An inductor (L = 0.03 H) and a resistor (R = 0.15 kΩ) are connected in series to a battery of 15V EM

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

An inductor opposes *changes* in current. After a long time in a DC circuit, the current becomes constant. Since the current isn't changing, the inductor produces no opposing EMF ($V = L \frac{di}{dt} = 0$), effectively acting like a plain wire.

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

The inductor stores energy in its magnetic field. When the battery is removed, this magnetic field collapses, inducing an EMF that keeps the current flowing in the same direction for a short time. It acts like a temporary battery!

Q: What is the significance of the time constant $L/R$?

The time constant $\tau = L/R$ tells us how fast the current decays. A larger inductance $L$ means more stored energy (slower decay), while a larger resistance $R$ means energy is dissipated faster (quicker decay).

Q: Why do we use $e^{-\frac{R}{L}t}$ instead of $1 - e^{-\frac{R}{L}t}$?

The formula $1 - e^{-\frac{R}{L}t}$ describes the *growth* of current when a battery is first connected. Since we are looking at the *decay* of current after the battery is removed, we use the decaying exponential $e^{-\frac{R}{L}t}$.

Q: I got a different answer. Did I mess up the units?

Most likely! A classic trap in this question is the units. The resistance is given in kilo-ohms ($0.15\text{ k}\Omega = 150\text{ }\Omega$) and time in milliseconds ($1\text{ ms} = 10^{-3}\text{ s}$). Always convert to standard SI units before calculating the exponent.

Question

An inductor ($L = 0.03	ext{ H}$) and a resistor ($R = 0.15	ext{ k}\Omega$) are connected in series to a battery of $15	ext{ V}$ EMF in a circuit shown below. The key $K_1$ has been kept closed for a long time. Then at $t = 0$, $K_1$ is opened and key $K_2$ is closed simultaneously. At $t = 1	ext{ ms}$, the current in the circuit will be ($e^5 \simeq 150$)

Options:

(A) $100	ext{ mA}$

(B) $67	ext{ mA}$

(C) $6.7	ext{ mA}$

(D) $0.67	ext{ mA}$

Sigma Percentile · Accepted Answer

### The Tale of the Decaying Current: A Journey Through an L-R Circuit

Imagine you are observing a bustling city's water supply system. The battery acts as the main pump, pushing water (current) through the pipes. The resistor is a narrow section of the pipe causing friction, and the inductor is a massive water wheel that takes time to spin up but, once spinning, wants to keep the water flowing. Let's dive into the physics of this fascinating circuit.

### The Setup: A Steady State

Initially, key $K_1$ is closed and $K_2$ is open. The battery is connected to the inductor and resistor for a long time. When a DC circuit with an inductor is left on for a long time, it reaches a **steady state**. The inductor's magnetic field is fully established, and it stops opposing the current. It behaves just like a simple connecting wire—a short circuit.

Because the inductor offers zero resistance in this steady state, the current reaches its maximum value, dictated purely by Ohm's Law:

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

Let's plug in our values. We have a $15	ext{ V}$ battery and a $0.15	ext{ k}\Omega$ resistor. A common trap here is forgetting to convert kilo-ohms to ohms!

$$R = 0.15	ext{ k}\Omega = 150	ext{ }\Omega$$

$$I_0 = \frac{15}{150} = 0.1	ext{ A} = 100	ext{ mA}$$

This $100	ext{ mA}$ is the current flowing through the inductor right before we flip the switches.

### The Switch: A Sudden Change

Now, at time $t = 0$, the magic happens. Key $K_1$ is opened, disconnecting the battery, and key $K_2$ is closed simultaneously. The inductor and resistor now form a closed loop on their own.

Without the battery, you might expect the current to instantly drop to zero. But remember our water wheel analogy? The inductor has energy stored in its magnetic field. As the field collapses, it induces an EMF that keeps the current flowing in the same direction. The energy starts to dissipate as heat through the resistor.

### The Math: Exponential Decay

The current decays exponentially according to the classic L-R decay formula:

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

We need to find the current at $t = 1	ext{ ms}$. Let's carefully substitute our values, ensuring all units are in standard SI format ($1	ext{ ms} = 10^{-3}	ext{ s}$):

$$I = 100 e^{-\frac{150}{0.03} 	imes 10^{-3}}$$

Let's simplify the exponent. Dividing $150$ by $0.03$ gives us $5000$. Multiplying $5000$ by $10^{-3}$ gives us exactly $5$. So, our exponent is simply $-5$.

$$I = 100 e^{-5}$$

### The Final Calculation

The problem kindly provides an approximation: $e^5 \simeq 150$. Therefore, $e^{-5} = \frac{1}{150}$.

$$I = 100 	imes \frac{1}{150}$$

$$I = \frac{2}{3}	ext{ mA} \simeq 0.67	ext{ mA}$$

And there we have it! The current in the circuit at $1	ext{ ms}$ is **$0.67	ext{ mA}$**. The beauty of exponential decay is perfectly captured in this elegant result.

Select Answer:

Visualized Solution

$t < 0$ Setup

Steady State Logic

Initial Current Setup

Calculate $I_{0}$

Switching at $t = 0$

Current Decay Logic

Decay Equation Setup

Exponent Calculation

Final Calculation

Conclusion

The Sigma Insight: Self and Mutual Inductance

The Tale of the Decaying Current

The Setup

The Switch

The Math

The Final Calculation

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