Question

A potentiometer wire of $10m$ length and $20ohm$ resistance is connected in series with a resistance $r$ ohms and a battery of emf $2V$, negligible internal resistance, Potential gradient the wire is $0.16$ millivolt/centimetre then $r$ is ohms:

• A. $50\Omega$
• B. $60\Omega$
• C. $230\Omega$
• D. $46\Omega$

Answer 1

The correct option is C $230\Omega$

Given that,

Resistance of the wire $R=20\Omega$

Potential meter length $L=10m$

e. m. f $E=2volt$

Potential gradient $K=0.16mv/cm$

Now, the current is

$I=\frac{E}{R+r}$

Potential across the wire is

$V=IR$

$V=\frac{ER}{R+r}$

Potential gradient is

$K=\frac{V}{L}$

$K=\frac{ER}{L(R+r)}....\left(I\right)$

Now, put the all values in equation (I)

$0.016=\frac{2\times 20}{10(20+r)}$

$3.2+0.16r=40$

$0.16r=40-3.2$

$0.16r=36.8$

$r=\frac{36.8}{0.16}$

$r=230\Omega$

Hence, the value of$r$ is $230\Omega$