LR Circuit
By An LR circuit consists of an inductor (L) and a resistor (R) connected in series to an AC voltage source. The behavior of an LR circuit is characterized by an exponential rise and decay of the current through the circuit, as the energy stored in the magnetic field of the inductor is transferred to the electrical energy stored in the resistor.
The current $i$ through the circuit varies with time as \begin{align} \frac{\mathrm{d}i}{\mathrm{d}t}=-\left(\frac{R}{L}\right)i. \end{align}
The inductive impedance of the inductor is $X_L=\omega L$. The impedance of the LR circuit is given by \begin{align} Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+\omega^2L^2}. \end{align}
The time constant of the circuit is given by \begin{align} \tau= \frac{L}{R}. \end{align}
In LR circuit, the current $i$ lags the voltage $V$ by a phase difference $\phi$ given by, \begin{align} \phi&=\tan^{-1}\left(\frac{X_L}{R}\right)=\tan^{-1}\left(\frac{\omega L}{R}\right) \end{align}
The growth of current in LR circuit is given by \begin{align} i=\frac{V}{R}\left[1-e^{-\frac{t}{L/R}}\right] \end{align} The current reaches the maximum value $i_\text{max}=V/R$ as $t\to\infty$. It reaches a value $i=0.63i_\text{max}$ at time $t=\tau=L/R$.
The decay of current in LR circuit is given by \begin{align} i=\frac{V}{R}e^{-\frac{t}{L/R}} \end{align} The current becomes zero as $t\to\infty$. It reaches a value $i=0.37i_\text{max}$ at time $t=\tau=L/R$.
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} &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{align} V_L &=L\frac{\mathrm{d}i}{\mathrm{d}t} \\ &=Ee^{-tR_2/L} \\ &=12e^{-t\times 2/0.4} \\ &=12e^{-5t}. \nonumber \end{align} 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} &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} & 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{align} i&=i_0 e^{-t(R_1+R_2)/L} \\ &=6e^{-t (2+2)/0.4}\\ &=6e^{-10t}. \nonumber \end{align}
