Question

An ideal battery of emf $2V$ and a series resistance $R$ are connected in the primary circuit of a potentiometer of length $1$ m and resistance $\Omega$. The value of $R$ to give a potential difference of $5$ mV across the 10cm of potentiometer wire is:

• A. $180\Omega$
• B. $190\Omega$
• C. $195\Omega$
• D. $200\Omega$

Answer 1

Answer: (C)

$195\Omega$

Let,

$L=1m=100cm$

$l=10cm$

Voltage drop across potentiometer wire is

$V=5\times {10}^{-3}\frac{100}{10}=0.05V$

Current will be

$I=\frac{2}{R+R^{{}^{\prime }}}=\frac{2}{R+5}$

We know that $V=I{R}^{{}^{\prime }}$

where, $R^{{}^{\prime }}$ is resistance of potentiometer wire

$0.05=\frac{2}{R+5}5$

$R+5=200$

$R=195\Omega$