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

Q: Why doesn't $R_1$ affect the current growth in the inductor when the switch is closed?

Because $R_1$ and the $L-R_2$ branch are in parallel across the ideal battery. The voltage across the $L-R_2$ branch is fixed at $E = 12\text{ V}$, completely independent of what happens in the $R_1$ branch.

Q: Why does the current through $R_1$ suddenly change direction when the switch is opened?

While the switch was closed, current flowed downwards through $R_1$ directly from the battery. When opened, the battery is gone, and the inductor acts as the new source. To maintain its downward current, the inductor pushes current leftwards along the bottom wire and upwards through $R_1$.

Q: Why is the steady-state current $6\text{ A}$? Doesn't the inductor have resistance?

An ideal inductor has zero resistance. In steady state (after a long time), the inductor acts like a plain wire (a short circuit). The only resistance limiting the current in that branch is $R_2 = 2\,\Omega$, so $i_0 = 12\text{ V} / 2\,\Omega = 6\text{ A}$.

Q: How do we know the new time constant is $0.1\text{ s}$?

The time constant for an L-R circuit is always $L / R_{eq}$. During discharge, the current flows through both $R_1$ and $R_2$ in series. So, $R_{eq} = 2\,\Omega + 2\,\Omega = 4\,\Omega$. Thus, $\tau' = 0.4\text{ H} / 4\,\Omega = 0.1\text{ s}$.

Q: What happens to the energy stored in the inductor?

The energy stored in the inductor's magnetic field ($\frac{1}{2}Li^2$) is gradually dissipated as heat in the resistors $R_1$ and $R_2$ during the discharging phase.

Question

An inductor of inductance $L = 400	ext{ mH}$ and resistors of resistances $R_1 = 2\,\Omega$ and $R_2 = 2\,\Omega$ are connected to a battery of emf $E = 12	ext{ V}$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at time $t = 0$.
What is the potential drop across $L$ as a function of time? After the steady state is reached, the switch is opened. What is the direction and the magnitude of current through $R_1$ as a function of time?

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a beautifully structured electrical circuit. On the left, we have a $12	ext{ V}$ ideal battery connected in series with a switch $S$. This main line then splits into two parallel branches. The middle branch contains a simple resistor $R_1 = 2\,\Omega$. The rightmost branch is slightly more complex; it houses an inductor $L = 400	ext{ mH}$ (or $0.4	ext{ H}$) in series with another resistor $R_2 = 2\,\Omega$.

When the switch $S$ is closed at $t=0$, the battery is directly connected across both parallel branches. Because they are in parallel, the voltage across the $L-R_2$ branch is exactly equal to the battery's electromotive force, $E = 12	ext{ V}$. The presence of $R_1$ in the neighboring branch does not affect the voltage across the inductor's branch at all.

### The Charging Phase: Switch Closed

Let's focus entirely on the right branch. This is a classic $L-R$ circuit connected to a constant DC voltage source. We know that an inductor opposes any sudden change in current. Therefore, the current doesn't instantly jump to its maximum value; instead, it grows exponentially over time.

The equation governing this growth is:
$$i(t) = i_0 (1 - e^{-t/	au_L})$$

Here, $i_0$ is the steady-state current, and $	au_L$ is the time constant of the branch. Let's calculate these parameters. The time constant is the ratio of inductance to resistance in that specific branch:
$$	au_L = \frac{L}{R_2} = \frac{0.4	ext{ H}}{2\,\Omega} = 0.2	ext{ s}$$

The steady-state current $i_0$ is reached after a long time when the inductor acts like a plain wire (zero resistance). It is simply determined by Ohm's law:
$$i_0 = \frac{E}{R_2} = \frac{12	ext{ V}}{2\,\Omega} = 6	ext{ A}$$

Plugging these values back into our growth equation, we get the current as a function of time:
$$i(t) = 6(1 - e^{-t/0.2}) = 6(1 - e^{-5t})	ext{ A}$$

### The Potential Drop Across the Inductor

Now, we need to find the potential drop across the inductor, $V_L$. According to Faraday's law of induction, the voltage across an inductor is proportional to the rate of change of current:
$$V_L = L \frac{di}{dt}$$

Let's differentiate our current expression with respect to time:
$$\frac{di}{dt} = \frac{d}{dt} [6(1 - e^{-5t})] = 6(0 - (-5)e^{-5t}) = 30e^{-5t}	ext{ A/s}$$

Multiplying this rate by the inductance $L = 0.4	ext{ H}$, we obtain the potential drop:
$$V_L = 0.4 	imes 30e^{-5t} = 12e^{-5t}	ext{ V}$$

This is our first answer. Notice how at $t=0$, the voltage across the inductor is exactly $12	ext{ V}$, meaning it initially blocks all current, taking the full voltage of the battery.

### The Discharging Phase: Switch Opened

After a long time, the circuit reaches a steady state. The current through the inductor is a constant $6	ext{ A}$, and the magnetic field inside it is fully established. Suddenly, we open the switch $S$!

Opening the switch completely disconnects the battery from the rest of the circuit. However, the inductor refuses to let the $6	ext{ A}$ current die instantly. It acts as a temporary battery, using its stored magnetic energy to push current through the only available closed path.

Look at the circuit diagram again. With the left branch broken, the middle branch ($R_1$) and the right branch ($L$ and $R_2$) form a new, isolated closed loop. The inductor was pushing current downwards through $R_2$. To maintain this flow, the current must travel leftwards along the bottom wire and then **upwards** through $R_1$. This gives us the direction of the current through $R_1$.

### The Final Calculation

This new loop is a discharging $L-R$ circuit. The total resistance opposing the current is now the series combination of both resistors:
$$R_{eq} = R_1 + R_2 = 2\,\Omega + 2\,\Omega = 4\,\Omega$$

The new time constant for this discharging phase, let's call it $	au_L'$, is:
$$	au_L' = \frac{L}{R_{eq}} = \frac{0.4	ext{ H}}{4\,\Omega} = 0.1	ext{ s}$$

In a discharging $L-R$ circuit, the current decays exponentially from its initial value $i_0$:
$$i'(t) = i_0 e^{-t/	au_L'}$$

Substituting our known values, we find the magnitude of the current through $R_1$:
$$i'(t) = 6 e^{-t/0.1} = 6 e^{-10t}	ext{ A}$$

And there we have it! The current through $R_1$ is $6 e^{-10t}	ext{ A}$, flowing in the upward direction. A beautiful demonstration of how inductors govern the transient behaviors of electrical circuits.

Visualized Solution

\text{Circuit Analysis}

\text{Switch } S \text{ is closed}

\text{Current Growth Equation}

\text{Calculating Parameters}

\text{Current in Inductor}

\text{Rate of Change of Current}

\text{Potential Drop Across } L

\text{Switch } S \text{ is opened}

\text{Discharging Loop}

\text{New Time Constant}

\text{Discharging Current}

\text{Final Answer}

The Sigma Insight: Self and Mutual Inductance

Analyzing the Setup

The Charging Phase

The Potential Drop Across the Inductor

The Discharging Phase

The Final Calculation

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