Question

The resistance for a unit length of the potentiometer wire is $1.2\,\Omega {m^{ - 1}}$. If the potential gradient is $0.009\,Vc{m^{ - 1}}$, the value of the current in ampere in the primary circuit will be:-
(A) $0.4$
(B) $0.75$
(C) $0.5$
(D) $0.25$

Answer 1

Calculate the value of the potential difference by substituting the potential gradient in the formula. With the help of this output, calculate the current flowing through the circuit by substituting the known parameters in the ohm’s law given below.
Useful formula:
(1) The potential difference developed along the circuit is given by
$V = \dfrac{K}{L}$
Where $V$ is the potential difference, $K$ is the potential gradient and $L$ is the length of the wire.
(2) The ohm’s law is given by
$V = IR$
Where $I$ is the current flowing through the circuit and $R$ is the resistance of the circuit.

Complete step by step solution:
It is given that the
Resistance along the unit length, $\dfrac{R}{L} = 1.2\,\Omega {m^{ - 1}}$
The potential gradient, $K = 0.009\,Vc{m^{ - 1}} = 0.9\,V{m^{ - 1}}$
By using the formula (1), the potential difference along the circuit is calculated.
$V = \dfrac{K}{L}$
Substituting the values in the above equation,
$V = \dfrac{{0.9}}{L}$ ------(1)
Using the formula (2),
$V = IR$
Substituting the (1) and $R = 1.2L$ in the above equation,
$\dfrac{{0.9}}{L} = I \times 1.2L$
By grouping the similar terms,
$I = \dfrac{{0.9}}{{1.2{L^2}}}$
Since the length of the wire is not given, it is taken as $1\,m$
$I = \dfrac{{0.9}}{{1.2}}$
$I = 0.75\,A$
Hence the current flows through the primary circuit is $0.75\,A$.

Thus the option (B) is correct.

Note: The resistance per unit length of the conductor is directly proportional to the ratio of the resistivity and the cross sectional area of the conductor. In this question, the resistivity value and the cross sectional area of the wire is not given, and hence the length is taken as $1\,m$.