Question

In figure, the potentiometer wire AB has a resistance of $5$ ohm and length $10 m$. The balancing length AJ for the cell of e.m.f. of $0.4 m$ is:

Answer 1

Hint: The potentiometer is based on the theory of potential gradient and null deflection. So, to work problems on the potentiometer, we should find a potential gradient and later for a null location. The galvanometer has a null point as the potential difference is equivalent to zero, and there is no current flow.

Complete step-by-step solution:
The potentiometer consists of L, along the resistive wire, and a battery of known EMF V, known as driver cell voltage. Consider a primary circuit arrangement by joining the two ends of L to the battery ends. One end of the primary circuit is joined to the cell whose EMF E is to be estimated, and the other end is joined to galvanometer G. This circuit is supposed to be a secondary circuit.
The practical principle depends on the potential to cross any portion of the wire, which is directly proportional to the wire length with a uniform cross-sectional area, and current movement is constant.
$E=KL$ ……(i)
L: length of potentiometer wire
E: EMF
K: potential gradient
Put $E=iR$ (ohm’s law) …...(ii)
Equate both equations, we get
$K=\dfrac{iR}{L}$ ……(iii)
R is the resistance of length of wire L.
i is the current.
$R =5 ohm$
$i=\dfrac{5}{4.5+5} = 0.526 A$
$L=10 m$
Put all values in (iii), we get
$K=\dfrac{0.526 \times 5}{10} = 0.236 $
Now we have to find the balancing length for $0.4 V$:
$E =KL$
$\implies 0.4 = 0.263 \times L$
$L = \dfrac{0.4}{0.263} = 1.52 m$
The balancing length AJ for the cell of e.m.f. of $0.4 V$ is $1.52 m$.

Note: The potentiometer is a simplistic device used to estimate electrical potentials. One form is a uniform high-resistance wire joined to insulating support. In use, a flexible regulated voltage source E, of higher magnitude than the potential to be estimated, is combined across the wire to transfer a steady current into it.