Animated Solution for Physics - Electromagnetic Induction: A coil of self inductance $10\text{ mH}$ and resistance $0.1\Omega$ is connected through a switch to a battery of internal resistance $0.9\Omega$. After the switch is closed, the time taken for the current to attain $80\%$ of the saturation value is [Take, $\ln 5=1.6$]

Welcome! Let's visualize the circuit we are dealing with. We have a battery with an internal resistance of $0.9\Omega$, and it is connected to a real inductor.

It is crucial to remember that a real inductor isn't perfect; it is made of a long wire, which means it has its own small resistance. In this case, the coil's resistance is $0.1\Omega$. The moment we close the switch, the inductor opposes the sudden change, and the current will start to flow and grow gradually rather than instantly jumping to its maximum value.

To find how the current grows over time, we use the standard $L-R$ circuit growth equation. The current $i$ at any time $t$ is given by:

Here, $i_0$ is the maximum saturation current that the circuit will eventually reach after a long time, and $R_{\text{eq}}$ is the total equivalent resistance of the circuit.

First, let's find the total equivalent resistance. Since the internal resistance of the battery and the resistance of the coil are connected in series, we simply add them up:

The question asks for the time when the current reaches $80\%$ of its maximum value. So, we substitute $i$ as $0.8i_0$ into our growth equation:

The $i_0$ on both sides cancels out beautifully. We are left with:

Rearranging this, we bring the exponential term to the left and subtract $0.8$ from $1$, which gives us $0.2$:

Notice that $0.2$ is exactly $\frac{1}{5}$. If we take the reciprocal of both sides, the negative sign in the exponent disappears:

Taking the natural logarithm on both sides, the exponent comes down:

We are almost there! Let's substitute the values and get the answer. The inductance $L$ is $10\text{ mH}$, which is $10\times 10^{-3}\text{ H}$. The problem gives us $\ln 5$ as $1.6$.

Multiplying these out, $10\times 1.6$ is $16$. So, the time $t$ is:

This matches option (d) perfectly!

Think about this: the quantity $\frac{L}{R}$ is called the time constant of the circuit, often denoted by $\tau$. It tells us how sluggish the circuit is. A larger inductance or a smaller resistance means the current takes much longer to reach its maximum value. In our circuit, $\tau =10\text{ ms}$. Understanding this helps you intuitively grasp how fast any $L-R$ circuit will respond to changes.