Question

The e.m.f of the driver cell of a potentiometer is $2V$ and its internal resistance is negligible. The length of the potentiometer wire is $100cm$ and resistance is $5\Omega$. How much resistance is to be connected in series with the potentiometer wire to have a potential gradient of $0.05mV{cm}^{-1}$?

• A. $2500\Omega$
• B. $1995\Omega$
• C. $2000\Omega$
• D. $1990\Omega$

Answer 1

The correct option is B

$1995\Omega$

Given Potential Gradient $k=0.05mV/cm=5mV/m$
Current in Potentiometer wire is
$I=\frac{V}{r}$ $(V=\text{potential drop across potentiometer wire}=kl)$
Where $l$ is length of potentiometer wire
$r$ is resistance of potentiometer wire

$I=\frac{kl}{r}=\frac{5\times {10}^{-3}\times 1}{5}={10}^{-3}A$

Now As a Resistance $R$ is connected in series to potentiometer wire, net resistance is $R+r$
$E=I(R+r)\Rightarrow 2={10}^{-3}(R+5)\Rightarrow R=1995\Omega$