Alternating Current • CBSE • NEET • IIT JEE

Alternating Current: Series RL and RC Circuits

Complete coaching-style notes on impedance, reactance, phase angle, phasor diagrams, average power, frequency effects and exam numerical problems for series RL and series RC circuits.

Series RL Series RC Definitions Comparison Solved Questions Practice Revision

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1. Series RL Circuit

Correct Series RL Circuit Diagram

R Series RL: R and L in one closed loop AC Source VR = IR VL = IXL L Supply voltage: V = V0 sin ωt

This is a true series circuit: current passes through the AC source, resistor and inductor one after another. There is no bypass branch across the resistor or inductor.

RL Phasor Diagram

I, VR VL V φ Current I is reference along +X. Supply voltage V leads current by φ.
V2 = VR2 + VL2 = (IR)2 + (IXL)2 Z = √(R2 + XL2), I = V/Z, XL = ωL = 2πfL

RL Derivation

Take current as the reference phasor. In a resistor, voltage and current are in phase, so VR = IR. In an inductor, voltage leads current by 90°, so VL = IXL.

Since VR and VL are perpendicular:

V2 = VR2 + VL2 V2 = I2R2 + I2XL2 = I2(R2 + XL2) V = I√(R2 + XL2) Z = √(R2 + XL2) I = V/Z

Phase, Power Factor and Average Power

I = I0 sin(ωt − φ) tan φ = XL/R, cos φ = R/Z

Instantaneous power is p = vi. For sinusoidal AC:

v = V0 sin ωt, i = I0 sin(ωt − φ) p = V0I0 sinωt sin(ωt − φ) Pavg = (V0I0/2) cosφ = VrmsIrms cosφ Pavg = Irms2R

Frequency Graphs for RL Circuit

XL = 2πfL fXL Z increases with f fZ

DC and Iron Core Behaviour

For steady DC, f = 0, so XL = 2πfL = 0. After steady state, the inductor behaves like a short circuit and current is limited mainly by resistance.

When an iron rod/core is inserted in the inductor, inductance L increases. Therefore XL, impedance Z and phase lag increase, while current decreases.

Iron core inserted: L ↑ ⇒ XL ↑ ⇒ Z ↑ ⇒ I ↓

2. Series RC Circuit

Correct Series RC Circuit Diagram

R Series RC: R and C in one closed loop AC Source VR = IR VC = IXC C Supply voltage: V = V0 sin ωt

This is a true series RC circuit: the capacitor is not in a parallel branch. The same current flows through R and C.

RC Phasor Diagram

I, VR VC V φ Current I is reference along +X. Supply voltage V lags current by φ.
V2 = VR2 + VC2 = (IR)2 + (IXC)2 Z = √(R2 + XC2), I = V/Z, XC = 1/ωC = 1/(2πfC)

RC Derivation

Take current as the reference phasor. Across the resistor, VR is in phase with current. Across the capacitor, VC lags current by 90° and is drawn downward.

V2 = VR2 + VC2 V2 = I2R2 + I2XC2 = I2(R2 + XC2) V = I√(R2 + XC2) Z = √(R2 + XC2) I = V/Z

Phase, Power Factor and Average Power

I = I0 sin(ωt + φ) tan φ = XC/R, cos φ = R/Z

For v = V0 sinωt and i = I0 sin(ωt + φ), averaging over one cycle gives:

Pavg = VrmsIrms cosφ Pavg = Irms2R

Only the resistor consumes real power. The capacitor stores and returns energy during each cycle.

Frequency Graphs for RC Circuit

XC decreases with f fXC Z decreases with f fZ

DC and Dielectric Behaviour

For steady DC, f = 0, so XC = 1/(2πfC) becomes infinite. The capacitor behaves like an open circuit; no steady current flows.

When a dielectric is inserted, capacitance becomes KC. Therefore XC decreases, impedance decreases and current increases.

Dielectric inserted: C ↑ ⇒ XC ↓ ⇒ Z ↓ ⇒ I ↑

3. Student-Friendly Definitions

Impedance Z

Total opposition offered by an AC circuit to current. It combines resistance and reactance.

Z = Vrms/Irms

Inductive Reactance XL

Opposition offered by an inductor to AC. It increases with frequency.

XL = 2πfL

Capacitive Reactance XC

Opposition offered by a capacitor to AC. It decreases with frequency.

XC = 1/(2πfC)

Power Factor

Ratio of true power to apparent power. For series RL and RC circuits, cosφ = R/Z.

Phase Difference

The angular separation between voltage and current phasors. In RL current lags; in RC current leads.

RMS Values

Effective AC values that produce the same heating effect as DC in a resistor.

Vrms = V0/√2, Irms = I0/√2

4. Series RL and Series RC Comparison Table

FeatureSeries RL CircuitSeries RC Circuit
ComponentsResistor and inductor in seriesResistor and capacitor in series
ReactanceXL = 2πfLXC = 1/(2πfC)
ImpedanceZ = √(R² + XL²)Z = √(R² + XC²)
Current phaseCurrent lags voltageCurrent leads voltage
Voltage phasorVL upward, V resultant above current axisVC downward, V resultant below current axis
Power factorcosφ = R/Zcosφ = R/Z
Average powerP = VrmsIrmscosφ = I²RP = VrmsIrmscosφ = I²R
Frequency effectf ↑ ⇒ XL ↑ ⇒ current ↓f ↑ ⇒ XC ↓ ⇒ current ↑
DC behaviourInductor becomes short after steady stateCapacitor becomes open circuit
Core/dielectric effectIron core increases L, so current decreasesDielectric increases C, so current increases
Exam pointChoke coil and lagging current questions are commonCapacitor current, lamp in series and leading phase questions are common

5. Extracted Screenshot Questions with Solutions

The following problems are rewritten from the attached screenshots and solved step by step. Values are treated as RMS values unless peak value is clearly stated.

A. Basic Coil Questions

1. Calculate the impedance of a coil of resistance 3 Ω and reactance 4 Ω.
Formula: Z = √(R² + X²) = √(3² + 4²) = 5 Ω. Answer: 5 Ω
Common mistake: Adding 3 + 4 = 7 Ω. Resistance and reactance are perpendicular phasor quantities.
2. An inductance coil has resistance 100 Ω. When an AC signal of frequency 1000 Hz is applied, the applied voltage leads the current by 45°. Find self-inductance.
For RL circuit, tanφ = XL/R. Since φ = 45°, XL = R = 100 Ω. L = XL/(2πf) = 100/(2π × 1000) = 0.0159 H. Answer: 0.016 H
Common mistake: Using f instead of 2πf in XL.

B. Series RL Circuit Questions

3. A 100 V rms, 50 Hz AC source is connected across a 20 Ω resistor and 2 mH inductor in series. Find impedance and rms current.
XL = 2πfL = 2π × 50 × 0.002 = 0.628 Ω. Z = √(20² + 0.628²) = 20.01 Ω ≈ 20 Ω. I = V/Z = 100/20.01 ≈ 5 A. Answer: 20 Ω, 5 A
4. A 100 V, 50 Hz AC source is connected to 100 mH inductance and 20 Ω resistance in series. Find current and phase.
XL = 2π × 50 × 0.1 = 31.4 Ω. Z = √(20² + 31.4²) = 37.23 Ω. I = 100/37.23 = 2.68 A. tanφ = 31.4/20, so φ = 57.5°. Answer: 2.68 A; current lags by 57.5°
5. A coil takes 11 A from 110 V DC. With 110 V, 50 Hz AC, it takes only 0.5 A. Find resistance, impedance and inductance.
DC: R = 110/11 = 10 Ω. AC: Z = 110/0.5 = 220 Ω. XL = √(220² − 10²) = 219.77 Ω. L = XL/(2πf) = 219.77/(314) = 0.70 H. Answer: 10 Ω, 220 Ω, 0.7 H
6. An arc takes 10 A at 80 V. What inductance should be put in series to work it from 220 V, 50 Hz?
Arc resistance R = 80/10 = 8 Ω. Required total impedance Z = 220/10 = 22 Ω. XL = √(22² − 8²) = 20.49 Ω. L = 20.49/(2π × 50) = 0.065 H. Answer: 0.065 H
7. In an LR circuit, I = 2.0 A, voltage across L is 240 V and across R is 100 V. Find source voltage and impedance.
V = √(240² + 100²) = 260 V. Z = V/I = 260/2 = 130 Ω. Answer: 260 V, 130 Ω
8. In a series circuit, L = 0.1 H, R = 30 Ω and V = 200 sin 400t volt. Find impedance and rms current.
ω = 400 rad s−1, so XL = ωL = 40 Ω. Z = √(30² + 40²) = 50 Ω. Vrms = 200/√2 V. Irms = (200/√2)/50 = 2√2 A. Answer: 50 Ω, 2√2 A
9. In an AC circuit, virtual current is 1.0 A, total voltage is 200 V and resistor voltage is 120 V. Find voltage across L, impedance and reactance.
VL = √(200² − 120²) = 160 V. Z = 200/1 = 200 Ω. XL = VL/I = 160 Ω. Answer: 160 V, 200 Ω, 160 Ω
10. R = 50 Ω and L = 1/π H are in series with a 200 V AC line of frequency 60 Hz. Find reactance, impedance, current and phase.
XL = 2πfL = 2π × 60 × 1/π = 120 Ω. Z = √(50² + 120²) = 130 Ω. I = 200/130 = 1.538 A. tanφ = 120/50 = 2.4. Answer: 120 Ω, 130 Ω, 1.538 A, φ = tan−1(2.4)
11. A 220 V, 50 Hz supply is connected across a 100 Ω resistor. What inductance in series will reduce current to half?
Original current = 220/100 = 2.2 A. New current = 1.1 A. Required Z = 220/1.1 = 200 Ω. XL = √(200² − 100²) = 173.2 Ω. L = 173.2/(314) = 0.55 H. Answer: 0.55 H
12. A solenoid takes 2 A from 12 V DC and 1 A from 12 V, 50 Hz AC. Find inductance.
R = 12/2 = 6 Ω. Z = 12/1 = 12 Ω. XL = √(12² − 6²) = 10.39 Ω. L = 10.39/(314) = 0.033 H. Answer: 33 mH
13. A choke coil and resistor are in series across 130 V AC. Voltage across resistance is 50 V. Find voltage across choke coil.
In an RL circuit, V² = VR² + VL². VL = √(130² − 50²) = 120 V. Answer: 120 V
14. E = 200 sin 377t volt is applied across an inductance L with resistance 1 Ω. Maximum current is 10 A. Find L.
Z = E0/I0 = 200/10 = 20 Ω. XL = √(20² − 1²) = 19.975 Ω. L = XL/ω = 19.975/377 = 0.053 H. Answer: 53 mH
15. A series LR circuit has impedance 50 Ω at 100 Hz and 100 Ω at 500 Hz. Find L and R.
50² = R² + (2π100L)² and 100² = R² + (2π500L)². Subtracting gives 7500 = (2πL)²(500² − 100²). Hence L = 0.02813 H. Then R = √[2500 − (2π100L)²] = 46.77 Ω. Answer: 28.13 mH, 46.77 Ω
16. In an RL circuit, R = 30 Ω, XL = 40 Ω and peak emf = 220 V. Find impedance, phase difference and peak current.
Z = √(30² + 40²) = 50 Ω. φ = tan−1(40/30) = 53.1°. I0 = 220/50 = 4.4 A. Answer: 50 Ω, 53.1°, 4.4 A

C. Series RC Circuit Questions

17. A circuit has a 20 Ω resistor and 0.1 μF capacitor in series connected to 230 V AC of angular frequency 100 rad s−1. Find impedance.
C = 0.1 μF = 10−7 F. XC = 1/(ωC) = 1/(100 × 10−7) = 105 Ω. Z ≈ 105 Ω because R is very small compared with XC. Answer: 105 Ω
18. R = 10 Ω, C = 0.1 μF. If 100 V, 50 Hz AC is applied, find current.
XC = 1/(2πfC) = 1/(2π × 50 × 10−7) = 31847 Ω. Z ≈ 31847 Ω. I = 100/31847 = 3.14 × 10−3 A. Answer: 3.14 mA
19. A 20 W, 50 V filament is connected in series with 250 V, 50 Hz AC mains. Find capacitance required to run the lamp.
Lamp current I = P/V = 20/50 = 0.4 A. Lamp resistance R = V/I = 125 Ω. Required impedance Z = 250/0.4 = 625 Ω. XC = √(625² − 125²) = 612.37 Ω. C = 1/(2πfXC) = 1/(2π × 50 × 612.37) = 5.2 μF. Answer: 5.2 μF
20. For an R-C series circuit, find impedance for DC and for AC frequency 10/π kHz. The printed answer is infinite and 32 Ω.
For DC, f = 0, so XC = ∞ and the capacitor blocks current. Thus impedance is infinite. For AC, use XC = 1/(2πfC) and Z = √(R² + XC²). With the figure values, the result is 32 Ω. Answer: DC: Infinite; AC: 32 Ω
Common mistake: Treating capacitor as a wire in DC. In steady DC, capacitor is open.
21. A 1 μF capacitor is connected to 220 V, 50 Hz AC. Find virtual current and peak voltage across the capacitor.
XC = 1/(2π × 50 × 10−6) = 3183 Ω. Irms = 220/3183 = 0.069 A ≈ 0.07 A. V0 = √2 × 220 = 311 V. Answer: 0.07 A, 311 V
22. A capacitor in series with 30 Ω resistance has capacitive reactance 40 Ω. Find phase difference.
For RC circuit, tanφ = XC/R = 40/30 = 4/3. Current leads supply voltage by φ. Answer: φ = tan−1(4/3)
23. A circuit has resistance 10 Ω and given impedance. Find reactance.
Use X = √(Z² − R²). The screenshot answer indicates X = 100 Ω, so the intended data corresponds to Z ≈ √(10² + 100²) = 100.5 Ω. Answer: 100 Ω
Common mistake: If a scanned value looks doubtful, always trust the phasor formula X = √(Z² − R²) rather than simple subtraction.

6. Exam-Style Practice Questions

CBSE Class 12

Q1. Why does current lag in a series RL circuit? Ans: Because inductor voltage leads current by 90° and supply voltage is the phasor sum.

Q2. R = 60 Ω, L = 0.2 H, f = 50 Hz. Find Z. Ans: XL = 62.8 Ω, Z = 87.0 Ω.

Q3. R = 30 Ω, C = 100 μF, f = 50 Hz. Find XC. Ans: 31.8 Ω.

NEET MCQs

1. In an RL circuit, frequency is doubled. XL becomes: A half B double C unchanged D zero. Ans: B.

2. In an RC circuit, frequency is increased. Current generally: A decreases B increases C becomes zero D unchanged. Ans: B.

3. Power consumed in ideal capacitor is: A VI B I²XC C zero D V²/XC. Ans: C.

JEE Main

Q1. Series RL has R = 8 Ω and XL = 6 Ω. For 100 V rms, find current and power factor. Ans: Z = 10 Ω, I = 10 A, cosφ = 0.8.

Q2. Series RC has R = 5 Ω and XC = 12 Ω. Find phase angle. Ans: φ = tan−1(12/5), current leads.

JEE Advanced

Q1. A coil takes 4 A on DC and 2 A on AC at same 100 V. Find XL. Ans: R = 25 Ω, Z = 50 Ω, XL = 43.3 Ω.

Q2. In a series RC circuit, current is maximum when frequency tends to high value. Explain. Ans: XC tends to zero, so Z tends to R.

7. Case Studies

Case 1: Choke Coil

A choke coil is used to reduce current without wasting much power. Questions: Why RL? What controls current? Why less heat? What happens with iron core?

Answers: Inductive reactance limits current; L controls XL; ideal inductor consumes no real power; iron core increases L and reduces current.

Case 2: Current Lags in RL

In a factory coil, current reaches maximum after voltage. Questions: Which element causes lag? Formula for phase? How to reduce lag? What is power factor?

Answers: Inductor; tanφ = XL/R; reduce L or frequency or add compensation; cosφ = R/Z.

Case 3: Current Leads in RC

A capacitor is used in an AC circuit and current leads voltage. Questions: Why lead? What is XC? What happens if f increases? DC behaviour?

Answers: Capacitor voltage lags current; XC=1/2πfC; XC decreases; capacitor blocks steady DC.

Case 4: Frequency Control

Same supply voltage is applied to RL and RC circuits while frequency changes. Questions: Effect on RL current? Effect on RC current? Which graph is straight? Which is hyperbola?

Answers: RL current decreases; RC current increases; XL-f is straight; XC-f is rectangular hyperbola.

Case 5: Real Power

In both circuits, students ask why reactance does not consume average power. Questions: Which part consumes power? Formula? Why cosφ? What if R = 0?

Answers: Only R; P = VrmsIrmscosφ = I²R; phase reduces real power; ideal L/C has zero average power.

8. Kumar Sir Exam Tips

RL: Current lags voltage RC: Current leads voltage Only R consumes real power P = VrmsIrmscosφ RL: tanφ = XL/R RC: tanφ = XC/R XL increases with f XC decreases with f Inductor is short for steady DC Capacitor is open for steady DC

9. Final Formula Sheet

Series RL Circuit

XL = ωL = 2πfL Z = √(R² + XL²) I = V/Z tanφ = XL/R cosφ = R/Z Pavg = VrmsIrmscosφ = Irms²R

Series RC Circuit

XC = 1/ωC = 1/(2πfC) Z = √(R² + XC²) I = V/Z tanφ = XC/R cosφ = R/Z Pavg = VrmsIrmscosφ = Irms²R
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