AC circuit analysis — impedance triangle, power triangle, resonance conditions

Question

In a series LCR circuit with $L = 0.1$ H, $C = 100 \mu F$ , and $R = 50 \Omega$ , connected to a 220 V, 50 Hz AC source: find the impedance, current, phase angle, and check if the circuit is at resonance.

(CBSE 12 + JEE Main)

Solution — Step by Step

Inductive reactance: $X_L = 2\pi f L = 2\pi \times 50 \times 0.1 = 31.4 \Omega$

Capacitive reactance: $X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 100 \times 10^{-6}} = 31.8 \Omega$

Since $X_L \neq X_C$ , the circuit is NOT at resonance (but very close).

Z = \sqrt{R^2 + (X_L - X_C)^2}

Z = \sqrt{50^2 + (31.4 - 31.8)^2} = \sqrt{2500 + 0.16} = \sqrt{2500.16} \approx \mathbf{50.0 \Omega}

The impedance is almost equal to $R$ because $X_L \approx X_C$ .

I = \frac{V}{Z} = \frac{220}{50} = \mathbf{4.4 \text{ A}}

Phase angle: $\tan\phi = \frac{X_L - X_C}{R} = \frac{-0.4}{50} = -0.008$

$\phi \approx -0.46°$ (nearly zero — almost purely resistive behavior).

Since $X_C > X_L$ , current leads voltage slightly — the circuit is capacitive.

At resonance: $X_L = X_C$ , which gives:

2\pi f_0 L = \frac{1}{2\pi f_0 C}

f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.1 \times 100 \times 10^{-6}}} = \frac{1}{2\pi \times 0.00316} \approx \mathbf{50.3 \text{ Hz}}

Very close to our 50 Hz source — that is why impedance is almost purely resistive.

flowchart TD
    A["AC Circuit Problem"] --> B["Calculate X_L = 2πfL"]
    A --> C["Calculate X_C = 1/(2πfC)"]
    B --> D{"Compare X_L and X_C"}
    C --> D
    D -- "X_L = X_C" --> E["RESONANCE: Z = R, max current"]
    D -- "X_L > X_C" --> F["Inductive: voltage leads current"]
    D -- "X_L < X_C" --> G["Capacitive: current leads voltage"]
    E --> H["Z = R"]
    F --> I["Z = √(R² + (X_L - X_C)²)"]
    G --> I
    H --> J["I = V/Z, Power = I²R"]
    I --> J

Why This Works

In AC circuits, inductors and capacitors oppose current changes in opposite ways — inductors resist current increase (creating a lagging voltage), capacitors resist voltage increase (creating a leading current). The impedance triangle captures these opposing effects geometrically.

At resonance, $X_L = X_C$ , so the inductor and capacitor effects cancel perfectly. The circuit behaves as if only the resistor exists, current is maximum, and power transfer is most efficient. This principle is used in radio tuning — adjusting $C$ to match the resonant frequency of the desired station.

Alternative Method

For JEE MCQs about resonance, remember these three facts: (1) At resonance, $Z = R$ (minimum impedance), (2) Current is maximum: $I_{max} = V/R$ , (3) Voltage across $L$ and $C$ can individually exceed the source voltage — they cancel each other. The quality factor $Q = \frac{1}{R}\sqrt{L/C}$ tells you how sharp the resonance peak is.

Common Mistake

Students add $X_L$ and $X_C$ in the impedance formula instead of subtracting. The impedance is $\sqrt{R^2 + (X_L - X_C)^2}$ , not $\sqrt{R^2 + X_L^2 + X_C^2}$ . Inductor and capacitor reactances oppose each other — they partially cancel. This is why at resonance ( $X_L = X_C$ ), impedance drops to just $R$ , not to $\sqrt{R^2 + 2X_L^2}$ .