An inductor (L = 100 mH), a resistor (R = 100 ohm) and a battery (E = 100 V) are initially connected

Q: Why does the inductor act as a short circuit initially?

An inductor opposes *changes* in current. After a long time connected to a steady DC source, the current becomes constant. Since the current isn't changing, the inductor produces no back EMF and behaves just like a regular wire with zero resistance.

Q: What happens to the battery when points A and B are short-circuited?

When A and B are connected by a zero-resistance wire, all the current prefers to flow through that path rather than the battery branch. The battery is effectively bypassed and isolated from the main L-R loop.

Q: Why doesn't the current drop to zero immediately when the battery is disconnected?

This is the core property of an inductor! It has 'electrical inertia'. The magnetic field built up around the inductor collapses, inducing an EMF that keeps the current flowing in the same direction, gradually dissipating the stored energy as heat in the resistor.

Q: How do we know to use the decay formula instead of the growth formula?

The growth formula $I = I_0(1 - e^{-t/\tau})$ is used when a source is *connected* and current is building up from zero. Here, we start with a maximum current $I_0$ and remove the source, so the current must decrease, which is described by the decay formula $I = I_0 e^{-t/\tau}$.

Q: What is the significance of the time constant $\tau$?

The time constant $\tau = L/R$ tells us how fast the current decays. Specifically, it's the time it takes for the current to drop to about $37\%$ (which is $1/e$) of its initial value. A larger inductance means a slower decay, while a larger resistance means a faster decay.

Question

An inductor ($L =100 	ext{ mH}$), a resistor ($R =100\Omega$) and a battery ($E =100	ext{ V}$) are initially connected in series as shown in the figure. After a long time, the battery is disconnected after short circuiting the points $A$ and $B$. The current in the circuit $1 	ext{ millisecond}$ after the short circuit is

Options:

(A) $1/e 	ext{ A}$

(B) $e 	ext{ A}$

(C) $0.1 	ext{ A}$

(D) $1 	ext{ A}$

Sigma Percentile · Accepted Answer

The journey of this problem begins with understanding the dual personality of an inductor. It's a component that fiercely resists change, but when things are calm, it's as docile as a simple wire.

### Analyzing the Setup

Imagine the circuit before any short-circuiting happens. The battery has been connected for a "long time." In the world of physics, a "long time" for a DC circuit means it has reached a **steady state**.

During this steady state, the current is no longer changing. Because an inductor's opposition (its back EMF) depends entirely on the *rate of change* of current, a constant current means zero back EMF. The inductor effectively becomes a plain, zero-resistance wire.

With the inductor acting as a short circuit, the only component resisting the flow of current is the $100\Omega$ resistor. We can easily find this initial steady current, $I_0$, using Ohm's Law:

$$I_0 = \frac{E}{R}$$

Substituting the given values:

$$I_0 = \frac{100}{100} = 1 	ext{ A}$$

This $1 	ext{ A}$ is the current flowing through the inductor right at the moment we decide to short the circuit.

### The Master Equation

Now, the critical moment arrives. We short-circuit points $A$ and $B$. This creates a zero-resistance path that completely bypasses the battery. Current always takes the path of least resistance, so the battery is effectively removed from the active circuit. We are left with a closed loop containing only the inductor and the resistor.

Without the battery pushing it, you might expect the current to instantly drop to zero. But here is where the inductor's "electrical inertia" shines. The collapsing magnetic field around the inductor induces an EMF that keeps the current flowing in the same direction.

This decaying current follows a beautiful exponential curve:

$$I(t) = I_0 e^{-\frac{t}{	au}}$$

Here, $	au$ is the **time constant** of the L-R circuit, which dictates how quickly the current fades away. It is given by the ratio of inductance to resistance:

$$	au = \frac{L}{R}$$

Let's calculate this time constant. We have $L = 100 	ext{ mH} = 100 	imes 10^{-3} 	ext{ H}$ and $R = 100 \Omega$:

$$	au = \frac{100 	imes 10^{-3}}{100} = 10^{-3} 	ext{ s}$$

So, our time constant is exactly $1 	ext{ millisecond}$.

### Final Calculation

The problem asks for the current exactly $1 	ext{ millisecond}$ after the short circuit. This means we need to evaluate our decay equation at $t = 10^{-3} 	ext{ s}$.

Let's substitute our known values into the master equation:

$$I(1 	ext{ ms}) = 1 	imes e^{-\frac{10^{-3}}{10^{-3}}}$$

The exponent simplifies beautifully to $-1$:

$$I = e^{-1} = \frac{1}{e} 	ext{ A}$$

And there we have it! The current has decayed to exactly $1/e$ of its initial value. This problem is a fantastic demonstration of how inductors store energy and release it, smoothing out what would otherwise be abrupt changes in the universe.

Select Answer:

Visualized Solution

Initial Steady State

Steady State Current

Substituting Values

Initial Current

Short Circuiting

Current Decay Formula

Time Constant

Current at $t = 1 ms$

Final Calculation

Conclusion

The Sigma Insight: Self and Mutual Inductance

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Initial Steady State

Steady State Current

Substituting Values

Initial Current

Short Circuiting

Current Decay Formula

Time Constant

Current at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=1 ms

Final Calculation

Conclusion

The Sigma Insight: Self and Mutual Inductance

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

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