Question

For the shown arrangement of potentiometer, Find the value of $x$ , if $\text{P}$ is a null point. Resistance of portion $\text{OB}$ is $20\Omega$

• A. $x=\frac{8}{5}\text{m}$
• B. $x=\frac{1}{6}\text{m}$
• C. $x=\frac{5}{8}\text{m}$
• D. $x=\frac{9}{10}\text{m}$

Answer 1

The correct option is C $x=\frac{5}{8}\text{m}$

Under the balanced state of potetniometer circuit the current in the primary and secondary circuit is as shown below,


The galvanometer will show zero deflection at balnaced condition.

Applying $\text{KVL}$ in the primary circuit,

$+100-5i_i-20i_1=0$

$\Rightarrow i_1=\frac{100}{25}=4A$

The potential difference across $\text{OB}$ is,

$\Delta V=i_1{R}_{OB}$

or, $\Delta V=4\times 20=80\text{V}$

Potential gradient of wire $\text{OB}$,

$\frac{dv}{dl}=\frac{80}{10}$

$\therefore \frac{dv}{dl}=8\text{V/m}$

Applying $\text{KVL}$ in secondary loop,

$+2+(i_2\times 2)+2i_2-8=0$

$\Rightarrow 4i_2=6$

$\therefore i_2=\frac{3}{2}\text{A}$

Now,

$V_C+2+2\left(i_2\right)=V_D$

$\Rightarrow V_D-V_C=2+(2\times \frac{3}{2})=5\text{volt}$

Now,

$|v_{OP}|=|v_{CD}|$

$\Rightarrow x\times \left(\frac{dv}{dl}\right)=5$

$\therefore x=\frac{5}{\left(\frac{dv}{dl}\right)}=\frac{5}{8}\text{m}$