RC, LR, and LC Circuits with DC and AC Sources
Problems from IIT JEE
Problem (IIT JEE 2001):
An inductor of inductance $L={400}\;\mathrm{mH}$ and resistors of resistance $R_1={2}\;\mathrm{\Omega}$ and $R_2={2}\;{\Omega}$ are connected to a battery of emf $E={12}\;\mathrm{V}$ as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at time $t=0$. What is the potential drop across $L$ as a function of time? After the steady state is reached, the switch is opened. What is the direction and the magnitude of current through $R_1$ as a function of time?
Solution:
Let $i$ be the current through the inductor $L$ and the resistor $R_2$ (see figure). The potential drop across $L$ is $L\frac{\mathrm{d}i}{\mathrm{d}t}$ and potential drop across $R_2$ is $iR_2$. Kirchhoff's loop law gives,
\begin{alignat}{2}
\label{ojb:eqn:1}
&L\frac{\mathrm{d}i}{\mathrm{d}t}+iR_2-E=0.
\end{alignat}
Integrate above equation and use initial condition $i=0$ at $t=0$ to get,
\begin{alignat}{2}
&i=\frac{E}{R_2}\left[1-e^{-tR_2/L}\right]. \nonumber
\end{alignat}
The potential drop across $L$ is given by,
\begin{alignat}{2}
&V_L=L\frac{\mathrm{d}i}{\mathrm{d}t}=Ee^{-tR_2/L}=12e^{-t\times 2/0.4}=12e^{-5t}. \nonumber
\end{alignat}
In the steady state ($t\to\infty$), the current through $L$ is $i_0={E}/{R_2}={6}\;\mathrm{A}$, $V_L=0$, and $V_{R_2}={12}\;\mathrm{V}$.
Let $i$ be the decaying current in the circuit when switch is opened (see figure). The potential across $L$ is $L\frac{\mathrm{d}i}{\mathrm{d}t}$, potential across $R_1$ is $iR_1$, and that across $R_2$ is $iR_2$. Kirchhoff's loop law gives,
\begin{alignat}{2}
\label{ojb:eqn:4}
&L\frac{\mathrm{d}i}{\mathrm{d}t}+iR_1+iR_2=0.
\end{alignat}
Integrate above equation with initial condition $i=i_0={6}\;\mathrm{A}$ at $t=0$ to get,
\begin{alignat}{2}
\label{ojb:eqn:5}
& i=i_0e^{-t(R_1+R_2)/L}.
\end{alignat}
Substitute $R_1={2}\;\mathrm{\Omega}$, $R_2={2}\;\mathrm{\Omega}$ and $L={0.4}\;\mathrm{H}$ in above equation to get current through $R_1$ as a function of time $t$,
\begin{alignat}{2}
&i=i_0 e^{-t(R_1+R_2)/L}=6e^{-t (2+2)/0.4}=6e^{-10t}. \nonumber
\end{alignat}