The circuit shown in the figure contains an inductor L, a capacitor C0, a resistor R0 and an ideal b

Q: Why is $I_1$ zero right after closing the switch?

An inductor fundamentally opposes any sudden change in current due to self-induction. Since the current was zero before closing the switch, it must remain zero at the exact instant $t=0^+$.

Q: Why does the inductor act as a short circuit in the steady state?

In a DC circuit, once the steady state is reached, the current becomes constant. Since the voltage across an inductor is $V = L(di/dt)$, a constant current means $di/dt = 0$, resulting in zero voltage drop, just like an ideal wire.

Q: What happens to the resistor in the final phase?

When switch $K_1$ is opened, the rightmost branch is broken. This completely disconnects the battery and the resistor from the top LC loop, leaving the inductor and capacitor isolated.

Q: Why do we equate maximum magnetic and electrical energies?

In an ideal LC circuit with no resistance, total energy is conserved. The energy continuously shifts back and forth entirely between the inductor's magnetic field and the capacitor's electric field.

Q: Does the battery affect the LC oscillation frequency?

No. Once $K_1$ is opened, the battery is no longer part of the active circuit. The oscillation frequency depends purely on the values of $L$ and $C_0$.

Question

The circuit shown in the figure contains an inductor L, a capacitor $C_0$, a resistor $R_0$ and an ideal battery. The circuit also contains two keys $K_1$ and $K_2$. Initially, both the keys are open and there is no charge on the capacitor. At an instant, key $K_1$ is closed and immediately after this the current in $R_0$ is found to be $I_1$. After a long time, the current attains a steady state value $I_2$. Thereafter, $K_2$ is closed and simultaneously $K_1$ is opened and the voltage across $C_0$ oscillates with amplitude $V_0$ and angular frequency $\omega_0$

List-I:

P. The value of $I_1$ in Ampere is

Q. The value of $I_2$ in Ampere is

R. The value of $\omega_0$ in kilo-radians/s

S. The value of $V_0$ in Volt is

List-II:

1. 0

2. 2

3. 4

4. 20

5. 200

Sigma Percentile · Accepted Answer

The beauty of physics often lies in how systems evolve over time. This problem is a perfect example of a transient circuit—a system that dances through different phases, each governed by its own set of rules. We are given a circuit with a resistor $R_0$, an inductor $L$, a capacitor $C_0$, and two switches $K_1$ and $K_2$.

Our mission is to track the behavior of this circuit across three distinct time phases. Let's dive into the journey of the current and voltage!

### Phase 1: The Inductor's Stubbornness

The story begins at $t = 0$. We close switch $K_1$ while keeping $K_2$ open. This creates a simple series circuit with the battery, the resistor $R_0$, and the inductor $L$.

But here is the catch: an inductor is fundamentally stubborn. It strongly opposes any sudden change in the current flowing through it. Before the switch was closed, the current was zero. Therefore, exactly at the instant $t = 0^+$, the inductor forces the current to remain zero.

$$I_1 = 0 	ext{ A}$$

Because the inductor acts like an open circuit at this very first instant, no current flows through the resistor either. This gives us our first match: **$I_1 = 0$**.

### Phase 2: Reaching the Steady State

Now, imagine we wait for a long time. The circuit reaches what we call a **steady state**.

In a DC circuit, once the steady state is achieved, the current stops changing. Since the voltage across an inductor is proportional to the rate of change of current ($V_L = L \frac{di}{dt}$), a constant current means zero voltage drop. The inductor effectively becomes a perfect, zero-resistance wire—a short circuit!

With the inductor acting as a short circuit, the only resistance left in our active loop is $R_0 = 5\,\Omega$. We can easily find the steady-state current using Ohm's Law:

$$I_2 = \frac{V}{R_0} = \frac{20}{5} = 4 	ext{ A}$$

This steady current of $4 	ext{ A}$ is now flowing happily through the inductor, storing energy in its magnetic field. This is our second match: **$I_2 = 4 	ext{ A}$**.

### Phase 3: The LC Oscillator Awakens

Here is where the magic happens. We simultaneously close $K_2$ and open $K_1$.

Opening $K_1$ completely severs the connection to the battery and the resistor. What remains is an isolated loop containing only the charged inductor and the uncharged capacitor $C_0$. We have just birthed an **LC oscillator**!

The inductor, carrying an initial current of $4 	ext{ A}$, will now start charging the capacitor. The energy will slosh back and forth between the magnetic field of the inductor and the electric field of the capacitor. The angular frequency of this harmonic dance is given by the classic formula:

$$\omega_0 = \frac{1}{\sqrt{LC_0}}$$

Let's plug in our given values: $L = 25 	imes 10^{-3} 	ext{ H}$ and $C_0 = 10 	imes 10^{-6} 	ext{ F}$.

$$\omega_0 = \frac{1}{\sqrt{25 	imes 10^{-3} 	imes 10 	imes 10^{-6}}} = \frac{1}{\sqrt{25 	imes 10^{-8}}}$$

$$\omega_0 = \frac{1}{5 	imes 10^{-4}} = 2000 	ext{ rad/s}$$

Since the question asks for the answer in kilo-radians per second, we convert it to get **$\omega_0 = 2 	ext{ krad/s}$**. This is our third match!

### Phase 4: The Peak of the Voltage Wave

Finally, we need to find the maximum voltage amplitude $V_0$ across the capacitor during these oscillations.

In an ideal LC circuit, the total energy is perfectly conserved. The maximum magnetic energy stored in the inductor (when the current is at its peak) must equal the maximum electrical energy stored in the capacitor (when the voltage is at its peak and current is momentarily zero).

$$\frac{1}{2} L I_2^2 = \frac{1}{2} C_0 V_0^2$$

We can rearrange this beautifully symmetric equation to solve for $V_0$:

$$V_0 = I_2 \sqrt{\frac{L}{C_0}}$$

Now, we substitute our known values into the equation:

$$V_0 = 4 \sqrt{\frac{25 	imes 10^{-3}}{10 	imes 10^{-6}}} = 4 \sqrt{2500}$$

$$V_0 = 4 	imes 50 = 200 	ext{ V}$$

The voltage across the capacitor will oscillate with a massive amplitude of **$200 	ext{ V}$**. This completes our final match, perfectly aligning all the pieces of this elegant puzzle!

List-I

List-II

Select Matching Pairs:

Visualized Solution

\text{Circuit Analysis Setup}

\text{Phase 1: Immediately after closing } K_1

\text{Calculating } I_1

\text{Phase 2: Steady State}

\text{Calculating } I_2

\text{Phase 3: LC Oscillations}

\text{Angular Frequency } \omega_0

\text{Calculating } \omega_0

\text{Amplitude of Voltage } V_0

\text{Calculating } V_0

The Sigma Insight: Alternating Current (AC) and Voltage

Phase 1

Phase 2

Phase 3

Phase 4

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