Current Electricity Ques 98
- An ideal battery of $4$ $V$ and resistance $R$ are connected in series in the primary circuit of a potentiometer of length $1$ $ m$ and resistance $5 $ $\Omega$. The value of $R$ to give a potential difference of $5$ $ mV$ across $10 $ $cm$ of potentiometer wire is
(2019 Main, 12 Jan I)
(a) $395 $ $\Omega$
(b) $495 $ $\Omega$
(c) $490 $ $\Omega$
(d) $480 $ $\Omega$
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Answer:
Correct Answer: 98.(a)
Solution:
Formula:
- Given, potential difference of $5$ $ mV$ is across $10$ $ m$ length of potentiometer wire. So potential drop per unit length is
$ =\frac{5 \times 10^{-3}}{10 \times 10^{-2}}=5 \times 10^{-2} (\frac{V}{m}) $
Hence, potential drop across $1 $ $m$ length of potentiometer wire is
$ V _{A B}=5 \times 10^{-2} (\frac{V}{m}) \times 1=5 \times 10^{-2} V $
Now, potential drop that must occurs across resistance $R$ is
$ V _R=4-5 \times 10^{-2}=\frac{395}{100} V $
Now, circuit current is
$ i=\frac{V}{R _{\text {total }}}=\frac{4}{R+5} $
Hence, for resistance $R$, using $V _R=i R$, we get
$ \begin{aligned} \frac{395}{100} & =\frac{4}{R+5} \times R \\ 395(R+5) & =400 R \\ 395 \times 5 & =(400-395) R \end{aligned} $
$ \Rightarrow \quad R=395$ $ \Omega $
BETA
