Animated Solution for Physics - Electromagnetic Induction: As shown in the figure, a battery of emf $\epsilon$ is connected to an inductor $L$ and resistance $R$ in series. The switch is closed at $t=0$. The total charge that flows from the battery, between $t=0$ and $t=t_c$ ($t_c$ is the time constant of the circuit) is

We have a battery providing an electromotive force (EMF) of $\epsilon$, an inductor with inductance $L$, and a resistor with resistance $R$, all connected in series. At the exact moment $t=0$, we close the switch $S$.

If this were a simple circuit with just a resistor, the current would instantly jump to its maximum value. But the inductor acts like the "inertia" of the circuit. It strongly opposes any sudden change in current. Because of this, the current doesn't just jump; it grows smoothly and exponentially over time.

The equation that governs this beautiful exponential growth of current $i(t)$ in an $L-R$ circuit is given by:

Here, $I_0$ is the maximum steady-state current that will eventually flow through the circuit after a long time. By Ohm's law, this maximum current is simply $I_0=\frac{\epsilon }{R}$.

The term $\frac{L}{R}$ is known as the time constant of the circuit, denoted by $t_c$. It gives us a measure of how fast the current grows. So, we can also write the exponent as $-\frac{t}{t_c}$.

We know from the fundamental definition of current that it is the rate of flow of charge, $i=\frac{dq}{dt}$. Therefore, to find the total charge, we must integrate the current with respect to time over the given interval:

Let's substitute our master equation into this integral:

Now, we perform the integration. The integral of $1$ is simply $t$, and the integral of the exponential function requires us to divide by the coefficient of $t$:

Simplifying the negative signs and bringing the fraction up, we get:

We first plug in the upper limit $t_c$ (which is equal to $\frac{L}{R}$), and then subtract the expression evaluated at the lower limit $0$:

Notice what happens in the exponent: $-\frac{R}{L}\times \frac{L}{R}=-1$. And we know that $e^0=1$. Substituting these in, the expression simplifies beautifully:

The $\frac{L}{R}$ terms cancel each other out perfectly! We are left with:

Finally, we substitute the value of the maximum current $I_0=\frac{\epsilon }{R}$ back into our equation:

Multiplying the terms together, we arrive at our final answer:

This elegant result shows exactly how much charge has been pushed through the circuit by the battery during the first time constant.