Q.A series LCR circuit with $R = 20\ \Omega$, $L = 1.5\ \text{H}$ and $C = 35\ \mu\text{F}$ is connected to a variable-frequency $200\ \text{V}$ ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Concept: Resonance in AC Circuits — at resonance, the impedance is purely resistive ($Z = R$), and the circuit behaves as if only the resistor is present.

Reasoning:

1. At resonance, the inductive and capacitive reactances cancel: $X_L = X_C$, so the impedance is minimum: $Z = R = 20\ \Omega$.
2. The RMS current is then $I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{200}{20} = 10\ \text{A}$. …

Why resonance simplifies everything

In an LCR series circuit, the impedance depends on frequency. The natural frequency — also called the resonant frequency — is where the inductive reactance $X_L = \omega L$ exactly cancels the capacitive reactance $X_C = 1/(\omega C)$. At this special frequency, the circuit offers minimum impedance, equal to just the resistance $R$.

The key consequence: voltage and current are in phase at resonance. That means the power factor is 1, and the average power over a cycle is simply the DC-like value $V_{\text{rms}}^2 / R$.

Step-by-step solution

1. Identify the condition.

The problem states: "when the frequency equals the natural frequency". That's the resonance condition. At resonance:

• $X_L = X_C$
• Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = R$
• Phase angle $\phi = 0$, so $\cos \phi = 1$

2. Recall the formula for average power in an AC circuit.

The average power transferred over one complete cycle is:

where $\cos \phi$ is the power factor.

3. Apply the resonance simplification.

Since $\cos \phi = 1$ at resonance:

But $I_{\text{rms}} = V_{\text{rms}} / Z = V_{\text{rms}} / R$ because $Z = R$. Substituting:

$P_{\text{av}} = V_{\text{rms}} \cdot \frac{V_{\text{rms}}}{R} = \frac{V_{\text{rms}}^2}{R}$ …

Method: Resonance Condition in Series LCR Circuit

This problem uses the Resonance Method — at resonance, the circuit behaves purely resistively, making power calculation straightforward.

Steps

Step 1: Identify the condition at resonance

At resonance, the inductive reactance equals the capacitive reactance:

The impedance becomes purely resistive:

Step 2: Recall the formula for average power

For an AC circuit, the average power over one complete cycle is:

At resonance, $\phi = 0$, so $\cos\phi = 1$. Hence:

Step 3: Find the rms current at resonance

Using Ohm's law for the rms values:

Given $V_{rms} = 200\ \text{V}$ and $R = 20\ \Omega$:

$I_{rms} = \frac{200}{20} = 10\ \text{A}$ …

Common Mistake #1: Forgetting that at resonance, $X_L = X_C$

Many students jump into calculating impedance using $Z = \sqrt{R^2 + (X_L - X_C)^2}$ without first simplifying.

Why it's wrong:

At resonance, $X_L = X_C$, so the reactive part cancels out. The impedance becomes purely resistive:

How to avoid:

Always first check if the frequency equals the natural frequency. If yes, immediately write:

No need to compute $X_L$ or $X_C$ individually.

Common Mistake #2: Using the wrong power formula

Students often use $P = V_{\text{rms}} I_{\text{rms}} \cos\phi$ but forget that at resonance $\cos\phi = 1$.

Why it's wrong:

At resonance, the circuit is purely resistive, so the power factor is 1. The formula simplifies to:

How to avoid:

Remember:

• At resonance: $\phi = 0$, $\cos\phi = 1$
• So $P = V_{\text{rms}} I_{\text{rms}}$ (no need for $\cos\phi$)

Common Mistake #3: Confusing peak voltage with RMS voltage

The problem gives $200\ \text{V}$ as the AC supply voltage. Many students treat this as peak voltage $V_0$.

Why it's wrong:

In AC circuit problems, unless stated as "peak voltage" or "$V_0$", the given voltage is RMS voltage ($V_{\text{rms}}$).

How to avoid:

Always check:

• "AC supply" → RMS value
• "Peak voltage" or "$V_0$" → peak value Here, $V_{\text{rms}} = 200\ \text{V}$ directly.

Common Mistake #4: Calculating current incorrectly

Students sometimes compute $I = V / Z$ but use $Z$ from a non-resonant calculation.

Why it's wrong:

At resonance, $Z = R$, so:

How to avoid:

Once you know $Z = R$, just divide $V_{\text{rms}}$ by $R$. No need for $L$ or $C$ values.

Common Mistake #5: Overcomplicating the power calculation …