Current Electricity 4 Question 5
5. 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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Solution:
- 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
$$ \begin{aligned} \frac{395}{100} & =\frac{4}{R+5} \times R \\ 395(R+5) & =400 R \\ 395 \times 5 & =(400-395) R \end{aligned} $$
$$ \Rightarrow \quad R=395 \Omega $$
