Figure shows a circuit that contains four identical resistors with resistance R = 2.0 ohm, two ident

Q: Why do inductors act as open circuits at $t=0$?

An inductor opposes any change in current due to its self-induced EMF ($V = -L \frac{di}{dt}$). Before the switch is closed, the current is zero. Immediately after closing, the inductor fights the sudden increase, keeping the current at zero for that exact instant, effectively acting as an open circuit.

Q: What happens to the resistors on the right side of the circuit?

Because the horizontal inductor acts as an open circuit, it breaks the path leading to the right side. Without a complete closed loop, no current can flow through those resistors, so they are effectively removed from the circuit at $t=0$.

Q: How do we know the two remaining resistors are in series and not parallel?

After removing the open branches, there is only one single path for the current to flow from the positive terminal of the battery, through the first resistor, then through the second resistor, and back to the negative terminal. Since the same current flows through both without splitting, they are in series.

Q: What would the circuit look like after a long time ($t \to \infty$)?

After a long time, the current becomes steady (constant). Since $\frac{di}{dt} = 0$, the inductor produces no opposing EMF and acts like a plain wire (a short circuit). The circuit would then be a complex parallel-series network.

Q: Does the ammeter have any resistance?

The problem states it's an "ideal battery", and typically, unless specified otherwise, ammeters in such textbook problems are considered ideal, meaning they have zero resistance and don't affect the current calculation.

Question

Figure shows a circuit that contains four identical resistors with resistance $R = 2.0 \Omega$, two identical inductors with inductance $L = 2.0 	ext{ mH}$ and an ideal battery with electromotive force $E = 9 	ext{ V}$. The current $i$ just after the switch $S$ is closed will be

Options:

(A) 2.25 A

(B) 3.0 A

(C) 3.37 A

(D) 9 A

Sigma Percentile · Accepted Answer

### The Illusion of Complexity

Imagine you are faced with a sprawling network of resistors and inductors. At first glance, this ladder-like circuit looks like a nightmare of Kirchhoff's laws.

But in physics, timing is everything. The question specifically asks for the current **just after** the switch is closed, at $t = 0$.

This tiny detail is our golden key. It transforms a complex differential equation problem into a beautifully simple conceptual puzzle.

### The Secret Weapon: Inductor's Inertia

To solve this, we must understand the soul of an inductor. An inductor is like a heavy mechanical flywheel; it strongly opposes any sudden change in its state of motion.

Mathematically, the induced EMF is given by $V = -L \frac{di}{dt}$. If the current were to jump instantly from zero to some value, the rate of change $\frac{di}{dt}$ would be infinite, creating an infinite opposing voltage.

Because nature abhors infinities, the current through an inductor cannot change instantaneously. Therefore, at the exact moment the switch is closed ($t=0$), the current through every inductor remains exactly zero. They behave as **perfect open circuits**.

### Pruning the Tree

Now, let's apply this powerful concept to our circuit. We can literally erase the inductors from our diagram.

The first inductor is in a vertical branch. Replacing it with an open circuit simply removes that branch, forcing any current to bypass it and continue horizontally.

The second inductor is in a horizontal branch. When this becomes an open circuit, it acts like a severed bridge. It completely disconnects the entire right half of the circuit!

### The Simplified Path

With the open branches removed, the intimidating network collapses into a single, elegant loop.

The current leaves the $9	ext{ V}$ battery, travels through the switch and the ammeter, and then has only one path to follow. It flows through the first horizontal resistor, turns the corner, and flows down through the second vertical resistor before returning to the battery.

Since the current has no junctions to split at, these two resistors are perfectly in series.

### Final Calculation

Calculating the equivalent resistance is now trivial. We simply add the resistances of the two active components:

$$R_{	ext{eq}} = R + R = 2R$$

Given that each resistor is $2.0 \Omega$, our total resistance is:

$$R_{	ext{eq}} = 2 	imes 2.0 = 4.0 \Omega$$

Finally, we apply Ohm's law to find the initial current $i$:

$$i = \frac{E}{R_{	ext{eq}}} = \frac{9}{4.0} = 2.25 	ext{ A}$$

The ammeter will read **$2.25 	ext{ A}$**, which perfectly matches option (a).

### Food for Thought

We conquered this problem by analyzing the $t=0$ state. But what if the switch was left closed for a very long time?

In the steady state ($t 	o \infty$), the current stops changing, meaning $\frac{di}{dt} = 0$. The inductors would stop opposing the flow and act as perfect short circuits (plain wires).

This would completely rewire the network, shorting out some resistors and putting others in parallel. I highly encourage you to draw that steady-state circuit and calculate the new current. It's a fantastic way to master LR circuits!

Select Answer:

Visualized Solution

$t = 0$ Circuit Analysis

Inductor Behavior at $t = 0$

Simplifying the Circuit

Tracing the Active Path

Equivalent Resistance

Calculating Initial Current

Final Answer

Steady State Analysis (Food for Thought)

The Sigma Insight: Alternating Current (AC) and Voltage

The Illusion of Complexity

The Secret Weapon

Pruning the Tree

The Simplified Path

Final Calculation

Food for Thought

Similar Questions

List-I

List-II

An ideal coil of $10 H$ is connected in series with a resistance of $5 Ω$ and a battery of $5 V$ . After $2 s$ , the connection is made, the current flowing (in ampere) in the circuit is

In the given circuit, the AC source has $ω = 100 rad/s$ . Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

Find the peak current and resonant frequency of the following circuit (as shown in figure).

In an AC-circuit, an inductor, a capacitor and a resistor are connected in series with $X_{L} = R = X_{C}$ . Impedance of this circuit is

In $L - C$ circuit, the inductance $L = 40$ mH and capacitance $C = 100 μ F$ . If a voltage $V (t) = 10 sin (314 t)$ is applied to the circuit, the current in the circuit is given as

In the circuit shown, $L = 1 μ H$ , $C = 1 μ F$ and $R = 1 k Ω$ . They are connected in series with an AC source $V = V_{0} sin ω t$ as shown. Which of the following options is/are correct?

In a series $L - C - R$ circuit, the inductive reactance ( $X_{L}$ ) is $10 Ω$ and the capacitive reactance ( $X_{C}$ ) is $4 Ω$ . The resistance ( $R$ ) in the circuit is $6 Ω$ . The power factor of the circuit is