MHT-CET Physics · AC Circuits

Series LR, RC and LCR: Impedance, Phase and Phasors

In a series circuit the resistor's voltage and the reactances' voltages are 90° apart, so they add like the sides of a right triangle: Z = √(R² + (X_L − X_C)²), and tan φ = (X_L − X_C)/R gives the phase.

All AC Circuits notes MHT-CET Physics strategy

Why this matters

33 PYQs, five HARD. Three shapes: an impedance and a current (including the coil that draws one current on d.c. and less on a.c.), a phase angle — or a component found from a given phase — and the voltages across R, L and C adding as phasors, never as numbers.

Concept 1 of 3: Impedance and Current

Impedance is the a.c. resistance of the whole circuit. Resistance and net reactance are at right angles, so they combine by Pythagoras. A real coil has both: on d.c. only its resistance shows; on a.c. its reactance joins in and the current drops.

Definition

$Z = \sqrt{R^2 + (X_L - X_C)^2}$ ; $I = \dfrac{V}{Z}$ . LR only: $Z = \sqrt{R^2 + 4\pi^2f^2L^2}$ .
A coil on d.c. then a.c.: $R = \dfrac{V}{I_{\text{dc}}}$ , $Z = \dfrac{V}{I_{\text{ac}}}$ , $X_L = \sqrt{Z^2 - R^2}$ , $L = \dfrac{X_L}{2\pi f}$ .
Adding a capacitor to an LR circuit reduces the net reactance (for $X_C < 2X_L$ ), so the current RISES.
Admittance $= \dfrac{1}{Z}$ .
With $|X_L - X_C| = R$ : $Z = \sqrt{2}R$ .

Impedance

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

Worked example

R = 30\,\Omega

and

X_L = 40\,\Omega

are in series on 100 V. Impedance and current?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift II · Q10Moderate

Example 1 · AC Circuits · Series LCR — Impedance, Phase and Phasors

In an

LR

circuit, the value of

L

\left( \frac{0 \cdot 3}{\pi} \right)

henry and the value of R is

40\Omega

, If in the circuit, an alternating e.m.f of 230 V at 50 cycles per second is connected, the impedance of the circuit and current will be respectively

Adding R and X directly

Z = R + X_L

is the tempting shortcut and is always wrong: 30 Ω and 40 Ω make 50 Ω, not 70 Ω.

Drill 12 more on impedance and current

Concept 2 of 3: The Phase Angle

The phase angle measures how far the circuit is from purely resistive. More inductive reactance tips the voltage ahead of the current; more capacitive tips it behind. Given the angle, the same equation runs backwards to find L, C or R.

Definition

$\tan\phi = \dfrac{X_L - X_C}{R}$ ; $\cos\phi = \dfrac{R}{Z}$ .
$X_L > X_C$ : voltage LEADS; $X_C > X_L$ : current leads. Current leading by $45^\circ$ ⇒ $X_L = X_C - R$ .
Voltage leading by $45^\circ$ : $\omega L - \dfrac{1}{\omega C} = R$ , so $L = \dfrac{1 + 2\pi fCR}{4\pi^2f^2C}$ .
From $Z$ and $\phi$ : $R = Z\cos\phi$ . 15 V, 0.5 A, $\phi = 60^\circ$ : $Z = 30$ , $R = 15\,\Omega$ .

Phase angle

\tan\phi = \frac{X_L - X_C}{R}

Worked example

R = 100\,\Omega

X_L = 250\,\Omega

X_C = 150\,\Omega

. Phase angle, and which leads?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 1 · Q7Moderate

Example 2 · AC Circuits · Series LCR — Impedance, Phase and Phasors

An inductance of

\frac{300}{\pi}

mH, a capacitance of

\frac{1}{\pi}

mF and a resistance of

20\,\Omega

are connected in series with an a.c. source of 240 V, 50 Hz. The phase angle of the circuit is

Inverting the tangent

\tan\phi = \frac{X}{R}

, reactance on top.

X_L = 400

R = 300

gives

\tan^{-1}\frac{4}{3}

;

\tan^{-1}\frac{3}{4}

sits beside it.

Drill 14 more on the phase angle

Concept 3 of 3: Voltages Add as Phasors

V_L and V_C point in opposite directions and both stand at right angles to V_R, so the supply voltage is the hypotenuse of V_R and the difference V_L − V_C. The same geometry makes the currents in a parallel L and C subtract.

Definition

$V^2 = V_R^2 + (V_L - V_C)^2$ — never $V = V_R + V_L + V_C$ .
$V_L$ and $V_C$ are $180^\circ$ apart; each is $90^\circ$ from $V_R$ ; neither is in phase with the source (except at resonance, where they cancel).
Parallel $L$ and $C$ : their currents are opposite, so the source supplies $|I_L - I_C|$ .

Series voltages

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

Worked example

In a series LCR circuit

V_R = 30

V_L = 90

V_C = 50

V. Supply voltage?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q24Moderate

Example 3 · AC Circuits · Series LCR — Impedance, Phase and Phasors

A series L-C-R circuit is connected to a source of alternating emf of 50 V and the potential difference across inductor and capacitor is 90 V and 60 V, respectively. The potential difference across the resistor is

Adding the voltages as numbers

50 V across a circuit can put 90 V on L and 60 V on C — more than the source — because they cancel in part. Only the phasor sum equals the supply.

Drill 4 more on voltages add as phasors

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Impedance and Current
Impedance
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
The Phase Angle
Phase angle
$\tan\phi = \frac{X_L - X_C}{R}$
Voltages Add as Phasors
Series voltages
$V = \sqrt{V_R^2 + (V_L - V_C)^2}$

Watch out for (3)

Adding R and X directly
→ Impedance and Current
Inverting the tangent
→ The Phase Angle
Adding the voltages as numbers
→ Voltages Add as Phasors

Test yourself on AC Circuits

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the AC Circuits chapter test Drill all 33 questions

Related notes

Power in AC Circuits — cos φ = R/Z →