The driver cell of a potentiometer has an e.m.f. of 2 V and negligible internal resistance. The potentiometer wire has a resistance of 3 Ω. Calculate the resistance needed in series with the wire if a pd 5 mV is required across the whole wire. The wire is 100 cm long and a balanced length of 60 cm is obtained for a thermocouple of emf E. What is the value of E?

Consider $AB$ is the potentiometer wire,

$V = 2\;V$

$R_{AB} = 3 \; \Omega$ 

Let $R$ be the resistance needed in the series, the the potential difference across $AB = 5\;mV = 5 * 10^{-3}\;V$

i.e. $I\;R_{AB} = 5\;mV$

or, $\frac{V}{R_{AB} + R} * R_{AB} = 5 * 10^{-3}$

or, $\frac{2}{3 \; + \; R} * 3 = 5 * 10^{-3}$

By Solving, we get

$R = 1197 \Omega$

Again,

$l_1 = 100 \; cm$

$V_1 = 5\; mV$

$l_2 = 60 \; cm$

$V_2 = ?$

We have,

$\frac{V_1}{V_2} = \frac{l_1}{l_2}$

$\frac{5}{V_2} = \frac{100}{60}$

$\therefore$  $V_2 = 3\;mV$

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