Question

Potentiometer wire of length \[1m\] is connected in series with \[490\Omega \] resistance and \[2V\] battery. If \[0.2mV/cm\] is the potential gradient, the resistance of the potentiometer wire is
A. \[4.9\Omega \]
B. \[7.9\Omega \]
C. \[5.9\Omega \]
D. \[6.9\Omega \]

Answer 1

Hint: The potential gradient is given. Use the relation between the potential gradient with the potential difference across the potentiometer and the length to find the potential difference.
We need to find the current from the given resistance and e.m.f of the battery.
The potential difference is the product of the current in the circuit and the resistance of the potentiometer.

Formula used:
The potential difference across the potentiometer, $P.D = K \times l$
$K$ is the potential gradient and $l$ is the length.
Again, $P.D = I \times {R_p}$
$I = $ the current flowing in the circuit and ${R_p}$ is the resistance of the potentiometer.
$I = \dfrac{E}{{R + {R_p}}}$
$E$ Is the voltage of the battery and $R$ is the external resistance.

Complete step by step answer:
The above problem states that a potentiometer of length $l$ (given $l = 1m$ ), resistance say ${R_p}$, and the potential gradient $K$ (given $K = 0.2mV/cm$ is connected to an external resistance $R$ ( given $R = 490\Omega $ ) and a battery of voltage $E$ (given $E = 2V$ ).
We have to find the resistance of the potentiometer ${R_p}$.
The circuit looks like,

The potential difference across the potentiometer, $P.D = K \times l$
$K = 0.2mV/cm = 0.2 \times {10^{ - 3}}V/cm$
$l = 1m = 100cm$
$\therefore P.D = 0.2 \times {10^{ - 3}} \times 100 = 0.02V$
Again, $P.D = I \times {R_p}$
$I = $ the current flowing in the circuit
$I = \dfrac{E}{{R + {R_p}}} = \dfrac{{0.02}}{{490 + {R_p}}}$
$\therefore \dfrac{{2{R_p}}}{{490 + {R_p}}} = 0.02$ [ equating the values of P.D]
$ \Rightarrow \dfrac{{{R_p}}}{{490 + {R_p}}} = \dfrac{1}{{100}}$
$ \Rightarrow 100{R_p} = 490 + {R_p}$
$ \Rightarrow {R_p} = 4.94\Omega $
So the resistance of the potentiometer $ \Rightarrow {R_p} = 4.94\Omega $

Hence, the correct answer is option (A).

Note: The Potential drop per length of the wire of the potentiometer is called the potential gradient which is represented by, \[{\text{Potential Gradient }} = \dfrac{{{\text{Potential Difference}}}}{{{\text{length of wire}}}}\]
The Potential gradient depends on the strength of the current and the resistance of the wire. So if the current and the resistance is fixed for a particular circuit, the Potential gradient will be constant for that circuit.
The negative potential gradient is equal to the electric field intensity.