Current Electricity 4 Question 2
2. In the circuit shown, a four-wire potentiometer is made of a $400 cm$ long wire, which extends between $A$ and $B$. The resistance per unit length of the potentiometer wire is $r=0.01$ $\Omega / cm$. If an ideal voltmeter is connected as shown with jockey $J$ at $50 cm$ from end $A$, the expected reading of the voltmeter will be
(a) $0.20 V$
(b) $0.75 V$
(c) $0.25 V$
(d) $0.50 V$
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Solution:
- In given potentiometer, resistance per unit length is $x=0.01 \Omega cm^{-1}$.
Length of potentiometer wire is $L=400 cm$
Net resistance of the wire $A B$ is
$R _{A B}=$ resistance per unit length $\times$ length of $A B$
$$ =0.01 \times 400 $$
$$ \Rightarrow \quad R _{A B}=4 \Omega $$
Net internal resistance of the cells connected in series,
$$ r=0.5+0.5=1 \Omega $$
$\therefore$ Current in given potentiometer circuit is
$$ \begin{aligned} I & =\frac{\text { Net emf }}{\text { Total resistance }}=\frac{\text { Net emf }}{r+R+R _{A B}} \\ & =\frac{3}{1+1+4}=0.5 A \end{aligned} $$
Reading of voltmeter when the jockey is at $50 cm\left(l^{\prime}\right)$ from one end $A$,
$$ \begin{aligned} V & =I R=I\left(x l^{\prime}\right) \\ & =0.5 \times 0.01 \times 50=0.25 V \end{aligned} $$
