In an ac circuit when the supply voltage frequency is equal to the resonant frequency, the current

Analyze the question: The question asks about the phase relationship between current and supply voltage in an AC circuit (series RLC) when the supply frequency equals the resonant frequency.

Explain the concept: In a series RLC circuit:

• Inductive reactance: X_L = 2πfL (increases with frequency)
• Capacitive reactance: X_C = 1/(2πfC) (decreases with frequency)
• Impedance: Z = √[R² + (X_L − X_C)²]
• Phase angle: tan φ = (X_L − X_C)/R

At resonance, the frequency is f₀ = 1/(2π√(LC)), and X_L = X_C.

Solution:

1. At resonance, X_L = X_C, so X_L − X_C = 0.
2. The impedance becomes Z = R, which is purely resistive and at its minimum.
3. The phase angle is tan φ = 0/R = 0, so φ = 0°.
4. A phase angle of 0° means the current and the supply voltage reach their peaks at the same time. They are in phase.
5. The current is also at its maximum, I = V/R.

Checking the options:

• A (leads by 90°): this is a purely capacitive circuit, which is not the case here.
• B (lags by 90°): this is a purely inductive circuit, which is not the case here.
• C (leads by 45°): this happens below resonance when X_C − X_L = R, which is not the case at resonance.
• D (in phase): correct, because the circuit behaves as a pure resistor at resonance.

Tip: Remember "Resonance = Resistive." At resonance, the reactances cancel, so the circuit acts like just a resistor. That gives you three facts to recall: the phase angle is zero, the impedance is minimum, and the current is maximum.