Current Electricity Ques 82
- 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 $ $V$
Show Answer
Answer:
Correct Answer: 82.(c)
Solution:
Formula:
- 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 $A B$
$ = \quad 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} $
BETA
