Animated Solution for Physics - Electromagnetic Induction: Consider the $L-R$ circuit shown in the figure. If the switch $S$ is closed at $t=0$, then the amount of charge that passes through the battery between $t=0$ and $t=\frac{L}{R}$ is

A Classic L-R Circuit

Imagine you are looking at a classic $L-R$ circuit. We have an inductor $L$ and a resistor $R$ connected in series with a battery of EMF $E$ and a switch $S$.

When the switch is closed at $t=0$, the battery tries to push current through the circuit. However, the inductor plays the role of electrical inertia. It opposes this sudden change in current by inducing a back EMF. Because of this, the current doesn't jump to its maximum value instantly; instead, it grows gradually over time.

Current Growth

The growth of current in an $L-R$ circuit is governed by a beautiful exponential equation. At any given time $t$, the current $i$ is given by:

Here, $I_0$ represents the maximum, or steady-state, current that will eventually flow through the circuit once the inductor stops opposing the flow. From Ohm's law, we know this maximum current is simply the battery's EMF divided by the resistance:

The question asks for the total amount of charge $q$ that passes through the battery between $t=0$ and $t=\frac{L}{R}$.

We know from the fundamental definition of current that it is the rate of flow of charge. Mathematically, $i=\frac{dq}{dt}$. Rearranging this gives us the infinitesimally small charge $dq$ that flows in a tiny time interval $dt$:

To find the total charge $q$ over the given time interval, we must integrate this expression from our starting time $t=0$ to our ending time $t=\frac{L}{R}$:

Executing the Integration

Now, let's roll up our sleeves and perform the integration. We can integrate the two terms inside the parenthesis separately.

The integral of $1$ with respect to $t$ is simply $t$. For the exponential term, the integral of $e^{-\frac{R}{L}t}$ is the same exponential function divided by the coefficient of $t$, which is $-\frac{R}{L}$.

Putting it all together, we get:

Simplifying the negative signs and the fraction, the expression becomes much cleaner:

Substituting Limits

It's time to substitute our upper and lower limits into the integrated expression.

First, we plug in the upper limit $t=\frac{L}{R}$. The $t$ becomes $\frac{L}{R}$, and the exponent becomes $-\frac{R}{L}\times \frac{L}{R}=-1$.

Next, we subtract the expression evaluated at the lower limit $t=0$. The $t$ becomes $0$, and the exponent becomes $0$, making $e^0=1$.

Notice what happens here! We have a $\frac{L}{R}$ from the upper limit and a $-\frac{L}{R}$ from the lower limit. These terms perfectly cancel each other out.

Finally, we substitute the value of our steady-state current $I_0=\frac{E}{R}$ back into the equation:

Since the mathematical constant $e$ (Euler's number) is approximately $2.72$, we can write our final answer as:

This perfectly matches option (b). A beautiful journey from circuit theory through calculus to a clean, elegant result!