Question

A potentiometer wire has a length 4 m and resistance 4Ω. What resistance must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient of ${10^{ - 3}}$V/cm on the wire?

Answer 1

To solve the question we will use the ohm’s law. The potentiometer is an instrument for measuring an electromotive force by balancing it against the potential difference produced by passing a known current through a known variable resistance.

Complete step by step answer:
Potentiometer wire length, I=10m
Resistance of the potentiometer, R=4Ω
Emf of the accumulator connected in series with the wire, E=2V
A resistance box in series is connected to the potentiometer wire.
Required potential gradient, k=${10^{ - 3}}$V/cm=0.1 V/m
Potential drop along the potentiometer wire, V=k.I=0.1 $\times$ 10=1=1 V
Therefore, current through the potentiometer wire
V= k.I = 0.1$\times$10 =1 = 1V
Now the current flow through the potentiometer wire,
$I = \dfrac{V}{R} = \dfrac{1}{4} = 0.25A$
Let the resistance box’s resistance is R”
Now the relation is using
$I = \dfrac{E}{R} - \dfrac{V}{{R''}} \\\ \Rightarrow 0.25 = \dfrac{2}{4} - \dfrac{1}{{R'}} \\\ \Rightarrow \dfrac{1}{{R'}} = 0.25 \\\ \Rightarrow R' = 4\Omega \\\$

Note: Ohm's Law states that the current flowing in a circuit is directly proportional to the applied potential difference and inversely proportional to the resistance in the circuit. In other words by doubling the voltage across a circuit the current will also double.