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 p.d. of 1.5 mV is required across the whole wire.

Given,

Emf $(E) = 2\;V$

Resistance of Potentiometer wire $(R) = 3\; \Omega$

Potential Difference $(V) = 1.5 \; mV = 1.5 * 10^{-3}\;V$

Let, $R'$ be the resistance to be connected in series.

$\therefore$ Total resistance of the circuit $(R_T) = R' + 3$

Here, current across the potentiometer circuit $(I) = \frac{E}{R_T} = \frac{2}{R'\; + \; 3}$

Also, we have, p.d. across potentiometer wire $= IR$

or, $1.5 * 10^{-3} = \frac{2}{R' \; + \; 3} * 3$

or, $R' + 3 = 4000$

$\therefore$ $R' = 3997 \; \Omega$

Hence resistance needed in series $= 3997 \; \Omega$

Return to Main Menu