Alternating Current: R, L and C Circuits
Complete AC roadmap for pure resistance, pure inductance and pure capacitance with formulas, circuit diagrams, phase relations, phasor diagrams, power factor, frequency graphs and solved screenshot questions.ac circuits with resistor inductor and capacitor
1. Pure Resistive AC Circuit
Equations
V = V0 sin ωt I = V/R = (V0/R) sin ωt I = I0 sin ωt, I0 = V0/RVoltage and current are in the same phase. The phase angle is zero.
Power factor = cos 0° = 1Waveform and Phasor
Average Power in Pure R
Pavg = (1/2π) ∫02π VI d(ωt) V = V0 sin ωt, I = I0 sin ωt Pavg = VrmsIrms Pavg = Irms2R Pavg = Vrms2/RR versus Frequency
Formula Summary
I0 = V0/R φ = 0°, cosφ = 1 Pavg = VrmsIrmsResistance does not depend on frequency. A pure resistor consumes real power.
2. Pure Inductive AC Circuit
Equations
V = V0 sin ωt V = L dI/dt I = I0 sin(ωt - π/2) I0 = V0/XL XL = ωL = 2πfLIn a pure inductor, current lags voltage by 90°.
Current Lags Voltage by 90°
Power and DC Behaviour
φ = 90°, cosφ = 0 Pavg = VrmsIrms cos90° = 0An ideal inductor consumes no average power. Energy is stored in the magnetic field and returned to the circuit.
For DC: f = 0 ⇒ XL = 0 Pure inductor behaves like short circuit for DC.XL versus Frequency
Formula Summary
XL = ωL = 2πfL I0 = V0/XL I = I0 sin(ωt - π/2)If L increases or f increases, XL increases and current decreases.
3. Pure Capacitive AC Circuit
Equations
V = V0 sin ωt q = CV, I = dq/dt I = I0 sin(ωt + π/2) I0 = V0/XC XC = 1/ωC = 1/(2πfC)In a pure capacitor, current leads voltage by 90°.
Current Leads Voltage by 90°
Power, Dielectric and DC Behaviour
φ = 90°, cosφ = 0 Pavg = VrmsIrms cos90° = 0If dielectric constant K is inserted, capacitance becomes KC. Therefore XC decreases by factor K and current increases.
For DC: f = 0 ⇒ XC = ∞ Capacitor blocks DC.XC versus Frequency
Formula Summary
XC = 1/ωC = 1/(2πfC) I0 = V0/XC I = I0 sin(ωt + π/2)If frequency increases, capacitive reactance decreases and current increases.
4. Comparison Table: Pure R, Pure L and Pure C Circuits
| Circuit | Current Equation | Phase Relation | Reactance / Resistance | Power Factor | Average Power | Frequency Dependence | DC Behaviour |
|---|---|---|---|---|---|---|---|
| Pure R | I = I0 sinωt | V and I in same phase | R | 1 | VrmsIrms | R independent of f | Conducts DC |
| Pure L | I = I0 sin(ωt - π/2) | I lags V by 90° | XL = 2πfL | 0 | 0 | XL increases with f | Short circuit ideally |
| Pure C | I = I0 sin(ωt + π/2) | I leads V by 90° | XC = 1/(2πfC) | 0 | 0 | XC decreases with f | Open circuit |
5. Extracted Questions with Complete Solutions
What is the inductive reactance of a coil if current through it is 800 mA and the voltage across it is 40 V?
Solution
Given: I = 800 mA = 0.8 A, V = 40 V.
Formula: XL = V/I
XL = 40/0.8 = 50 Ω.
Final answer: 50 Ω
Common mistake: Not converting mA into A.
Find the value of current through an inductance of 2.0 H and negligible resistance, when connected to an AC source of 150 V and 50 Hz.
Solution
Given: L = 2.0 H, Vrms = 150 V, f = 50 Hz.
Formula: XL = 2πfL and I = V/XL
XL = 2 × 3.14 × 50 × 2 = 628 Ω.
I = 150/628 = 0.239 A.
Final answer: 0.239 A
Common mistake: Using L directly as resistance.
An inductance of negligible resistance, whose reactance is 22 Ω at 200 Hz, is connected to a 220 V, 50 Hz power line. What is the value of inductance and reactance?
Solution
Given: XL1 = 22 Ω at f1 = 200 Hz. Find L and XL2 at f2 = 50 Hz.
Formula: XL = 2πfL
L = XL1/(2πf1) = 22/(2 × 3.14 × 200) = 0.0175 H.
XL2 = 2πf2L = 2 × 3.14 × 50 × 0.0175 = 5.5 Ω.
Final answer: L = 0.0175 H, XL = 5.5 Ω
Common mistake: Keeping reactance same when frequency changes.
A coil of self-inductance has inductive reactance 88 Ω. Calculate the self-inductance of the coil if frequency is 50 Hz.
Solution
Given: XL = 88 Ω, f = 50 Hz.
Formula: L = XL/(2πf)
L = 88/(2 × 3.14 × 50) = 0.28 H.
Final answer: 0.28 H
Common mistake: Forgetting factor 2π.
Find the maximum value of current when an inductance of one henry is connected to an AC source of 200 volts, 50 Hz.
Solution
Given: L = 1 H, Vrms = 200 V, f = 50 Hz.
Formula: XL = 2πfL, Irms = Vrms/XL, I0 = √2 Irms
XL = 2 × 3.14 × 50 × 1 = 314 Ω.
Irms = 200/314 = 0.637 A.
I0 = 1.414 × 0.637 = 0.9 A.
Final answer: 0.9 A
Common mistake: Reporting RMS current instead of maximum current.
Calculate the frequency at which the inductive reactance of 0.7 H inductor is 220 Ω.
Solution
Given: L = 0.7 H, XL = 220 Ω.
Formula: f = XL/(2πL)
f = 220/(2 × 3.14 × 0.7) = 50 Hz.
Final answer: 50 Hz
Common mistake: Writing f = 2πLXL.
What is the capacitive reactance of a 5 µF capacitor when it is a part of a circuit whose frequency is (i) 50 Hz, (ii) 106 Hz?
Solution
Given: C = 5 µF = 5 × 10-6 F.
Formula: XC = 1/(2πfC)
At 50 Hz: XC = 1/(2 × 3.14 × 50 × 5 × 10-6) = 636.6 Ω.
At 106 Hz: XC = 1/(2 × 3.14 × 106 × 5 × 10-6) = 3.18 × 10-2 Ω.
Final answer: 636.6 Ω, 3.18 × 10-2 Ω
Common mistake: Not converting microfarad into farad.
A capacitor has capacitance of 1/π µF. Find its reactance for a frequency of (i) 50 Hz and (ii) 106 Hz.
Solution
Given: C = (1/π) µF = 1/(π × 106) F.
Formula: XC = 1/(2πfC)
Since C = 10-6/π F, XC = 1/[2πf(10-6/π)] = 1/(2f × 10-6).
At 50 Hz: XC = 1/(100 × 10-6) = 10,000 Ω.
At 106 Hz: XC = 1/(2 × 106 × 10-6) = 0.5 Ω.
Final answer: 10,000 Ω, 0.5 Ω
Common mistake: Misreading 1/π µF as π µF.
A 1.5 µF capacitor has capacitive reactance of 12 Ω. What is the frequency of the source? If the frequency of the source is doubled, what will be the capacitive reactance?
Solution
Given: C = 1.5 × 10-6 F, XC = 12 Ω.
Formula: f = 1/(2πCXC)
f = 1/(2 × 3.14 × 1.5 × 10-6 × 12) = 8846 Hz.
Since XC ∝ 1/f, if frequency is doubled, XC becomes half.
New XC = 12/2 = 6 Ω.
Final answer: 8846 Hz, 6 Ω
Common mistake: Thinking XC increases when frequency increases.
A capacitor of capacitance 10 µF is connected to an oscillator giving an output voltage E = 10 sin ωt volt. If ω = 10 rad s-1, find the peak current in the circuit.
Solution
Given: C = 10 µF = 10 × 10-6 F, V0 = 10 V, ω = 10 rad s-1.
Formula: I0 = V0/XC = V0ωC
I0 = 10 × 10 × 10 × 10-6 = 1000 × 10-6 A = 1.0 mA.
Final answer: 1.0 mA
Common mistake: Using frequency f instead of angular frequency ω.
A capacitor has a reactance of 100 Ω at 50 Hz. What will be its reactance at 125 Hz?
Solution
Given: XC1 = 100 Ω, f1 = 50 Hz, f2 = 125 Hz.
Formula: XC ∝ 1/f, so XC2 = XC1 f1/f2
XC2 = 100 × 50/125 = 40 Ω.
Final answer: 40 Ω
Common mistake: Multiplying by 125/50 instead of 50/125.
6. Exam Practice Section
Why does a pure inductor consume zero average power in AC?
Answer
Voltage and current differ in phase by 90°. During one quarter cycle energy is stored in the magnetic field and in another quarter cycle it is returned. Hence net average power is zero.
Formula: Pavg = VrmsIrmscos90° = 0.
In a pure capacitor, current: A) lags voltage by 90° B) leads voltage by 90° C) is in phase D) is opposite phase
Answer
Correct answer: B. Current leads voltage by 90°.
A 0.5 H inductor is connected to 100 V, 50 Hz AC source. Find RMS current.
Answer
XL = 2πfL = 2 × 3.14 × 50 × 0.5 = 157 Ω. I = 100/157 = 0.637 A.
A capacitor C is connected to AC supply. If frequency is tripled and capacitance is doubled, how does peak current change for same peak voltage?
Answer
I0 = V0ωC. If frequency triples, ω triples. If C doubles, current becomes 3 × 2 = 6 times.
Explain why AC can pass through a capacitor but DC cannot.
Answer
For DC, f = 0 and XC = 1/(2πfC) becomes infinite, so capacitor blocks DC. For AC, frequency is non-zero, so finite reactance allows alternating current.
Which graph is a straight line through origin: R vs f, XL vs f or XC vs f?
Answer
XL vs f is a straight line through origin because XL = 2πfL.
7. NEET and IIT-JEE PYQ-Style MCQ Bank
These questions are written in actual exam pattern style: formula selection, phase relation, reactance-frequency traps, RMS/peak confusion and graph interpretation. Each question includes four choices and a short solution.
NEET Level MCQs
A pure inductor of 0.5 H is connected to a 100 V, 50 Hz AC source. The RMS current is nearly:
A) 0.32 A B) 0.64 A C) 1.28 A D) 3.14 A
Answer and Solution
Answer: B) 0.64 A. Formula: XL = 2πfL = 2×3.14×50×0.5 = 157 Ω. I = V/XL = 100/157 = 0.64 A.
Common mistake: Using L as resistance.
In a pure capacitive AC circuit, the current:
A) lags voltage by 90° B) leads voltage by 90° C) is in phase D) differs by 180°
Answer and Solution
Answer: B) leads voltage by 90°. In capacitor, I = I0 sin(ωt + π/2).
Common mistake: Mixing capacitor with inductor phase relation.
If frequency of AC source is doubled, inductive reactance becomes:
A) half B) double C) four times D) unchanged
Answer and Solution
Answer: B) double. Since XL = 2πfL, XL ∝ f.
Common mistake: Applying capacitor relation XC ∝ 1/f.
If frequency of AC source is doubled, capacitive reactance becomes:
A) half B) double C) four times D) unchanged
Answer and Solution
Answer: A) half. XC = 1/(2πfC), so XC ∝ 1/f.
Common mistake: Thinking both XL and XC increase with frequency.
An ideal inductor connected to DC source behaves as:
A) open circuit B) short circuit after steady state C) infinite resistance D) capacitor
Answer and Solution
Answer: B) short circuit after steady state. For DC, f = 0, hence XL = 0.
Common mistake: Saying inductor always opposes all current.
An ideal capacitor connected to DC source behaves as:
A) short circuit always B) open circuit after charging C) pure resistor D) pure inductor
Answer and Solution
Answer: B) open circuit after charging. For DC, f = 0, so XC = ∞.
Common mistake: Confusing initial charging current with steady DC current.
Power factor of an ideal capacitor is:
A) 1 B) 0 C) 1/2 D) √2
Answer and Solution
Answer: B) 0. For pure C, φ = 90°, so cosφ = 0.
Common mistake: Using power factor 1 for all AC circuits.
A 10 µF capacitor is connected to 50 Hz AC. Its capacitive reactance is approximately:
A) 31.8 Ω B) 318 Ω C) 636 Ω D) 3.18 Ω
Answer and Solution
Answer: B) 318 Ω. XC = 1/(2πfC) = 1/(2×3.14×50×10×10-6) = 318 Ω.
Common mistake: Not converting µF to F.
JEE Main Level MCQs
A coil has negligible resistance and inductive reactance 100 Ω at 100 Hz. Its reactance at 250 Hz is:
A) 40 Ω B) 100 Ω C) 250 Ω D) 625 Ω
Answer and Solution
Answer: C) 250 Ω. XL ∝ f, so XL2 = 100 × 250/100 = 250 Ω.
Common mistake: Using inverse frequency relation.
A capacitor has reactance 200 Ω at 60 Hz. Its reactance at 120 Hz is:
A) 50 Ω B) 100 Ω C) 200 Ω D) 400 Ω
Answer and Solution
Answer: B) 100 Ω. XC ∝ 1/f. Doubling frequency halves reactance.
Common mistake: Treating capacitor like inductor.
For V = V0 sinωt across pure inductor, current is proportional to:
A) sinωt B) cosωt C) -cosωt D) -sinωt
Answer and Solution
Answer: C) -cosωt. I = I0 sin(ωt - π/2) = -I0 cosωt.
Common mistake: Writing capacitor current expression.
For V = V0 sinωt across pure capacitor, current is proportional to:
A) sinωt B) cosωt C) -cosωt D) -sinωt
Answer and Solution
Answer: B) cosωt. Since I = C dV/dt = ωCV0 cosωt.
Common mistake: Forgetting derivative of voltage.
A 2 µF capacitor is connected to 200 V, 50 Hz AC. RMS current is approximately:
A) 0.126 A B) 0.0126 A C) 1.26 A D) 12.6 A
Answer and Solution
Answer: A) 0.126 A. I = V/XC = V(2πfC) = 200×2×3.14×50×2×10-6 = 0.126 A.
Common mistake: Calculating peak current while V is RMS.
In a pure inductor, instantaneous power averaged over one cycle is zero because:
A) current is zero B) voltage is zero C) energy is alternately stored and returned D) resistance is infinite
Answer and Solution
Answer: C. Ideal inductor has no real power loss; magnetic energy is returned to source.
Common mistake: Thinking zero average power means no current.
JEE Advanced Level Problems
A pure capacitor connected to V = V0 sinωt has peak current I0. If frequency is doubled and capacitance is halved, peak current becomes:
A) I0/2 B) I0 C) 2I0 D) 4I0
Answer and Solution
Answer: B) I0. I0 = V0ωC. New product = (2ω)(C/2) = ωC.
Common mistake: Changing only frequency and ignoring capacitance.
A pure inductor has RMS current I at angular frequency ω. Keeping voltage same, angular frequency is made 3ω. New current is:
A) I/3 B) I C) 3I D) 9I
Answer and Solution
Answer: A) I/3. I = V/XL = V/(ωL). Tripling ω triples XL, current becomes one-third.
Common mistake: Saying current increases with frequency in inductor.
For a pure capacitor, choose the correct statements: (i) current leads voltage, (ii) average power is zero, (iii) capacitor blocks high frequency AC, (iv) XC decreases with frequency.
A) i, ii only B) i, ii, iv C) ii, iii, iv D) all four
Answer and Solution
Answer: B) i, ii, iv. Capacitor does not block high frequency AC; its reactance decreases with frequency.
Common mistake: Remembering only "capacitor blocks DC" and extending it incorrectly to all AC.
A source of fixed RMS voltage is connected separately to R, ideal L and ideal C. Which circuit consumes non-zero average power?
A) only R B) only L C) only C D) all three
Answer and Solution
Answer: A) only R. Pure L and pure C have phase angle 90°, so average power is zero.
Common mistake: Confusing current flow with power consumption.
8. Case Study Section
Case 1: Resistor Consumes Power
In a pure resistor, V and I are in phase. Positive power is delivered throughout both half cycles.
Pavg = Irms2RCase 2: Ideal Inductor
An ideal inductor stores energy in its magnetic field and returns it to the source.
Pavg = 0Case 3: Capacitor Blocks DC
For DC, f = 0 and XC = ∞. Hence no steady DC current flows through an ideal capacitor.
XC = 1/(2πfC)Case 4: Frequency Effect
As frequency increases, XL increases while XC decreases. This is a common graph-based exam concept.
XL ∝ f, XC ∝ 1/f