An inductor of inductance L = 400 mH and resistors of resistances R1 = 4 ohm and R2 = 2 ohm are conn

Q: Why doesn't $R_1$ affect the current in the inductor?

Because $R_1$ is in a parallel branch directly connected to the ideal battery. The voltage across the $L-R_2$ branch is fixed at 12V, completely independent of $R_1$.

Q: What happens to the inductor exactly at $t = 0$?

At $t = 0$, the inductor strongly opposes any change in current, so it acts like an open circuit. The entire 12V drops across the inductor initially, and the current is zero.

Q: What happens to the inductor after a long time ($t \to \infty$)?

After a long time, the current reaches a steady state and stops changing. Since $V_L = L \frac{dI}{dt}$ and $\frac{dI}{dt} = 0$, the inductor acts like a plain wire (short circuit), and the voltage drop across it becomes zero.

Q: Can I use $V_L = L \frac{dI}{dt}$ instead of KVL?

Absolutely! Differentiating the current expression $I(t)$ and multiplying by $L$ will give you the exact same result for $V_L$. It's a great way to verify your answer.

Q: Why is the time constant $\tau = L/R_2$ and not $L/(R_1 + R_2)$?

The time constant depends only on the equivalent resistance in series with the inductor within its specific loop. Since $R_1$ is in a parallel branch, it does not affect the time constant of the $L-R_2$ branch.

Question

An inductor of inductance $L = 400 	ext{ mH}$ and resistors of resistances $R_1 = 4 \Omega$ and $R_2 = 2 \Omega$ are connected to battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t = 0$. The potential drop across $L$ as a function of time is

Options:

(A) $6 e^{-5t} 	ext{ V}$

(B) $\frac{12}{t} e^{-3t} 	ext{ V}$

(C) $6(1 - e^{-t/0.2}) 	ext{ V}$

(D) $12 e^{-5t} 	ext{ V}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a water pipe system where the flow is suddenly turned on. In our circuit, the $12	ext{ V}$ battery acts as the pump, and the switch $S$ is the valve.

When the switch is closed at $t = 0$, the current splits into two parallel branches. The middle branch contains only the resistor $R_1$, which immediately draws a steady current.

However, the rightmost branch contains an inductor $L$ in series with a resistor $R_2$. The inductor acts like a heavy water wheel; it opposes any sudden change in current, causing the current in this branch to grow gradually.

### The Master Equation

Because the $L-R_2$ branch is connected directly in parallel with the ideal $12	ext{ V}$ battery, it experiences the full $12	ext{ V}$ independently of the $R_1$ branch.

The current in an $L-R$ circuit grows exponentially according to the master equation:

$$I_2(t) = I_0 \left(1 - e^{-t/	au}ight)$$

Here, $I_0$ is the maximum steady-state current, and $	au$ is the time constant that dictates how quickly the current reaches that maximum.

### Finding the Time Constant and Steady State

Let's determine the steady-state current $I_0$. After a long time ($t 	o \infty$), the inductor behaves like a perfectly conducting wire, offering zero resistance.

The current is then limited only by the resistor $R_2$:

$$I_0 = \frac{E}{R_2} = \frac{12}{2} = 6 	ext{ A}$$

Next, we calculate the time constant $	au$. This is the ratio of the inductance to the resistance in that specific branch:

$$	au = \frac{L}{R_2} = \frac{400 	imes 10^{-3}}{2} = 0.2 	ext{ s}$$

Substituting these values back into our master equation, we get the instantaneous current:

$$I_2(t) = 6 \left(1 - e^{-t/0.2}ight) = 6 \left(1 - e^{-5t}ight) 	ext{ A}$$

### Final Calculation

Now, we need to find the potential drop across the inductor, $V_L$. We can apply Kirchhoff's Voltage Law (KVL) to the loop containing the battery, the inductor, and $R_2$.

The sum of the voltage drops must equal the battery's EMF:

$$E - V_L - I_2 R_2 = 0 \implies V_L = E - I_2 R_2$$

Let's substitute our expression for $I_2(t)$ into this KVL equation:

$$V_L = 12 - 6 \left(1 - e^{-5t}ight) 	imes 2$$

Simplifying the expression, we distribute the 2:

$$V_L = 12 - 12 \left(1 - e^{-5t}ight)$$

$$V_L = 12 - 12 + 12 e^{-5t} = 12 e^{-5t} 	ext{ V}$$

**Alternatively**, you could use the fundamental inductor equation $V_L = L \frac{dI_2}{dt}$. Differentiating the current expression yields the exact same elegant result. The potential drop decays exponentially as the current stabilizes!

Select Answer:

Visualized Solution

Circuit Analysis

Current in the $L - R$ Branch

Calculating Steady-State Current

Calculating Time Constant

Current Equation

Potential Drop Across Inductor

Final Calculation

The Sigma Insight: Self and Mutual Inductance

Analyzing the Setup

The Master Equation

Finding the Time Constant and Steady State

Final Calculation

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Select Answer:

Visualized Solution

Circuit Analysis

Current in the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L−R Branch

Calculating Steady-State Current

Calculating Time Constant

Current Equation

Potential Drop Across Inductor

Final Calculation

The Sigma Insight: Self and Mutual Inductance

Analyzing the Setup

The Master Equation

Finding the Time Constant and Steady State

Final Calculation

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