JEE Mains Physics · Alternating Current

Series LCR: Impedance, Phase and Power

In a series LCR circuit the resistance and the net reactance combine at right angles: Z = √(R² + (ωL − 1/ωC)²), the power factor is R/Z, and only the resistance takes power.

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Why this matters

Thirty PYQs, twenty of them multiple choice, and five from 2026 alone. Twelve find an impedance, a current or the voltage across one part. Twelve find a phase angle or a power factor, often after the frequency changes or a part is added. Six find the average power or test wattless current and the choke coil. One right-angled triangle of R and X answers nearly all of them.

Concept 1 of 3: Impedance and the voltage triangle

The voltage across R is in step with the current, the voltage across L is 90° ahead, and the voltage across C is 90° behind. So

V_L

and

V_C

pull in opposite directions and cancel in part, and what is left sits at right angles to

V_R

. Add them like the sides of a right triangle, never as plain numbers. Divide every side by the current and you get the impedance triangle.

Definition

$Z = \sqrt{R^{2} + (X_L - X_C)^{2}}$ ; $I = \dfrac{V}{Z}$ , peak with peak and rms with rms.
$V_R = IR$ , $V_L = IX_L$ , $V_C = IX_C$ , and $V^{2} = V_R^{2} + (V_L - V_C)^{2}$ .
R and L only: $Z = \sqrt{R^{2} + X_L^{2}}$ and $V^{2} = V_R^{2} + V_L^{2}$ . R and C only: the same with $X_C$ .
A new frequency changes both reactances the opposite way: at $k\omega$ , $X_L \to kX_L$ and $X_C \to \dfrac{X_C}{k}$ . Recompute Z after scaling both.
A lamp rated P at V in series: its resistance is $\dfrac{V^{2}}{P}$ and its current at full brightness is $\dfrac{P}{V}$ . The rest of the supply voltage sits at right angles across the L or C.
The sign of $X_L - X_C$ does not change Z, but it does decide whether the current leads or lags.

Impedance and voltages

Z = \sqrt{R^{2} + (X_L - X_C)^{2}} \qquad V^{2} = V_R^{2} + (V_L - V_C)^{2}

Worked example

R =

40\ \Omega

, L = 0.2 H and C =

50\ \mu\text{F}

are in series across

v = 100\sin(250t)\ \text{V}

. Find Z, the peak current and the peak voltage across each part.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 27 July 2022 · Q104Moderate

Example 1 · Alternating Current · Series LCR: Impedance, Phase and Power

A series

LCR

circuit has

L = 0.01H,R = 10\Omega

and

C = 1\mu F

and it is connected to ac voltage of amplitude

\left( V_{m} \right)50\text{ }V

. At frequency

60\%

lower than resonant frequency, the amplitude of current will be approximately:

Adding the voltages as numbers

V_R + V_L + V_C

is not the supply voltage. The parts are out of phase, so use

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

Changing only one reactance with the frequency

When ω changes,

X_L

and

X_C

both change, in opposite directions. Scale both before finding Z.

Mixing peak and rms

Peak voltage over Z gives the peak current; rms over Z gives rms. A 'percent lower' frequency means ω times the remaining fraction: 30% lower is 0.7ω.

Drill 11 more on impedance and the voltage triangle

Concept 2 of 3: Phase angle and power factor

The phase angle is the angle of the impedance triangle: net reactance up, resistance across. Its cosine, R/Z, is the power factor. If the inductive side wins, the current lags; if the capacitive side wins, it leads; if they balance, the power factor is 1. Since only R and the reactances set the triangle, the source's amplitude never changes the power factor.

Definition

$\tan\phi = \dfrac{X_L - X_C}{R}$ ; power factor $\cos\phi = \dfrac{R}{Z}$ .
$X_L > X_C$ : inductive, current lags. $X_C > X_L$ : capacitive, current leads. $X_L = X_C$ : $\cos\phi = 1$ .
From the equations: if $v = V_0\sin(\omega t + \alpha)$ and $i = I_0\sin(\omega t + \beta)$ , then $\phi = |\alpha - \beta|$ .
Remove one part and read the angle: an RL circuit lagging by $45^{\circ}$ has $X_L = R$ ; an RC circuit leading by $45^{\circ}$ has $X_C = R$ .
New frequency: from the old angle get $X_L/R$ (or $X_C/R$ ), scale it, then find the new $\cos\phi$ .
Adding a capacitor to an RL circuit cancels part of $X_L$ and raises the power factor.

Phase angle and power factor

\tan\phi = \frac{X_L - X_C}{R} \qquad \cos\phi = \frac{R}{Z}

Worked example

An RL circuit has power factor 0.6 at

\omega = 500\ \text{rad/s}

. Find its power factor at

\omega = 1000\ \text{rad/s}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 30 Jan 2024 · Q20Moderate

Example 2 · Alternating Current · Series LCR: Impedance, Phase and Power

A series L, R circuit connected with an ac source

E = (25\sin1000t)V

has a power factor of

\frac{1}{\sqrt{2}}

. If the source of emf is changed to

E = (20\sin2000

t)V

, the new power factor of the circuit will be :

Using sin φ for the power factor

The power factor is cos φ = R/Z. The ratio of the net reactance to Z is sin φ.

Losing lead or lag

cos φ is the same for +φ and −φ. Decide lead or lag from which reactance is bigger, or from which equation has the larger phase constant.

Letting the amplitude change the power factor

A new source amplitude changes the current, not the angle. Only a new frequency or a new part changes the power factor.

Drill 11 more on phase angle and power factor

Concept 3 of 3: Average power, wattless current and the choke coil

Over a cycle an inductor or a capacitor takes energy and gives all of it back, so only the resistance uses power. That is why the average power carries the factor cos φ. A current that is 90° out of step with the voltage does no work at all: it is called wattless. A choke uses this to limit a current without wasting power.

Definition

$P = V_{rms}I_{rms}\cos\phi = \dfrac{1}{2}V_0I_0\cos\phi = I_{rms}^{2}R$ .
From $v = V_0\sin\omega t$ and $i = I_0\sin(\omega t \pm \phi)$ : $P = \dfrac{V_0I_0}{2}\cos\phi$ . Watch the units: mA times V gives mW.
Pure L, pure C, or L with C and no R: $\cos\phi = 0$ , so $P = 0$ . The current is wattless.
Zero net reactance needs no empty circuit: $X_L = X_C$ cancel each other.
A choke coil has a large L and a tiny R. It cuts the current in a tube-light circuit with almost no power loss, where a series resistor would waste $I^{2}R$ .

Average power

P = V_{rms}I_{rms}\cos\phi = \frac{1}{2}V_0I_0\cos\phi = I_{rms}^{2}R

Worked example

A voltage

v = 50\sin(100t)\ \text{V}

drives a current

i = 4\sin\left(100t - \dfrac{\pi}{6}\right)\ \text{A}

. Find the average power.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q95Moderate

Example 3 · Alternating Current · Series LCR: Impedance, Phase and Power

An AC voltage

V = 20\sin200\pi t

is applied to a series LCR circuit which drives a current

I = 10\sin\left( 200\pi t +\frac{\pi}{3} \right)

. The average power dissipated is:

Peak values in the rms formula

V_{rms}I_{rms}\cos\phi

needs rms values. With peak values, halve the product:

\dfrac{1}{2}V_0I_0\cos\phi

Using Z in I²R

Power is

I_{rms}^{2}R

. Writing

I^{2}Z

charges the inductor and capacitor for power they never keep.

Drill 5 more on average power, wattless current and the choke coil

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 the voltage triangle
Impedance and voltages
$Z = \sqrt{R^{2} + (X_L - X_C)^{2}} \qquad V^{2} = V_R^{2} + (V_L - V_C)^{2}$
Phase angle and power factor
Phase angle and power factor
$\tan\phi = \frac{X_L - X_C}{R} \qquad \cos\phi = \frac{R}{Z}$
Average power, wattless current and the choke coil
Average power
$P = V_{rms}I_{rms}\cos\phi = \frac{1}{2}V_0I_0\cos\phi = I_{rms}^{2}R$

Watch out for (8)

Adding the voltages as numbers
→ Impedance and the voltage triangle
Changing only one reactance with the frequency
→ Impedance and the voltage triangle
Mixing peak and rms
→ Impedance and the voltage triangle
Using sin φ for the power factor
→ Phase angle and power factor
Losing lead or lag
→ Phase angle and power factor
Letting the amplitude change the power factor
→ Phase angle and power factor
Peak values in the rms formula
→ Average power, wattless current and the choke coil
Using Z in I²R
→ Average power, wattless current and the choke coil

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