Question

A potentiometer wire of length 200cm has a resistance of 20 ohm it is connected in a series with resistance of 10 ohm and an accumulator of emf 6V having negligible internal resistance. A source of 2.4V is balanced against a length L of the potentiometer wire. The value of L is?
A. 100cm
B. 120cm
C. 110cm
D. 140cm

Answer 1

Hint: We will take the help of the formula for the help of potential gradients to find the unknown resistance. For the potential gradient is voltage divided by the length of the wire. By putting the known value from the equations in the formula we will get the value of resistance which is to be connected in series with the wire of potentiometer and an accumulator.

Formula Used: $I = \dfrac{V}{R}$

Complete step by step answer:
If the current in the potentiometer is increased, it means the voltage it also increased and potential gradient increases but the value of EMF is constant so the balancing length should decrease.

Firstly we can find the current,
$I = \dfrac{V}{R}$
V=6V
$I = \dfrac{6}{{10 + 20}} = \dfrac{6}{{30}} = 1.5A$=0.2A
Total length of the potentiometer=200cm
R=20ohm
So, the resistance between A to C is,
$200cm \to 20ohm$
$1cm \to \dfrac{{20}}{{200}}ohm$/cm
$lcm \to \dfrac{{20}}{{200}} \times l = \dfrac{{20l}}{{200}} = \dfrac{l}{{10}}$
$I = 0.2A$
Resistance (for length l) =$\dfrac{l}{{10}}ohm$
E=2.4V
$\mathop V\nolimits_{AC} = E = I \times \dfrac{i}{{10}}$
$2.4 = 0.2 \times \dfrac{l}{{10}}$
$ \Rightarrow l = \dfrac{{2.4 \times 10}}{{0.2}} = 120cm$
Then, a source of 2.4V is balanced against a length L of the potentiometer wire is 120cm
Therefore, the value of $l = 120cm$

So, the correct answer is option(b).

Note:- The potential drop per unit length of the potentiometer wire is known as the potential gradient of the potentiometer for that circuit. Its S.I unit is $\mathop {Vm}\nolimits^{ - 1} $. The potentiometer can also be used to compare the emf of two primary cells and the internal resistance of the primary cell.