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

Question

A series LCR circuit with $R = 20\ \Omega$, $L = 1.5\ 	ext{H}$ and $C = 35\ \mu	ext{F}$ is connected to a variable-frequency $200\ 	ext{V}$ 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?

Accepted Answer

> [!TLDR]
> At resonance, the circuit behaves purely resistively, so the average power is simply $V^2/R$. With $V = 200\ 	ext{V}$ and $R = 20\ \Omega$, the average power is $2000\ 	ext{W}$.

## 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 $V_{	ext{rms}}^2 / R$.

> [!WARNING]
> A common mistake is to forget that the supply voltage given (200 V) is the **rms** value, not the peak value. For AC power calculations, always use rms values unless told otherwise.

## 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:
$$P_{	ext{av}} = V_{	ext{rms}} I_{	ext{rms}} \cos \phi$$
where $\cos \phi$ is the power factor.

**3. Apply the resonance simplification.**  
Since $\cos \phi = 1$ at resonance:
$$P_{	ext{av}} = V_{	ext{rms}} I_{	ext{rms}}$$

But $I_{	ext{rms}} = V_{	ext{rms}} / Z = V_{	ext{rms}} / R$ because $Z = R$. Substituting:
$$P_{	ext{av}} = V_{	ext{rms}} \cdot \frac{V_{	ext{rms}}}{R} = \frac{V_{	ext{rms}}^2}{R}$$ …

Concept understanding — Resonance in AC Circuits

Resonance in AC Circuits

The Competing Reactances

The Resonance Condition

Physical Picture

Sharpness and the Q-factor

Why It Matters …

Why this formula?

Resonance in AC Circuits: Why the Key Formulas Hold

1. The Core Condition for Resonance

2. Deriving the Resonant Frequency

3. What Happens at Resonance — Key Consequences

(a) Impedance is Minimum (Purely Resistive)

(b) Current is Maximum

(c) Voltage Across L and C Can Be Very Large

Why resonance simplifies everything

Step-by-step solution

Method: Resonance Condition in Series LCR Circuit

Steps

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

Common Mistake #2: Using the wrong power formula

Common Mistake #3: Confusing peak voltage with RMS voltage

Common Mistake #4: Calculating current incorrectly

Common Mistake #5: Overcomplicating the power calculation …

Concept

Solution

Why resonance simplifies everything

Step-by-step solution

🎓Unlock everything free for 14 days

Unlock everything free for 14 days