LR, CR, LCR Circuits

Q: What is the impedance of a series LCR circuit?

It is the total opposition to alternating current, Z = √(R² + (XL - XC)²), where XL = ωL and XC = 1/(ωC). The reactances subtract because the voltages across the inductor and capacitor are opposite in phase. Impedance is measured in ohm and I = V/Z.

Q: What is the difference between an LR and a CR circuit?

In an LR circuit the voltage leads the current, tan φ = XL/R, and the impedance rises with frequency. In a CR circuit the voltage lags the current, tan φ = XC/R, and the impedance falls with frequency. A CR circuit passes no steady DC; an LR circuit does.

Q: What is the power factor and why is it important?

Power factor is cos φ = R/Z, the fraction of the apparent power VrmsIrms that is actually consumed. A low power factor means a larger current is needed for the same useful power, which raises heating losses in the supply lines.

Q: What is wattless current?

It is the component of the current, Irms sin φ, that is perpendicular to the voltage in the phasor diagram. It flows back and forth but consumes no average power. In a pure inductor or capacitor the whole current is wattless, since the phase difference is 90 degrees.

Q: What does the Q-factor of a resonant circuit mean?

The quality factor measures how sharp the resonance is: Q = ω₀/(Δω) = ω₀L/R, where Δω is the bandwidth between the half-power frequencies. A high Q means a narrow, tall peak and good selectivity, useful in radio tuning circuits.

PhysicsAlternating CurrentsFor JEE aspirants

LR, CR and LCR circuits are series AC circuits in which a resistor is combined with an inductor, a capacitor or both. Their voltages add as phasors, giving the impedance $Z = R^{2} + (X_{L} - X_{C})^{2}$ and a phase angle between current and voltage. This page covers LR, CR and LCR circuits, power factor, resonance and the Q-factor, the choke coil and LC oscillations, all high-scoring topics in JEE Main and NEET.

On this page1LR circuit2CR circuit3LC circuit4LCR circuit5Power factor6Resonance7Q-factor8Choke coil9LC oscillations

Key Formulas - Quick Reference

Series LCR: $Z = R^{2} + (X_{L} - X_{C})^{2}$ , $V = V_{R}^{2} + (V_{L} - V_{C})^{2}$ , $i_{0} = \frac{V _{0}}{Z}$
★ Must learnPhase angle: $tan ϕ = \frac{X _{L} - X _{C}}{R}$ (voltage leads current if $X_{L} > X_{C}$ ); LR: $tan ϕ = \frac{X _{L}}{R}$ ; CR: $tan ϕ = \frac{X _{C}}{R}$ (voltage lags)
★ Must learnAverage power $P = V_{r m s} I_{r m s} cos ϕ = I_{r m s}^{2} R$ ; power factor $cos ϕ = \frac{R}{Z}$
Resonance: $ω_{0} = \frac{1}{L C}$ , $f_{0} = \frac{1}{2 π L C}$ ; $Z_{min} = R$ , $i_{ma x} = \frac{V}{R}$ , $cos ϕ = 1$
Q-factor $Q = \frac{ω _{0}}{ω _{2} - ω _{1}} = \frac{ω _{0} L}{R} = \frac{1}{R} \frac{L}{C}$ ; bandwidth $Δ ω = \frac{R}{L}$
At resonance $V_{L} = V_{C} = Q V$ (can exceed the source voltage)
★ Must learnLC oscillations: $ω = \frac{1}{L C}$ , $q = q_{0} cos ω t$ , $i = q_{0} ω sin ω t$ , total energy $\frac{q _{0}^{2}}{2 C}$
Apparent power $= V_{r m s} I_{r m s}$ ; wattless current $= I_{r m s} sin ϕ$ ; admittance $= \frac{1}{Z}$ , susceptance $= \frac{1}{X}$

1. Series LR Circuit

A resistor $R$ and an inductor $L$ are in series with an ac source. The same current $i = i_{0} sin ω t$ flows through both, so we draw the current phasor along the $x$ -axis and place each voltage relative to it:

$V_{R} = i_{0} R$ is in phase with the current (along $+ x$ ).
$V_{L} = i_{0} X_{L}$ leads the current by $9 0^{\circ}$ (along $+ y$ ).

The instantaneous source voltage is the sum $v (t) = i_{0} R sin ω t + i_{0} X_{L} sin (ω t + \frac{π}{2})$ . Adding the phasors like vectors:

V_{0} = V_{R}^{2} + V_{L}^{2} = i_{0} R^{2} + X_{L}^{2}, v = i_{0} R^{2} + X_{L}^{2} sin (ω t + ϕ), tan ϕ = \frac{X _{L}}{R}

$Z = R^{2} + X_{L}^{2} = R^{2} + ω^{2} L^{2}$ is the impedance of the LR circuit, and $ϕ = tan^{- 1} \frac{ω L}{R}$ : the voltage leads the current by $ϕ$ (between $0$ and $9 0^{\circ}$ ).

Figure 1: Series LR (values of Solved Example 2).

V_{R} = 120 V

along

i

V_{L} = 50 V

9 0^{\circ}

ahead;

V = V_{R}^{2} + V_{L}^{2} = 130 V

leads

i

ϕ = tan^{- 1} \frac{X _{L}}{R} = 22. 6^{\circ}

2. Series CR Circuit

Now a resistor $R$ and a capacitor $C$ are in series. With the current $i = i_{0} sin ω t$ along $+ x$ , $V_{R} = i_{0} R$ is along the current, but $V_{C} = i_{0} X_{C}$ lags the current by $9 0^{\circ}$ (along $- y$ ):

v (t) = V_{R 0} sin ω t + V_{C 0} sin (ω t - \frac{π}{2}) = V_{0} sin (ω t - ϕ)

V_{0} = V_{R 0}^{2} + V_{C 0}^{2} = i_{0} R^{2} + X_{C}^{2}, Z = R^{2} + \frac{1}{ω ^{2} C ^{2}}, tan ϕ = \frac{X _{C}}{R}

In a CR circuit the voltage lags the current by $ϕ$ (the current leads).

Figure 2: Series CR (lamp circuit of Solved Example 4).

V_{R} = 100 V

V_{C} = 173 V

9 0^{\circ}

behind

i

;

V = 200 V

lags

i

ϕ = tan^{- 1} \frac{X _{C}}{R} = 6 0^{\circ}

. The rms voltages add as vectors, not as numbers (

100 + 173 \neq = 200

LR circuit

$Z = R^{2} + ω^{2} L^{2}$ rises with frequency. Voltage leads current, $tan ϕ = \frac{ω L}{R}$ . On DC the current is $\frac{V}{R}$ .

CR circuit

$Z = R^{2} + \frac{1}{ω ^{2} C ^{2}}$ falls with frequency. Voltage lags current, $tan ϕ = \frac{1}{ω C R}$ . On DC no steady current flows.

Key idea

Series voltages add as phasors, not as numbers:

V = V_{R}^{2} + V_{X}^{2}

, so the meter readings across the parts add up to more than the source voltage.

3. Series LC Circuit

With an inductor and a capacitor in series (no resistance) and $i = i_{0} sin ω t$ :

v (t) = i_{0} X_{L} sin (ω t + \frac{π}{2}) + i_{0} X_{C} sin (ω t - \frac{π}{2}) = i_{0} (X_{L} - X_{C}) cos ω t

$V_{L}$ and $V_{C}$ are exactly opposite, so they subtract. The impedance is $Z = ∣ X_{L} - X_{C} ∣$ and the phase angle is $9 0^{\circ}$ : if $X_{L} > X_{C}$ the voltage leads the current by $\frac{π}{2}$ , if $X_{L} < X_{C}$ it lags by $\frac{π}{2}$ . The power factor is $cos ϕ = 0$ , so no power is consumed. The peak voltages are $V_{C 0} = i_{0} X_{C}$ and $V_{L 0} = i_{0} X_{L}$ .

4. Series LCR Circuit

A resistor, an inductor and a capacitor are in series with an ac source. The current phasor is along $+ x$ ; then $V_{R}$ is along $+ x$ , $V_{L}$ along $+ y$ (leads by $9 0^{\circ}$ ) and $V_{C}$ along $- y$ (lags by $9 0^{\circ}$ ). $V_{L}$ and $V_{C}$ first cancel partly, leaving $V_{L} - V_{C}$ ; adding $V_{R}$ at right angles:

V = V_{R}^{2} + (V_{L} - V_{C})^{2}

Figure 3: Series LCR (Solved Example 5).

V_{L}

and

V_{C}

point in opposite directions and partly cancel.

V_{0} = V_{R}^{2} + (V_{L} - V_{C})^{2} = 6^{2} + 8^{2} = 10 V

, leading

i

ϕ = 5 3^{\circ}

because

V_{L} > V_{C}

★ Must learn

Impedance of a series LCR circuit. Dividing each voltage by the current:

Z = R^{2} + (X_{L} - X_{C})^{2} = R^{2} + (ω L - \frac{1}{ω C})^{2}

tan ϕ = \frac{X _{L} - X _{C}}{R}, cos ϕ = \frac{R}{Z}

$Z$ is measured in ohm; $V_{0} = i_{0} Z$ and $V_{r m s} = I_{r m s} Z$ .

Figure 4: The three triangles are similar, so they share

ϕ = 53. 1^{\circ}

and

cos ϕ = 0.6

. Divide the voltage triangle (peak values of Solved Example 5) by

i_{0} = 2 A

to get the impedance triangle,

Z = 5 Ω

. Multiply that by

I_{r m s}^{2} = 2 A^{2}

to get the power triangle: true power

P = 6 W

, reactive power

Q = 8 var

, apparent power

S = V_{r m s} I_{r m s} = 10 VA

4.1 Instantaneous voltages

If the current is $i = i_{0} sin ω t$ :

Quantity	Instantaneous value	Peak value	Phase relative to $i$
Source voltage	$v = i_{0} Z sin (ω t + ϕ)$	$V_{0} = i_{0} Z$	Leads by $ϕ$ (if $X_{L} > X_{C}$ )
Across $L$	$v_{L} = i_{0} X_{L} sin (ω t + \frac{π}{2})$	$V_{0 L} = i_{0} X_{L}$	Leads by $\frac{π}{2}$
Across $C$	$v_{C} = i_{0} X_{C} sin (ω t - \frac{π}{2})$	$V_{0 C} = i_{0} X_{C}$	Lags by $\frac{π}{2}$
Across $R$	$v_{R} = i_{0} R sin ω t$	$V_{0 R} = i_{0} R$	In phase

Figure 5: For

X_{L} > X_{C}

(Solved Example 5) the current reaches each peak

ϕ = 5 3^{\circ}

after the voltage: the circuit is inductive. The current amplitude is

i_{0} = \frac{V _{0}}{Z} = 2 A

(drawn to a smaller scale).

4.2 Three special cases

$X_{L} > X_{C}$ ( $V_{L} > V_{C}$ ): the emf leads the current by $ϕ$ , $tan ϕ = \frac{X _{L} - X _{C}}{R}$ . The circuit is inductive.
$X_{L} < X_{C}$ ( $V_{L} < V_{C}$ ): the current leads the emf by $ϕ$ , $tan ϕ = \frac{X _{C} - X _{L}}{R}$ . The circuit is capacitive.
$X_{L} = X_{C}$ ( $V_{L} = V_{C}$ ): $ϕ = 0$ , emf and current are in phase and $Z = R$ . The circuit is purely resistive (resonance, Section 6), with $I_{0} = \frac{E _{0}}{R}$ and $I_{r m s} = \frac{E _{r m s}}{R}$ .

Figure 6: The sign of

X_{L} - X_{C}

decides everything.

X_{L} > X_{C}

: inductive, voltage leads.

X_{L} < X_{C}

: capacitive, voltage lags.

X_{L} = X_{C}

V_{L}

and

V_{C}

cancel,

Z = R

ϕ = 0

(resonance).

Susceptance is the reciprocal of the reactance of an AC circuit, $\frac{1}{X}$ . Admittance is the reciprocal of its impedance, $\frac{1}{Z}$ . Both are measured in siemens ( $Ω^{- 1}$ ) and are convenient for parallel circuits, where admittances add.

Exam Trick

Put the numbers on the impedance triangle. Most answers are Pythagorean triples: $(3, 4, 5)$ , $(6, 8, 10)$ , $(5, 12, 13)$ , $(8, 15, 17)$ . $R$ is the base, $∣ X_{L} - X_{C} ∣$ the height, $Z$ the hypotenuse; $cos ϕ = \frac{R}{Z}$ is the power factor. LR and CR circuits are just $X_{C} = 0$ and $X_{L} = 0$ .

Quick Recall: tap to check

In a series LCR circuit which voltage is in phase with the current?

The voltage across the resistor,

V_{R}

Can the peak voltage across the inductor be greater than the peak voltage of the source?

Yes.

V_{L}

and

V_{C}

cancel partly, so each can exceed

V

; at resonance

V_{L} = V_{C} = Q V

What is the impedance of a series LC circuit?

Z = ∣ X_{L} - X_{C} ∣

5. Power in an AC Circuit

For steady current, power is $P = V i$ . In an AC circuit both $v$ and $i$ change, so the work done by the source in time $d t$ is $d W = v i d t$ . Let the current lead the voltage by $ϕ$ :

$v = V_{0} sin ω t$ , $i = i_{0} sin (ω t + ϕ)$ , so
$d W = V_{0} i_{0} (sin^{2} ω t cos ϕ + sin ω t cos ω t sin ϕ) d t$
Over one cycle $\int_{0}^{T} sin^{2} ω t d t = \frac{T}{2}$ and $\int_{0}^{T} sin ω t cos ω t d t = 0$ , so $W = \frac{1}{2} V_{0} i_{0} T cos ϕ$ .
Average power:
$P = \frac{W}{T} = \frac{V _{0} i _{0}}{2} cos ϕ = \frac{V _{0}}{2} \cdot \frac{i _{0}}{2} cos ϕ = V_{r m s} I_{r m s} cos ϕ$

★ Must learn

Power factor $cos ϕ = \frac{R}{Z}$ . It is called leading if the current leads the voltage and lagging if it lags; a power factor of $0.5$ lagging means the current lags by $6 0^{\circ}$ . Since $V_{r m s} = I_{r m s} Z$ , the average power is also $P = I_{r m s}^{2} Z \cdot \frac{R}{Z} = I_{r m s}^{2} R$ : only the resistance consumes power.

The product $V_{r m s} I_{r m s}$ is the apparent power (unit volt-ampere, VA); the true power is apparent power $\times$ power factor (unit watt). Multiplying each side of the impedance triangle by $I_{r m s}^{2}$ gives the power triangle (third panel of the triangles figure in Section 4): true power $P = I_{r m s}^{2} R$ along the base, reactive power $Q = I_{r m s}^{2} (X_{L} - X_{C})$ (unit var) up the side, and apparent power $S = I_{r m s}^{2} Z = V_{r m s} I_{r m s}$ along the hypotenuse. So $S^{2} = P^{2} + Q^{2}$ and $cos ϕ = \frac{P}{S}$ .

Figure 7: With a phase difference

ϕ = 6 0^{\circ}

p = v i

dips below zero for part of each half cycle (energy returned by

L

C

). The average power is

V_{r m s} I_{r m s} cos ϕ

, here half of what it would be at

ϕ = 0

Circuit	$ϕ$	Power factor $cos ϕ$	Average power
Pure $R$	$0$	1	$V_{r m s} I_{r m s}$ (maximum)
Pure $L$ or pure $C$	$9 0^{\circ}$	0	0 (wattless)
LR, CR or LCR	$0$ to $9 0^{\circ}$	$\frac{R}{Z}$	$V_{r m s} I_{r m s} cos ϕ$
LCR at resonance	$0$	1	$\frac{V _{r m s}^{2}}{R}$

5.1 Wattless current

Resolve the current phasor into two parts: $I_{r m s} cos ϕ$ along the voltage and $I_{r m s} sin ϕ$ perpendicular to it. Only the first carries power, $V_{r m s} (I_{r m s} cos ϕ)$ . The component $I_{r m s} sin ϕ$ gives zero average power and is called the wattless current. When $ϕ = 9 0^{\circ}$ (a pure $L$ or $C$ ) the whole current is wattless.

Exam Trick

For power, use $I_{r m s}^{2} R$ . It needs no phase angle and cannot go wrong: inductors and capacitors never consume average power. Low power factor is costly: for the same power, a smaller $cos ϕ$ means a larger current and more $I^{2} R$ loss in the supply lines, so factories add capacitors to raise $cos ϕ$ towards $1$ .

Figure 8: Every series AC problem follows one route: reactances, impedance, current, then the sign of

X_{L} - X_{C}

for the phase, and

P = I_{r m s}^{2} R

for the power. LR and CR circuits are the special cases

X_{C} = 0

and

X_{L} = 0

Key idea

P = V_{r m s} I_{r m s} cos ϕ = I_{r m s}^{2} R

;

cos ϕ = \frac{R}{Z}

1

for pure

R

and at resonance,

0

for pure

L

C

6. Resonance in a Series LCR Circuit

A series LCR circuit is at resonance when the current through it is maximum. The current amplitude is

i_{0} = \frac{V _{0}}{R ^{2} + ( ω L - \frac{1}{ω C} ) ^{2}}

It tends to zero both as $ω \to 0$ (capacitor blocks) and as $ω \to \infty$ (inductor blocks), and is largest when the bracket is zero:

★ Must learn

ω_{0} L = \frac{1}{ω _{0} C} \Rightarrow ω_{0} = \frac{1}{L C}, f_{0} = \frac{1}{2 π L C}

At resonance $X_{L} = X_{C}$ , $Z_{min} = R$ , $i_{0, ma x} = \frac{V _{0}}{R}$ , $ϕ = 0$ and the power factor $cos ϕ = \frac{R}{R} = 1$ . The circuit behaves as a pure resistor, and the heat produced is maximum. The resonant frequency does not depend on $R$ .

Figure 9:

L = 0.1 H

C = 10 μ F

V_{0} = 10 V

: resonance at

ω_{0} = \frac{1}{L C} = 1000 rad s^{- 1}

for any

R

. Smaller

R

gives a taller, sharper peak. For

R = 10 Ω

the current falls to

\frac{i _{ma x}}{2}

ω_{1} = 951

and

ω_{2} = 1051 rad s^{- 1}

: bandwidth

\frac{R}{L} = 100 rad s^{- 1}

Q = \frac{ω _{0}}{Δ ω} = 10

Figure 10: Same circuit (

R = 10 Ω

Z = R^{2} + (X_{L} - X_{C})^{2}

is smallest, and equal to

R

, where the

X_{L}

and

X_{C}

curves cross. Below

ω_{0}

the capacitor dominates, above it the inductor.

Voltage magnification. At resonance $V_{L} = i_{0} X_{L}$ and $V_{C} = i_{0} X_{C}$ are equal and opposite, and each can be many times the source voltage. That is why a series resonant circuit is used to pick out and magnify one frequency, for example in the tuning circuit of a radio.

6.1 Sharpness of resonance: the Q-factor

The Q-factor of a series resonant circuit is the ratio of the resonant frequency to the difference of the two frequencies on either side of resonance at which the current amplitude falls to $\frac{1}{2}$ of its resonant value (the half-power points $ω_{1}$ and $ω_{2}$ ):

Q = \frac{ω _{0}}{ω _{2} - ω _{1}} = \frac{resonant frequency}{bandwidth} = \frac{X _{L}}{R} = \frac{ω _{0} L}{R} = \frac{1}{R} \frac{L}{C}

The bandwidth is $Δ ω = ω_{2} - ω_{1} = \frac{R}{L}$ . A small $R$ gives a large $Q$ and a sharp, tall peak (good selectivity); a large $R$ flattens the curve. $Q$ is also the voltage magnification: $V_{L} = V_{C} = Q V$ at resonance.

JEE Advanced

Exact half-power frequencies. Setting $ω L - \frac{1}{ω C} = R$ gives $ω_{1, 2} = \frac{R ^{2}}{4 L ^{2}} + \frac{1}{L C} \mp \frac{R}{2 L}$ . So $ω_{2} - ω_{1} = \frac{R}{L}$ exactly and $ω_{1} ω_{2} = ω_{0}^{2}$ : the resonant frequency is the geometric mean of the half-power frequencies. Maximum voltage across $C$ or $L$ occurs slightly off resonance (below $ω_{0}$ for $C$ , above for $L$ ) when $R$ is not small. Parallel LC (tank) circuits behave oppositely: at $ω_{0}$ their impedance is maximum and the line current minimum.

Quick Recall: tap to check

Does the resonant frequency of a series LCR circuit depend on R?

No.

ω_{0} = \frac{1}{L C}

;

R

only sets the height and sharpness of the peak.

What is the power factor at resonance?

1

, since

Z = R

and

ϕ = 0

How does the Q-factor change if R is doubled?

It halves,

Q = \frac{ω _{0} L}{R}

; the bandwidth doubles.

7. Choke Coil

A choke coil is an inductor with a large inductance and a small resistance, used to reduce the current in an AC circuit without much loss of energy.

Principle: in a coil whose resistance $r$ is tiny compared with $ω L$ , the current lags the voltage across the coil by nearly $\frac{π}{2}$ . The choke's own power factor $\frac{r}{r ^{2} + ω ^{2} L ^{2}}$ is nearly zero, so the power it uses is nearly zero.
Construction: many turns of insulated copper wire on a soft-iron core. The core is laminated to reduce eddy-current losses. Low-frequency chokes use an iron core; high-frequency chokes use an air core.
Working: the choke is put in series with an appliance of resistance $R$ that needs less than the supply voltage. The current is $I_{r m s} = \frac{V _{r m s}}{( R + r ) ^{2} + ω ^{2} L ^{2}}$ , set mainly by $ω L$ . The voltage across the choke is almost at $9 0^{\circ}$ to the current, so the choke itself uses only $I_{r m s}^{2} r \approx 0$ . The power factor of the whole circuit is $\frac{R + r}{Z}$ , which is not zero: the appliance still takes its full power.
Use: a resistance used to limit current wastes energy as heat; a choke limits the current by its reactance and wastes almost nothing. Fluorescent tube lights have long used chokes this way: on $220 V$ mains a tube that needs $100 V$ gets it with the choke taking about $196 V$ in quadrature, while a series resistor would have to drop $120 V$ and burn $1.2$ times the lamp's own power.

Figure 11: The tube needs

V_{R} = 100 V

from the

220 V

mains. The choke takes the rest as

V_{L} = 22 0^{2} - 10 0^{2} \approx 196 V

9 0^{\circ}

i

, so it uses almost no power (its own resistance

r ≪ ω L

); the whole circuit has

ϕ \approx 6 3^{\circ}

. A series resistor would have to drop

120 V

in phase with

i

and would waste

1.2

times the lamp's power as heat.

8. Oscillations in an LC Circuit

If a charged capacitor $C$ is connected across an inductor $L$ , the charge and current oscillate simple harmonically. With zero resistance and no radiation, no energy is lost: it passes back and forth between the electric field of the capacitor and the magnetic field of the inductor, and the total stays constant. This is the electrical twin of a spring-mass system.

The capacitor is charged to $q_{0} = C V$ . At $t = 0$ the switch is closed and it starts discharging; at time $t$ its charge is $q$ and the current is $i = - \frac{d q}{d t}$ (the charge is decreasing), so $\frac{d i}{d t} = - \frac{d ^{2} q}{d t ^{2}}$ .
The potential difference across the capacitor equals that across the inductor: $\frac{q}{C} = L \frac{d i}{d t} = - L \frac{d ^{2} q}{d t ^{2}}$ .
So
$\frac{d ^{2} q}{d t ^{2}} = - \frac{1}{L C} q = - ω^{2} q, ω = \frac{1}{L C}$
the equation of SHM, $\frac{d ^{2} x}{d t ^{2}} = - ω^{2} x$ .
Solution with $q = q_{0}$ at $t = 0$ : $q = q_{0} cos ω t$ ; then $i = - \frac{d q}{d t} = q_{0} ω sin ω t$ and $\frac{d i}{d t} = q_{0} ω^{2} cos ω t$ .

$q$ , $i$ and $\frac{d i}{d t}$ all oscillate with the same $ω$ , with amplitudes $q_{0}$ , $q_{0} ω$ and $q_{0} ω^{2}$ ; the phase difference between $q$ and $i$ , and between $i$ and $\frac{d i}{d t}$ , is $\frac{π}{2}$ .

Figure 12: In an ideal LC circuit the energy sloshes between the electric field of

C

(

U_{E} = q^{2} /2 C

, red) and the magnetic field of

L

(

U_{B} = L i^{2} /2

, blue) every quarter period; the total stays

q_{0}^{2} /2 C

Energy. Energy in the capacitor $U_{E} = \frac{q ^{2}}{2 C} = \frac{q _{0}^{2}}{2 C} cos^{2} ω t$ ; energy in the inductor $U_{B} = \frac{1}{2} L i^{2} = \frac{q _{0}^{2}}{2 C} sin^{2} ω t$ (using $L ω^{2} = \frac{1}{C}$ ). Each swings between zero and $\frac{q _{0}^{2}}{2 C}$ at twice the frequency of $q$ , and $U_{E} + U_{B} = \frac{q _{0}^{2}}{2 C}$ at every instant.

Figure 13:

q = q_{0} cos ω t

and

i = q_{0} ω sin ω t

are

\frac{π}{2}

out of phase. The energies

U_{E} \propto cos^{2} ω t

and

U_{B} \propto sin^{2} ω t

oscillate at

2 ω

(period

\frac{T}{2}

); their sum is constant.

Spring-mass system	LC circuit
Displacement $x$	Charge $q$
Velocity $v$	Current $i$
Mass $m$	Inductance $L$
Force constant $k$	$\frac{1}{C}$
$ω = \frac{k}{m}$	$ω = \frac{1}{L C}$
Kinetic energy $\frac{1}{2} m v^{2}$	Magnetic energy $\frac{1}{2} L i^{2}$
Potential energy $\frac{1}{2} k x^{2}$	Electric energy $\frac{q ^{2}}{2 C}$

JEE Advanced

RC and LR circuits with a DC source (the other half of the syllabus line). Charging a capacitor through $R$ from a cell of emf $E$ : $q = C E (1 - e^{- t / R C})$ , time constant $τ = R C$ ; discharging: $q = q_{0} e^{- t / R C}$ . Growth of current in an LR circuit: $i = \frac{E}{R} (1 - e^{- R t / L})$ , $τ = \frac{L}{R}$ ; decay: $i = i_{0} e^{- R t / L}$ . At $t = 0$ a capacitor acts as a short circuit and an inductor as an open circuit; at $t \to \infty$ the reverse. An LC circuit with small resistance gives damped oscillations, $q = q_{0} e^{- R t /2 L} cos ω^{'} t$ .

Key idea

LC oscillation is SHM of charge:

ω = \frac{1}{L C}

U_{E} ⇌ U_{B}

every quarter period, total

\frac{q _{0}^{2}}{2 C}

conserved.

Figure 14: Revision map for series AC circuits, power, resonance and LC oscillations.

9. Solved Examples

Solved Example 1

When

100 V

DC is applied across a coil, a current of

1 A

flows. When

100 V

AC of

50 Hz

is applied to the same coil, only

0.5 A

flows. Find the resistance and inductance of the coil.

Solution:

A coil is an LR circuit: $i = \frac{V}{Z}$ with $Z = R^{2} + ω^{2} L^{2}$ .

With DC, $ω = 0$ and $Z = R$ : $R = \frac{100}{1} = 100 Ω$ .

With AC, $Z = \frac{100}{0.5} = 200 Ω$ , so $ω^{2} L^{2} = Z^{2} - R^{2} = 20 0^{2} - 10 0^{2} = 3 \times 1 0^{4}$ and $ω L = 173.2 Ω$ .

$L = \frac{173.2}{2 π \times 50} = \frac{3}{π}$ .

Answer: $R = 100 Ω$ , $L = \frac{3}{π} H \approx 0.55 H$ .

Solved Example 2

12 Ω

resistance and an inductance of

\frac{0.05}{π} H

with negligible resistance are connected in series across a

130 V

50 Hz

supply. Find the current and the potential differences across the resistance and the inductance.

Solution:

$X_{L} = 2 π f L = 2 π (50) (\frac{0.05}{π}) = 5 Ω$ ; $Z = 1 2^{2} + 5^{2} = 13 Ω$ .

$i = \frac{V}{Z} = \frac{130}{13} = 10 A$ ; $V_{R} = i R = 120 V$ ; $V_{L} = i X_{L} = 50 V$ .

Answer: $10 A$ ; $120 V$ and $50 V$ . Check: $12 0^{2} + 5 0^{2} = 130 V$ , while $120 + 50 = 170 V$ would be wrong.

Solved Example 3

An AC source of angular frequency

ω

is connected across a resistor

R

and a capacitor

C

in series, and the current is

i

. When the frequency is changed to

\frac{ω}{3}

(same voltage), the current is halved. Find the ratio of reactance to resistance at the original frequency.

Solution:

At $ω$ : $i = \frac{V}{R ^{2} + X ^{2}}$ with $X = \frac{1}{ω C}$ . At $\frac{ω}{3}$ the reactance becomes $3 X$ : $\frac{i}{2} = \frac{V}{R ^{2} + 9 X ^{2}}$ .

Dividing: $2 = \frac{R ^{2} + 9 X ^{2}}{R ^{2} + X ^{2}}$ , so $4 R^{2} + 4 X^{2} = R^{2} + 9 X^{2}$ , i.e. $3 R^{2} = 5 X^{2}$ .

Answer: $\frac{X}{R} = \frac{3}{5} \approx 0.77$ .

Solved Example 4

50 W

100 V

lamp is to be run on

200 V

50 Hz

AC mains. What capacitance must be put in series with it?

Solution:

Lamp resistance $R = \frac{V ^{2}}{P} = \frac{10 0 ^{2}}{50} = 200 Ω$ ; rated current $i = \frac{100}{200} = 0.5 A$ .

On $200 V$ the impedance must be $Z = \frac{200}{0.5} = 400 Ω$ , so $X_{C}^{2} = Z^{2} - R^{2} = 40 0^{2} - 20 0^{2} = 1.2 \times 1 0^{5}$ and $X_{C} = 346.4 Ω$ .

$C = \frac{1}{2 π f X _{C}} = \frac{1}{2 π ( 50 ) ( 346.4 )}$ .

Answer: $C \approx 9.2 μ F$ .

Solved Example 5

A series LCR circuit is connected to

V = 10 sin (ω t + \frac{π}{4})

volt, with

X_{L} = 10 Ω

X_{C} = 6 Ω

and

R = 3 Ω

. Find

Z

i_{0}

I_{r m s}

V_{r m s}

, the peak and rms voltages across

L

C

and

R

, the phase angle

ϕ

, and

i (t)

V_{L} (t)

V_{C} (t)

V_{R} (t)

Solution:

$V_{0} = 10 V$ , $V_{r m s} = \frac{10}{2} = 7.07 V$ . $Z = 3^{2} + (10 - 6)^{2} = 5 Ω$ .

$i_{0} = \frac{V _{0}}{Z} = 2 A$ , $I_{r m s} = 2 A = 1.41 A$ .

Peak: $V_{0 L} = i_{0} X_{L} = 20 V$ , $V_{0 C} = i_{0} X_{C} = 12 V$ , $V_{0 R} = i_{0} R = 6 V$ . RMS: $\frac{20}{2} = 14.1 V$ , $\frac{12}{2} = 8.49 V$ , $\frac{6}{2} = 4.24 V$ .

$cos ϕ = \frac{R}{Z} = \frac{3}{5}$ , so $ϕ = 5 3^{\circ}$ ; $X_{L} > X_{C}$ , so the current lags the voltage by $5 3^{\circ}$ . With $\frac{π}{4} = 4 5^{\circ}$ :

$i (t) = 2 sin (ω t + 4 5^{\circ} - 5 3^{\circ}) = 2 sin (ω t - 8^{\circ}) A$

$V_{L} (t) = 20 sin (ω t - 8^{\circ} + 9 0^{\circ}) = 20 sin (ω t + 8 2^{\circ}) V$

$V_{C} (t) = 12 sin (ω t - 8^{\circ} - 9 0^{\circ}) = 12 sin (ω t - 9 8^{\circ}) V$

$V_{R} (t) = 6 sin (ω t - 8^{\circ}) V$

Answer: $Z = 5 Ω$ , $i_{0} = 2 A$ , $ϕ = 5 3^{\circ}$ (current lagging). Note $V_{0 L} = 20 V$ is twice the source peak.

Solved Example 6

A resistor of

16 Ω

, an inductor and a capacitor are in series with an AC supply. At the supply frequency

X_{L} = 24 Ω

and

X_{C} = 12 Ω

. If the current is

5 A

, find (a) the potential differences across

R

L

and

C

, (b) the impedance, (c) the supply voltage, (d) the phase angle.

Solution:

(a) $V_{R} = 5 \times 16 = 80 V$ , $V_{L} = 5 \times 24 = 120 V$ , $V_{C} = 5 \times 12 = 60 V$ .

(b) $Z = 1 6^{2} + (24 - 12)^{2} = 20 Ω$ .

(d) $ϕ = tan^{- 1} \frac{24 - 12}{16} = tan^{- 1} (0.75)$ .

Answer: $ϕ = 36.8 7^{\circ}$ (about $3 7^{\circ}$ ), voltage leading.

Solved Example 7

A series circuit has

R = 15 Ω

L = 0.08 H

and

C = 30 μ F

. The applied voltage has angular frequency

500 rad s^{- 1}

. Does the current lead or lag the voltage, and by what angle?

Solution:

$X_{L} = ω L = 500 \times 0.08 = 40 Ω$ ; $X_{C} = \frac{1}{500 \times 30 \times 1 0 ^{- 6}} = 66.7 Ω$ .

$tan ϕ = \frac{X _{L} - X _{C}}{R} = \frac{40 - 66.7}{15} = - 1.78$ , so $ϕ = - 60. 6^{\circ}$ .

Answer: $X_{C} > X_{L}$ , so the circuit is capacitive and the current leads the voltage by about $60. 6^{\circ}$ .

Solved Example 8

A current of

4 A

flows in a coil connected to a

12 V

DC source. On a

12 V

50 rad s^{- 1}

AC source the current is

2.4 A

. Find the inductance of the coil, and the power developed if a

2500 μ F

capacitor is connected in series with the coil on the same AC source.

Solution:

DC: $R = \frac{12}{4} = 3 Ω$ . AC: $Z = \frac{12}{2.4} = 5 Ω$ , so $(50 L)^{2} = 5^{2} - 3^{2} = 16$ , $50 L = 4 Ω$ and $L = 0.08 H$ .

With the capacitor: $X_{C} = \frac{1}{50 \times 2500 \times 1 0 ^{- 6}} = 8 Ω$ , $Z = 3^{2} + (4 - 8)^{2} = 5 Ω$ , $I_{r m s} = 2.4 A$ , $cos ϕ = \frac{3}{5} = 0.6$ .

$P = V_{r m s} I_{r m s} cos ϕ = 12 \times 2.4 \times 0.6$ (check: $I_{r m s}^{2} R = 2. 4^{2} \times 3$ ).

Answer: $L = 0.08 H$ ; $P = 17.28 W$ .

Solved Example 9

20 V

5 W

lamp is to be run on

220 V

50 Hz

AC mains. Find (i) the capacitance, (ii) the inductance to be put in series with it, (iii) the pure resistance that would do the same job, and (iv) which arrangement is most economical, and why.

Solution:

Lamp current $i = \frac{5}{20} = 0.25 A$ ; resistance $R = \frac{20}{0.25} = 80 Ω$ . The total impedance must be $Z = \frac{220}{0.25} = 880 Ω$ .

(i) $X_{C} = 88 0^{2} - 8 0^{2} = 876.4 Ω$ , so $C = \frac{1}{2 π ( 50 ) ( 876.4 )} = 3.63 \times 1 0^{- 6} F$ .

(ii) $X_{L} = 876.4 Ω$ , so $L = \frac{876.4}{2 π ( 50 )} = 2.79 H$ .

(iii) $\frac{220}{80 + r} = 0.25$ , so $80 + r = 880$ and $r = 800 Ω$ .

(iv) The capacitor or inductor: they consume no average power, whereas the $800 Ω$ resistor would waste $I^{2} r = (0.25)^{2} (800) = 50 W$ , ten times the lamp's own power.

Answer: $C \approx 3.6 μ F$ , $L \approx 2.8 H$ , $r = 800 Ω$ ; $L$ or $C$ is economical.

Solved Example 10

In a series LCR circuit

X_{L} = X_{C} = 20 Ω

R = 2 Ω

and

V = 10 sin ω t

volt. Find

Z

i (t)

and the peak voltages across

C

and

L

Solution:

$X_{L} = X_{C}$ , so the circuit is at resonance: $Z = Z_{min} = R = 2 Ω$ .

$i_{0} = \frac{10}{2} = 5 A$ and $i (t) = 5 sin ω t A$ (in phase with $V$ ).

$V_{0 C} = i_{0} X_{C} = 100 V$ ; $V_{0 L} = i_{0} X_{L} = 100 V$ .

Answer: $Z = 2 Ω$ , $i = 5 sin ω t$ , $V_{0 C} = V_{0 L} = 100 V$ . The circuit magnifies the $10 V$ source to $100 V$ across $L$ and $C$ ( $Q = 10$ ), which is why it can be used as a voltage amplifier.

Solved Example 11

A series LCR circuit with

R = 20 Ω

L = 1.5 H

and

C = 35 μ F

is connected to a variable-frequency

200 V

AC supply. When the supply frequency equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Solution:

At the natural frequency ( $ω_{0} = \frac{1}{1.5 \times 35 \times 1 0 ^{- 6}} = 138 rad s^{- 1}$ ) the circuit resonates: $Z = R = 20 Ω$ .

$I_{r m s} = \frac{200}{20} = 10 A$ ; $P = V_{r m s} I_{r m s} cos 0^{\circ} = 200 \times 10 \times 1$ .

Answer: $2000 W$ .

Solved Example 12

A capacitor of

25 μ F

is charged to

300 V

and then connected across a

10 mH

inductor; the resistance is negligible. (a) Find the frequency of oscillation. (b) Find the potential difference across the capacitor and the current

1.2 ms

after connection. (c) Find the magnetic and electric energy at

t = 0

and

t = 1.2 ms

Solution:

(a) $ω = \frac{1}{L C} = \frac{1}{( 1 0 ^{- 2} ) ( 25 \times 1 0 ^{- 6} )} = 2000 rad s^{- 1}$ , so $f = \frac{ω}{2 π} = \frac{1 0 ^{3}}{π} = 318.3 Hz$ .

(b) $q_{0} = C V_{0} = (25 \times 1 0^{- 6}) (300) = 7.5 \times 1 0^{- 3} C$ . At $t = 1.2 \times 1 0^{- 3} s$ , $ω t = 2.4 rad$ :

$q = q_{0} cos ω t = (7.5 \times 1 0^{- 3}) cos 2.4 = - 5.53 \times 1 0^{- 3} C$ (the plates have reversed polarity), so $∣ V ∣ = \frac{5.53 \times 1 0 ^{- 3}}{25 \times 1 0 ^{- 6}} = 221 V$ .

$∣ i ∣ = q_{0} ω sin ω t = (7.5 \times 1 0^{- 3}) (2000) sin 2.4 = 10.13 A$ .

(c) At $t = 0$ : $i = 0$ , so $U_{B} = 0$ and $U_{E} = \frac{q _{0}^{2}}{2 C} = \frac{( 7.5 \times 1 0 ^{- 3} ) ^{2}}{2 ( 25 \times 1 0 ^{- 6} )} = 1.125 J$ . At $t = 1.2 ms$ : $U_{B} = \frac{1}{2} L i^{2} = \frac{1}{2} (1 0^{- 2}) (10.13)^{2} = 0.513 J$ and $U_{E} = 1.125 - 0.513 = 0.612 J$ (check: $\frac{q ^{2}}{2 C} = 0.612 J$ ).

Answer: (a) $318 Hz$ ; (b) $221 V$ , $10.1 A$ ; (c) $U_{B} = 0$ , $U_{E} = 1.125 J$ at $t = 0$ ; $U_{B} = 0.513 J$ , $U_{E} = 0.612 J$ at $1.2 ms$ .

Solved Example 13

A series LCR circuit with

L = 5.0 H

C = 80 μ F

and

R = 40 Ω

is connected to a

230 V

variable-frequency supply. Find the resonant angular frequency, the rms current at resonance and the rms voltage across the inductor at resonance.

Solution:

$ω_{0} = \frac{1}{5 \times 80 \times 1 0 ^{- 6}} = \frac{1}{0.02} = 50 rad s^{- 1}$ .

$I_{r m s} = \frac{230}{40} = 5.75 A$ . $X_{L} = ω_{0} L = 250 Ω$ , so $V_{L} = I_{r m s} X_{L} = 5.75 \times 250$ .

Answer: $50 rad s^{- 1}$ ; $5.75 A$ ; $V_{L} = V_{C} = 1437.5 V$ , over six times the supply voltage ( $Q = 6.25$ ), while $V_{R} = 230 V$ .

Solved Example 14

A series LCR circuit has

R = 10 Ω

L = 0.1 H

and

C = 10 μ F

. Find its resonant angular frequency, Q-factor and bandwidth.

Solution:

$ω_{0} = \frac{1}{0.1 \times 1 0 ^{- 5}} = 1000 rad s^{- 1}$ ; $Q = \frac{ω _{0} L}{R} = \frac{1000 \times 0.1}{10} = 10$ ; $Δ ω = \frac{R}{L} = 100 rad s^{- 1}$ .

Answer: $1000 rad s^{- 1}$ , $Q = 10$ , bandwidth $100 rad s^{- 1}$ (half-power points $951$ and $1051 rad s^{- 1}$ , the same circuit as the resonance-curve figure).

Solved Example 15

A coil has

R = 30 Ω

and

X_{L} = 40 Ω

at the supply frequency. Its power factor is
(A)

0.8

(B)

0.6

(C)

0.75

(D)

1

Solution:

Answer: (B). $Z = 3 0^{2} + 4 0^{2} = 50 Ω$ and $cos ϕ = \frac{R}{Z} = \frac{30}{50} = 0.6$ . Option (A) is $sin ϕ$ and (C) is $\frac{R}{X _{L}}$ .

Solved Example 16

A circuit draws

5 A

(rms) from

220 V

mains with the current lagging by

6 0^{\circ}

. Find the true power, the apparent power and the wattless component of the current.

Solution:

Apparent power $= V_{r m s} I_{r m s} = 220 \times 5 = 1100 VA$ . True power $= 1100 \times cos 6 0^{\circ} = 550 W$ .

Wattless current $= I_{r m s} sin ϕ = 5 \times sin 6 0^{\circ} = 4.33 A$ ; power component $= 5 cos 6 0^{\circ} = 2.5 A$ .

Answer: $550 W$ , $1100 VA$ , $4.33 A$ .

Solved Example 17

In an ideal LC circuit the capacitor is fully charged at

t = 0

. The electric and magnetic energies first become equal at
(A)

T /4

(B)

T /8

(C)

T /2

(D)

T /6

Solution:

Answer: (B). $U_{E} \propto cos^{2} ω t$ and $U_{B} \propto sin^{2} ω t$ are equal when $ω t = \frac{π}{4}$ , i.e. $t = \frac{π /4}{2 π / T} = \frac{T}{8}$ . Then $q = \frac{q _{0}}{2}$ .

Practice Questions

Can the peak voltage across the inductor be greater than the peak voltage of the source in a series LCR circuit?Answer: Yes; $V_{L}$ and $V_{C}$ partly cancel, and at resonance $V_{L} = V_{C} = Q V$
A $200 V$ (rms) source is connected to $R = 30 Ω$ and $X_{L} = 40 Ω$ in series. Find the current and the phase angle.Answer: $4 A$ ; $5 3^{\circ}$ , voltage leading
Find the resonant frequency of a series circuit with $L = 1 mH$ and $C = 1 μ F$ .Answer: $f_{0} \approx 5.03 kHz$
What is the power factor of a series LCR circuit at resonance?Answer: $1$
An LC circuit has $L = 20 mH$ , $C = 50 μ F$ and an initial charge of $10 mC$ . Find the angular frequency and the total energy.Answer: $1000 rad s^{- 1}$ ; $1 J$
In a series RC circuit, what happens to the rms current if the frequency is increased at constant voltage?Answer: It increases, because $X_{C}$ and hence $Z$ decrease
Why is a choke coil preferred to a resistor for limiting current in an AC circuit?Answer: The choke's own power factor is nearly zero ( $r ≪ ω L$ ), so it limits the current while wasting almost no energy as heat

Common Mistakes to Avoid

Watch out

Adding the voltages across $R$ , $L$ and $C$ as numbers. They add as phasors: $V = V_{R}^{2} + (V_{L} - V_{C})^{2}$ .
Writing $tan ϕ = \frac{R}{X _{L}}$ for an LR circuit. It is $\frac{X _{L}}{R}$ : reactance over resistance.
Using $Z = X_{L} - X_{C}$ for an LC circuit and getting a negative impedance. Impedance is $∣ X_{L} - X_{C} ∣$ .
Using $P = V_{r m s} I_{r m s}$ without the power factor. Average power is $V_{r m s} I_{r m s} cos ϕ$ , or simply $I_{r m s}^{2} R$ .
Thinking the resonant frequency depends on $R$ . $ω_{0} = \frac{1}{L C}$ ; $R$ changes only the height and width of the peak.
Getting the sign of $ϕ$ wrong. Inductive ( $X_{L} > X_{C}$ ): voltage leads; capacitive ( $X_{C} > X_{L}$ ): current leads.
Using $f$ in $ω_{0} = \frac{1}{L C}$ . That gives angular frequency in $rad s^{- 1}$ ; divide by $2 π$ for hertz.
Assuming the energies in an LC circuit oscillate at the same frequency as $q$ . $U_{E}$ and $U_{B}$ oscillate at $2 ω$ .

Frequently Asked Questions

What is the impedance of a series LCR circuit?

It is the total opposition to alternating current, $Z = R^{2} + (X_{L} - X_{C})^{2}$ , where $X_{L} = ω L$ and $X_{C} = \frac{1}{ω C}$ . The reactances subtract because the voltages across the inductor and capacitor are opposite in phase. Impedance is measured in ohm and $I = \frac{V}{Z}$ .

What is the difference between an LR and a CR circuit?

In an LR circuit the voltage leads the current, $tan ϕ = \frac{X _{L}}{R}$ , and the impedance rises with frequency. In a CR circuit the voltage lags the current, $tan ϕ = \frac{X _{C}}{R}$ , and the impedance falls with frequency. A CR circuit passes no steady DC; an LR circuit does.

What is resonance in a series LCR circuit?

Resonance occurs when the inductive and capacitive reactances are equal, at $ω_{0} = \frac{1}{L C}$ . The impedance then drops to its minimum value R, the current is maximum, the current and voltage are in phase and the power factor is 1.

What is the power factor and why is it important?

Power factor is $cos ϕ = \frac{R}{Z}$ , the fraction of the apparent power $V_{r m s} I_{r m s}$ that is actually consumed. A low power factor means a larger current is needed for the same useful power, which raises heating losses in the supply lines.

What is wattless current?

It is the component of the current, $I_{r m s} sin ϕ$ , that is perpendicular to the voltage in the phasor diagram. It flows back and forth but consumes no average power. In a pure inductor or capacitor the whole current is wattless, since the phase difference is 90 degrees.

What does the Q-factor of a resonant circuit mean?

The quality factor measures how sharp the resonance is: $Q = \frac{ω _{0}}{Δ ω} = \frac{ω _{0} L}{R}$ , where $Δ ω$ is the bandwidth between the half-power frequencies. A high Q means a narrow, tall peak and good selectivity, useful in radio tuning circuits.

Which LCR circuit questions come in NEET?

NEET usually asks for the impedance and current of a series LCR circuit, the resonant frequency $\frac{1}{2 π L C}$ , the power factor $\frac{R}{Z}$ , average power at resonance, and the frequency of LC oscillations. Questions are short numericals built on Pythagorean triples.

What LCR and LC topics are important for JEE Main and Advanced?

JEE tests phasor diagrams with voltage readings across each element, the Q-factor and bandwidth, voltage magnification at resonance, energy exchange and phase in LC oscillations, the choke coil, and in JEE Advanced, transients in RC and LR circuits with DC sources.

Previous year questions on LR, CR, LCR Circuits

23 questions from past papers, each with a step-by-step solution.

Show all 23 questions

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