Question

In series with $20\Omega$ resistor, a $5H$ inductor is placed. To the combination, an emf of $5V$ is applied. What will be the rate of increase of current at $t = 0.25s$ ?
(A) $e$
(B) ${e^{ - 2}}$
(C) ${e^{ - 1}}$
(D) None of these

Answer 1

Here, a battery is connected to a series combination of a resistor and an inductor. As you can see that the resulting circuit will be an LR circuit. Initially at $t = 0$, the current will start to build up. An emf will generate across the inductor due to self-inductance. Apply Kirchhoff’s loop law and try to obtain an expression for the rate of increase of current.

Complete step by step answer:
First, we will derive general expression for an LR circuit.Consider a resistor of resistance $R$ and an inductor of inductance $L$. Connect them in series. Now connect a battery of emf equal to $\xi$. We have built an LR circuit. The emf generated across the inductor due to self-inductance is proportional to the rate of change of current passing through the inductor. Mathematically, the emf induced is given by$\left( { - L\dfrac{{di}}{{dt}}} \right)$.
Now, if we apply Kirchhoff’s loop law, we get, $\xi - L\dfrac{{di}}{{dt}} - iR = 0$. Here, the constants are $\xi ,L\& R$.
Let us rearrange the equation so that we can integrate the equation in order to obtain the current in the circuit as a function of time.

$$\xi - iR = L\dfrac{{di}}{{dt}} \\\ \Rightarrow\dfrac{1}{L}dt = \dfrac{{di}}{{\xi - iR}} \\\$$

You can see that the equation is ready to be integrated, let us integrate with the initial conditions as follows: at $t = 0,i = 0$

$$\int\limits_0^t {\dfrac{1}{L}dt} = \int\limits_0^i {\dfrac{{di}}{{\xi - iR}}} \\\ \Rightarrow\dfrac{1}{L}\left. t \right|_0^t = \left. {\dfrac{{\ln \left( {\xi - iR} \right)}}{{ - R}}} \right|_0^i \\\ \Rightarrow- \dfrac{R}{L}t = \ln \left( {\xi - iR} \right) - \ln \left( {\xi - 0R} \right) \\\ \Rightarrow- \dfrac{R}{L}t = \ln \left( {\xi - iR} \right) - \ln \left( \xi \right) \\\ \Rightarrow- \dfrac{R}{L}t = \ln \left( {\dfrac{{\xi - iR}}{\xi }} \right) \\\ \therefore i = \dfrac{\xi }{R}\left( {1 - {e^{ - \dfrac{{Rt}}{L}}}} \right) \\\$$

In order to get the rate of increase of current, you need to differentiate the current with respect to time.

$$i = \dfrac{\xi }{R}\left( {1 - {e^{ - \dfrac{{Rt}}{L}}}} \right) \\\ \Rightarrow\dfrac{{di}}{{dt}} = \dfrac{{d\left( {\dfrac{\xi }{R}\left( {1 - {e^{ - \dfrac{{Rt}}{L}}}} \right)} \right)}}{{dt}} \\\ \Rightarrow\dfrac{{di}}{{dt}} = \dfrac{\xi }{R}\dfrac{{d\left( {1 - {e^{ - \dfrac{{Rt}}{L}}}} \right)}}{{dt}} \\\ \Rightarrow\dfrac{{di}}{{dt}} = \dfrac{\xi }{R}\dfrac{R}{L}{e^{ - \dfrac{{Rt}}{L}}} \\\ \Rightarrow\dfrac{{di}}{{dt}} = \dfrac{\xi }{L}{e^{ - \dfrac{{Rt}}{L}}} \\\$$

Now, we will substitute the given values,
$\therefore\dfrac{{di}}{{dt}} = \dfrac{5}{5}{e^{ - \dfrac{{\left( {20} \right)\left( {0.25} \right)}}{5}}} = {e^{ - 1}}$
Therefore, the rate of increase of current at $t = 0.25s$ will be ${e^{ - 1}}$

Hence, option C is correct.

Note: You should know how to apply Kirchhoff’s loop law in various situations. Keep in mind the way we applied Kirchhoff’s loop law to our circuit. Also keep in mind that the emf induced due to self-inductance of an inductor is given as $\left( { - L\dfrac{{di}}{{dt}}} \right)$. The minus sign indicates that it opposes the flow of current.