\times \frac{40}{100}$
$R+10=\frac{8}{10\times {10}^{-3}}$
$\Rightarrow R+10=800$
$\therefore R=790\Omega$