Q.Suppose the frequency of the source in the previous example can be varied.

Using the previous example's data: $R=3\ \Omega$, $L=25.48\text{ mH}$, $C=796\ \mu\text{F}$, $V_{rms}=200\text{ V}$.

1. Resonant frequency:

   $f_0=\frac{1}{2\pi\sqrt{LC}}=\frac{1}{2\pi\sqrt{(25.48\times10^{-3})(796\times10^{-6})}}\approx35.4\text{ Hz}$

2. At resonance the reactances cancel, so $Z=R$: $Z=R=3\ \Omega,\qquad I=\frac{V}{Z}=\frac{200}{3}\approx66.7\text{ A}$ …

At resonance the inductive and capacitive reactances of the series LCR circuit become equal and cancel, leaving a purely resistive impedance. This is where the impedance is smallest and the current is largest.

(a) Resonant frequency

Resonance requires $X_L=X_C$, i.e. $\omega_0 L=1/\omega_0 C$, giving $\omega_0=1/\sqrt{LC}$ and

Substituting $L=25.48\times10^{-3}\text{ H}$ and $C=796\times10^{-6}\text{ F}$:

(b) Impedance, current and power at resonance

Since $X_L=X_C$, the net reactance is zero and

The rms current is then maximum:

$I=\frac{V}{Z}=\frac{200}{3}\approx66.7\text{ A}.$ …

Method: Resonance Condition in Series RLC Circuit

This method applies when an AC source is connected to a series combination of a resistor ($R$), an inductor ($L$), and a capacitor ($C$). Resonance occurs when the inductive reactance equals the capacitive reactance.

Step 1 — Write the resonance condition

At resonance:

Where:

• $X_L = 2\pi f L$ (inductive reactance)
• $X_C = \frac{1}{2\pi f C}$ (capacitive reactance)

Step 2 — Solve for resonant frequency $f_0$

Set $X_L = X_C$:

Multiply both sides:

Thus:

Answer (a): The resonant frequency is $\boxed{f_0 = \frac{1}{2\pi \sqrt{LC}}}$

Step 3 — Impedance at resonance

At resonance, $X_L = X_C$, so they cancel each other. The total impedance is purely resistive:

Impedance at resonance: $\boxed{Z = R}$

Step 4 — Current at resonance

Using Ohm’s law for AC circuits:

Where $V$ is the RMS voltage of the source.

Current at resonance: $\boxed{I = \frac{V}{R}}$

Step 5 — Power dissipated at resonance …

Here are the common mistakes students make on resonance in AC circuits, along with how to avoid each — tailored for Indian exam accuracy (JEE, NEET, CBSE).

1. Using the Wrong Formula for Resonant Frequency

Mistake:

Students often confuse the formula for resonance in a series LCR circuit with that of a parallel circuit, or they forget the square root.

Correct formula (series LCR):

How to avoid:

• Memorise: Resonance occurs when $X_L = X_C$.
• Derive quickly:

• Never write $f_0 = \frac{1}{2\pi} \sqrt{\frac{L}{C}}$ — that’s wrong.

2. Forgetting That Impedance is Minimum (Not Maximum) at Resonance

Mistake:

Thinking $Z$ is maximum at resonance (confusing with parallel resonance or voltage across L/C).

Correct:

At resonance, $X_L = X_C$, so:

Impedance is minimum and purely resistive.

How to avoid:

• Remember: Resonance = minimum opposition to current.
• In a series circuit, current is maximum → impedance must be minimum.

3. Calculating Current Without Using the Correct Impedance

Mistake:

Using $I = V / (X_L - X_C)$ or forgetting that $Z = R$ at resonance.

Correct:

At resonance:

How to avoid:

• Always first find $Z$ at resonance.
• If $f = f_0$, then $Z = R$ — no reactance left.

4. Power Dissipation Formula Error

Mistake:

Using $P = V_{\text{rms}} I_{\text{rms}} \cos\phi$ but forgetting that at resonance $\cos\phi = 1$.

Correct:

At resonance, $\phi = 0$, so:

How to avoid:

• At resonance, circuit is purely resistive → power factor = 1.
• So $P = V_{\text{rms}}^2 / R$ also works.

5. Mixing Up RMS and Peak Values

Mistake:

Using peak voltage $V_0$ in formulas meant for RMS values, or vice versa.

Correct approach:

• If source voltage is given as $V = V_0 \sin(\omega t)$, then $V_{\text{rms}} = V_0 / \sqrt{2}$.
• Use RMS values for power and current calculations unless asked for peak.

How to avoid:

• Check the problem statement: “220 V” usually means RMS.
• Write explicitly:

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