Question

An inductor of $20H$ and a resistance of $10\Omega$, are connected to a battery of $5V$ in series, then initial rate of charge of current is:
(A) $0.5A{s^{ - 1}}$
(B) $2A{s^{ - 1}}$
(C) $2.5A{s^{ - 1}}$
(D) $0.25A{s^{ - 1}}$

Answer 1

Hint
We are given that an inductor and a resistor are connected in series to a battery of a given potential and we are asked to find the initial rate of change of current. Thus, clearly we are given with an $LR$circuit. Hence, we will discuss the decay current in an $LR$ circuit and then differentiate it to find the change.
$\Rightarrow i = {i_0}(1 - {e^{\dfrac{{ - Rt}}{L}}})$
Where, $i$ is the decay current of the circuit, ${i_0}$ is the steady current, $e$ is the Euler’s constant, $R$ is the resistance of the circuit, $L$ is the inductance of the circuit and $t$ is the time at which we want the decay current at.

Complete step by step answer
We know,
$\Rightarrow i = {i_0}(1 - {e^{\dfrac{{ - Rt}}{L}}})$
Then differentiating both sides with respect to time in order to find the change in current,
$\Rightarrow \dfrac{{di}}{{dt}} = \dfrac{d}{{dt}}[{i_0}(1 - {e^{\dfrac{{ - Rt}}{L}}})] \Rightarrow \dfrac{{di}}{{dt}} = \dfrac{d}{{dt}}[{i_0} - {i_0}{e^{\dfrac{{ - Rt}}{L}}})]$
Now, Implementing the fundamentals rule of differentiation, we get
$\Rightarrow \dfrac{{di}}{{dt}} = \dfrac{{d{i_0}}}{{dt}} - \dfrac{d}{{dt}}{i_0}{e^{\dfrac{{ - Rt}}{L}}}$
Now,
${i_0}$ as we know is the steady current flowing through the circuit which is a constant at any point of time. Also, we know that the value of differentiation of any constant is zero.
Thus, we can say
$\Rightarrow \dfrac{{di}}{{dt}} = 0 - \dfrac{d}{{dt}}{i_0}{e^{\dfrac{{ - Rt}}{L}}}$
Now, The next term is in the form of a differentiation of a constant multiplied by a varying term. That means it is of the form $\dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}k{y_1}$, where $k$ is a constant and ${y_1}$ is a varying term.
Thus, applying this and the chain rule of differentiation, we get
$\Rightarrow \dfrac{{di}}{{dt}} = - {i_o}(\dfrac{{ - R}}{L})({e^{\dfrac{{ - Rt}}{L}}})$
Now, We were asked to find the initial current change.
Thus, we take $t = 0$
And putting the given values $L = 20H$ and $R = 10\Omega$
And,
$\Rightarrow {i_0} = \dfrac{V}{R} = \dfrac{5}{{10}} = \dfrac{1}{2}A$
Thus, we put this value and we get
$\Rightarrow \dfrac{{di}}{{dt}} = \dfrac{1}{2} \times \dfrac{{10}}{{20}} \times {e^{\dfrac{{ - 10 \times 0}}{{20}}}}$
Thus, we get
$\Rightarrow \dfrac{{di}}{{dt}} = \dfrac{1}{4} = 0.25A{s^{ - 1}}$
Hence, the correct option is (D).

Note
Here we discussed the decay current of an $LR$ circuit and we went through the above formula and through a calculation and arrived at an answer. But if we were to discuss an $LC$ or $RC$ circuit, then the fundamental formula of the decay current changes and hence changes the total workflow and calculation and hence the final answer.