In an L-C-R circuit, capacitance is changed from C to 2C. For the resonant frequency to remains unch

Q: Why does resonance occur exactly when $X_L = X_C$?

At this specific frequency, the voltage drop across the inductor and the voltage drop across the capacitor are equal in magnitude but exactly $180^\circ$ out of phase. They perfectly cancel each other out, leaving the circuit to behave as if it were purely resistive.

Q: Does changing $L$ and $C$ affect the power dissipated at resonance?

No. At resonance, the total impedance is purely resistive ($Z=R$). As long as the resistance $R$ and the source voltage remain the same, the current and the power dissipated remain completely unchanged.

Q: What happens to the quality factor when $C$ is doubled and $L$ is halved?

The quality factor is given by $Q = \frac{1}{R}\sqrt{\frac{L}{C}}$. If $L$ becomes $L/2$ and $C$ becomes $2C$, the term inside the square root becomes $\frac{L/2}{2C} = \frac{1}{4}\frac{L}{C}$. Taking the square root, the new $Q$ becomes exactly half of the original $Q$. The resonance peak becomes wider and less sharp.

Q: Can we change the resonant frequency by changing the resistor?

No, the resonant frequency $\omega_0 = \frac{1}{\sqrt{LC}}$ is completely independent of the resistance $R$. The resistor only affects the sharpness (Quality Factor) of the resonance, not the frequency at which it occurs.

Q: Why do we use the product $LC = \text{constant}$ instead of just plugging into the formula twice?

It's a mathematical shortcut that prevents silly mistakes. Since $\omega_0$ is constant, squaring both sides and rearranging immediately shows that $LC$ must be constant, saving calculation time and reducing the chance of algebraic errors.

Question

In an $L-C-R$ circuit, capacitance is changed from $C$ to $2C$. For the resonant frequency to remains unchanged, the inductance should be changed from $L$ to

Options:

(A) $4L$

(B) $2L$

(C) $L/2$

(D) $L/4$

Sigma Percentile · Accepted Answer

### The Magic of Resonance

Imagine you are pushing a child on a swing. If you push at random intervals, the swing won't go very high. But if you time your pushes perfectly to match the swing's natural rhythm, the swing goes higher and higher with very little effort. This phenomenon is called **resonance**.

In the world of electronics, an $L-C-R$ circuit behaves exactly like that swing. The inductor ($L$) acts like the mass or inertia of the swing, storing energy in a magnetic field. The capacitor ($C$) acts like the spring or gravity, storing energy in an electric field. The resistor ($R$) is the friction, slowly draining energy away as heat.

### The Master Equation

When an alternating voltage is applied to this circuit, the energy sloshes back and forth between the inductor and the capacitor. Resonance occurs when the opposition to current flow from the inductor (inductive reactance, $X_L$) perfectly cancels out the opposition from the capacitor (capacitive reactance, $X_C$).

Mathematically, this happens when:
$$X_L = X_C$$
$$\omega L = \frac{1}{\omega C}$$

Solving for the angular frequency $\omega$, we get the famous resonant frequency formula:
$$\omega_0 = \frac{1}{\sqrt{LC}}$$

### Analyzing the Setup

The problem presents us with a fascinating constraint: we are going to change the capacitor, but we demand that the resonant frequency $\omega_0$ remains absolutely unchanged.

If $\omega_0$ is constant, then the term $\frac{1}{\sqrt{LC}}$ must be constant. Squaring both sides tells us that the product of the inductance and capacitance must remain constant:
$$LC = 	ext{constant}$$

This is a powerful realization. It means that $L$ and $C$ are inversely proportional if we want to maintain the same resonant frequency. If one goes up, the other must come down by the exact same factor.

### Final Calculation

Let's set up our before-and-after equation. Initially, we have an inductance $L_1 = L$ and a capacitance $C_1 = C$.

We are told the new capacitance is doubled, so $C_2 = 2C$. We need to find the new inductance, $L_2$.

$$L_1 C_1 = L_2 C_2$$
$$L \cdot C = L_2 \cdot (2C)$$

Now, we simply isolate $L_2$:
$$L_2 = \frac{L \cdot C}{2C}$$
$$L_2 = \frac{L}{2}$$

And there we have it! To keep the circuit perfectly tuned to the same frequency after doubling the capacitance, we must cut the inductance in half. It is a beautiful, symmetric dance of physical parameters.

Animated Solution for Physics - Alternating Current: In an $L - C - R$ circuit, capacitance is changed from $C$ to $2 C$ . For the resonant frequency to remains unchanged, the inductance should be changed from $L$ to

Select Answer:

Visualized Solution

Visualizing the $L - C - R$ Circuit

The Condition for Resonance

Establishing the Constant Relationship

Substituting the Given Values

Calculating the New Inductance

Analyzing the Aftermath

The Sigma Insight: AC Circuits and Power in AC Circuits

The Magic of Resonance

The Master Equation

Analyzing the Setup

Final Calculation

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