Z = √(R² + (XL - XC)²). Resonance when XL = XC.

Impedance of an LCR Circuit

Question

A series LCR circuit has resistance $R = 100\ \Omega$ , inductance $L = 0.5\ \text{H}$ , and capacitance $C = 10\ \mu\text{F}$ . The circuit is connected to an AC source of frequency $f = 100\ \text{Hz}$ . Find the impedance of the circuit.

Solution — Step by Step

X_L = 2\pi f L = 2\pi \times 100 \times 0.5 = 100\pi \approx 314\ \Omega

We’re converting frequency to angular frequency $\omega = 2\pi f$ because the reactance formulas work with $\omega$ , not $f$ directly.

X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 100 \times 10 \times 10^{-6}} = \frac{1}{2\pi \times 10^{-3}} \approx 159\ \Omega

Notice that $X_C$ decreases as frequency increases — exactly the opposite of $X_L$ . This is what makes resonance possible.

X = X_L - X_C = 314 - 159 = 155\ \Omega

Since $X_L \gt X_C$ , the circuit is inductive — current lags the voltage. If $X_C$ had been larger, it would be capacitive.

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{(100)^2 + (155)^2}

Z = \sqrt{10000 + 24025} = \sqrt{34025} \approx \mathbf{184.5\ \Omega}

Why This Works

The reason we can’t just add $R$ , $X_L$ , and $X_C$ directly is that they don’t point in the same direction on a phasor diagram. Resistance is always in phase with current. Inductive reactance leads by 90°. Capacitive reactance lags by 90° — meaning $X_L$ and $X_C$ are literally opposite each other, so they partially cancel.

What remains after that cancellation is $(X_L - X_C)$ , which is perpendicular to $R$ . Two perpendicular quantities add via the Pythagorean theorem — that’s exactly where the $\sqrt{R^2 + (X_L - X_C)^2}$ formula comes from. It’s not magic; it’s just vector addition on a phasor diagram.

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

where $X_L = \omega L$ and $X_C = \dfrac{1}{\omega C}$ , with $\omega = 2\pi f$

Phase angle: $\tan\phi = \dfrac{X_L - X_C}{R}$

Alternative Method — Using Phasor Diagram

Draw voltage phasors: $V_R$ along the x-axis, $V_L$ pointing up (90° ahead), $V_C$ pointing down (90° behind). The net voltage is:

V = \sqrt{V_R^2 + (V_L - V_C)^2}

Divide everything by current $I$ (same throughout a series circuit), and you get $Z = V/I$ , which gives the same formula. This approach is useful in JEE problems that ask you to draw or interpret phasor diagrams — it’s the same calculation, just visualised differently.

Common Mistake

Adding all three instead of subtracting: A very common error is writing $Z = \sqrt{R^2 + X_L^2 + X_C^2}$ . This treats all three as if they point in different directions, which is wrong. $X_L$ and $X_C$ point in opposite directions, so they subtract first. The correct form is always $Z = \sqrt{R^2 + (X_L - X_C)^2}$ .

This exact trap appeared in JEE Main 2024 Shift 1 as a conceptual MCQ — students who memorised the formula without understanding the phasor diagram lost that mark.

Resonance check: At resonance, $X_L = X_C$ , so $(X_L - X_C) = 0$ , and impedance reduces to just $Z = R$ . This is the minimum possible impedance, giving maximum current. In numerical problems, if the question gives you a frequency where $\omega = 1/\sqrt{LC}$ , the circuit is at resonance — you don’t need to calculate $Z$ beyond $R$ .

Scoring note: In CBSE Class 12 boards, this derivation is a standard 3-mark question. The phasor diagram is worth 1 mark, the formula derivation is 1 mark, and substitution with the final numerical answer is 1 mark. Don’t skip the diagram — examiners specifically look for it.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Phasor Diagram

Common Mistake

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