NEET UG Physics · Chapter 18

Alternating Current

rms values, reactance, impedance, LCR circuits, resonance, power factor and transformers.

NCERT Class 12NEET: 1–2 Qs per paper 12 min read

1. Chapter at a Glance

Alternating current (AC) changes its direction periodically, usually sinusoidally. The electricity supplied to our homes is AC at 50 Hz. This chapter covers rms values, the behaviour of a resistor, inductor and capacitor in an AC circuit, the series LCR circuit with its impedance and resonance, power and power factor, and the transformer. NEET asks one or two questions from here, mostly numerical.

What NEET asks most
  • Peak and rms values; reading frequency from V = V₀ sin ωt.
  • Inductive and capacitive reactance and phase relations.
  • Impedance, phase angle and current in a series LCR circuit.
  • Resonant frequency and quality factor.
  • Average power, power factor, wattless current and transformers.

2. AC Voltage and rms Values

v = V₀ sin ωt    ω = 2πf    Vrms = V₀/√2 ≈ 0.707V₀    Irms = I₀/√2
  • The average of a sinusoidal current over a full cycle is zero, so we use the root-mean-square (rms) value, which gives the same heating as a steady DC current.
  • Household supply in India is 220 V rms at 50 Hz; its peak is 220√2 ≈ 311 V.
  • AC ammeters and voltmeters read rms values.

3. R, L and C in AC Circuits

ElementOppositionPhase of currentAverage powerNEET Note
Resistor RRIn phase with VVrmsIrmsDoes not depend on f
Inductor LXL = ωLLags V by π/2ZeroBlocks high f; passes DC
Capacitor CXC = 1/(ωC)Leads V by π/2ZeroBlocks DC; passes high f
Memory aid: "CIVIL"

In a Capacitor, I leads V; in an inductor (L), V leads I. So C-I-V and V-I-L.

4. Series LCR Circuit and Resonance

Z = √(R² + (XL − XC)²)    tan φ = (XL − XC)/R    I = V/Z
  • The voltages across R, L and C must be added as phasors, not as ordinary numbers: V = √(VR² + (VL − VC)²).
  • If XL > XC, the circuit is inductive (current lags); if XC > XL, it is capacitive (current leads).
  • Resonance: XL = XC, so Z = R (minimum) and the current is maximum. ω₀ = 1/√(LC), f₀ = 1/(2π√(LC)).
  • At resonance, VL and VC are equal and opposite and can each be much larger than the supply voltage.
  • Quality factor Q = ω₀L/R = 1/(ω₀CR) measures the sharpness of resonance. A small R gives a sharp, high-Q resonance, which is used in radio tuning.
small R (high Q)large R ω₀ωI
Current against frequency in a series LCR circuit. The peak is at ω₀ = 1/√(LC). A smaller resistance gives a taller, sharper peak.

5. Power in AC Circuits

Pavg = VrmsIrms cos φ = Irms²R    power factor cos φ = R/Z
  • For a pure resistor cos φ = 1; for a pure inductor or capacitor cos φ = 0, so no power is used.
  • A current that consumes no power (in a purely reactive circuit) is called a wattless current. The component I sin φ is the wattless part.
  • At resonance cos φ = 1, so the power is maximum.
  • A choke coil (an inductor with small resistance) controls current in AC circuits with very little power loss, unlike a resistor.

6. Transformer

Vs/Vp = Ns/Np = Ip/Is (ideal)    efficiency η = Pout/Pin
  • A transformer works on mutual induction, so it works only with AC, not with steady DC.
  • Step-up: Ns > Np, voltage increases and current decreases. Step-down: the reverse.
  • Energy losses: copper loss (I²R in windings), eddy-current loss (reduced by lamination), hysteresis loss (reduced by soft iron) and flux leakage.
  • Power is transmitted at high voltage so that the current, and hence the I²R loss in the lines, is small.

7. Formula Master Table

QuantityFormula
rms valueVrms = V₀/√2
Inductive reactanceXL = ωL = 2πfL
Capacitive reactanceXC = 1/(2πfC)
ImpedanceZ = √(R² + (XL − XC)²)
Phase angletan φ = (XL − XC)/R
Resonant frequencyf₀ = 1/(2π√LC)
Quality factorQ = ω₀L/R
Average powerP = VrmsIrms cos φ
Power factorcos φ = R/Z
TransformerVs/Vp = Ns/Np

8. Master Summary Table

ConceptKey ideaNEET Note
rms valueV₀/√2Meters read rms
InductorXL ∝ fCurrent lags by 90°
CapacitorXC ∝ 1/fCurrent leads by 90°
ResonanceXL = XCZ = R, current maximum
Q factorω₀L/RSharpness of tuning
Power factorR/ZZero for pure L or C
TransformerMutual inductionDoes not work on DC

9. Mind Map

Alternating Current
Basics
  • V₀ sin ωt
  • rms = peak/√2
Elements
  • R in phase
  • L lags
  • C leads
LCR
  • Impedance
  • Resonance
  • Q factor
Power
  • cos φ
  • Wattless current
  • Transformer

10. Tips and Tricks

Shortcuts
  • At f = 50 Hz, ω = 100π ≈ 314 rad s⁻¹.
  • If VR, VL, VC are given, the supply voltage is √(VR² + (VL − VC)²), not their sum.
  • Doubling f doubles XL and halves XC.
  • For an ideal transformer, VpIp = VsIs.
Common NEET traps
  • Adding voltages of R, L and C arithmetically.
  • Using peak values where rms values are needed for power.
  • Thinking a step-up transformer increases power.
  • Saying a capacitor passes DC.

11. Important One-liners

  • The average value of AC over a complete cycle is zero.
  • The rms value of AC is 0.707 times its peak value.
  • An inductor offers zero reactance to DC; a capacitor offers infinite reactance to DC.
  • At resonance, a series LCR circuit behaves as a pure resistor.
  • The power factor of a pure inductor or capacitor is zero.
  • A transformer cannot work on DC.
  • Choke coils are preferred over resistors because they waste little power.

12. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Click to reveal each answer and working.

Average level

N1Average

An AC voltage is v = 170 sin(120πt) V. Find its peak value, rms value and frequency.

Show answer & solution

Concept: compare with V₀ sin(2πft).

  1. V₀ = 170 V.
  2. Vrms = 170/√2 ≈ 120 V.
  3. 2πf = 120π ⇒ f = 60 Hz.

Answer: 170 V; about 120 V; 60 Hz

N2Average

Find the reactance of a 0.5 H inductor at 50 Hz.

Show answer & solution

Concept: XL = 2πfL.

  1. XL = 2π × 50 × 0.5 = 50π.
  2. XL ≈ 157 Ω.

Answer: about 157 Ω

N3Average

Find the reactance of a 20 μF capacitor at 50 Hz. What is its reactance for DC?

Show answer & solution

Concept: XC = 1/(2πfC).

  1. XC = 1/(2π × 50 × 20 × 10⁻⁶) = 1/(6.28 × 10⁻³).
  2. XC ≈ 159 Ω.
  3. For DC, f = 0, so XC is infinite (the capacitor blocks DC).

Answer: about 159 Ω; infinite for DC

N4Average

An ideal transformer steps 220 V down to 11 V. Its primary has 2000 turns. Find the number of secondary turns. If the secondary current is 4 A, find the primary current.

Show answer & solution

Concept: Vs/Vp = Ns/Np = Ip/Is.

  1. Ns = 2000 × 11/220 = 100 turns.
  2. Ip = Is × Vs/Vp = 4 × 11/220 = 0.2 A.

Answer: 100 turns; 0.2 A

N5Average

A series LC circuit has L = 0.4 H and C = 25 μF. Find its resonant angular frequency and frequency.

Show answer & solution

Concept: ω₀ = 1/√(LC).

  1. LC = 0.4 × 25 × 10⁻⁶ = 10⁻⁵ s².
  2. ω₀ = 1/√(10⁻⁵) ≈ 316 rad s⁻¹.
  3. f₀ = ω₀/2π ≈ 50.3 Hz.

Answer: about 316 rad s⁻¹; about 50 Hz

Advanced level

N6Advanced

A series LCR circuit has R = 40 Ω, XL = 70 Ω and XC = 40 Ω, and is connected to 250 V rms. Find the impedance, current, power factor, phase angle and average power.

Show answer & solution

Concept: Z = √(R² + (XL − XC)²).

  1. XL − XC = 30 Ω; Z = √(1600 + 900) = 50 Ω.
  2. I = 250/50 = 5 A.
  3. cos φ = R/Z = 0.8; φ = tan⁻¹(30/40) ≈ 37°, current lagging (inductive).
  4. P = VI cos φ = 250 × 5 × 0.8 = 1000 W (= I²R = 25 × 40 ✓).

Answer: 50 Ω; 5 A; 0.8; about 37° lagging; 1000 W

N7Advanced

A series LCR circuit has R = 10 Ω, L = 0.2 H and C = 20 μF. Find its resonant angular frequency and quality factor.

Show answer & solution

Concept: ω₀ = 1/√(LC), Q = ω₀L/R.

  1. LC = 0.2 × 20 × 10⁻⁶ = 4 × 10⁻⁶ ⇒ ω₀ = 1/(2 × 10⁻³) = 500 rad s⁻¹.
  2. Q = 500 × 0.2/10 = 10.
  3. At resonance, VL = VC = Q × supply voltage, i.e. 10 times the supply.

Answer: 500 rad s⁻¹; Q = 10

N8Advanced

A lamp rated 100 V, 500 W is to be run from a 200 V, 50 Hz supply. What inductance must be connected in series with it?

Show answer & solution

Concept: the lamp must still get 5 A; the inductor supplies the extra impedance without wasting power.

  1. Lamp current I = 500/100 = 5 A; lamp resistance R = 100/5 = 20 Ω.
  2. Required Z = 200/5 = 40 Ω.
  3. XL = √(Z² − R²) = √(1600 − 400) ≈ 34.6 Ω.
  4. L = XL/(2πf) = 34.6/314 ≈ 0.11 H.

Answer: about 0.11 H

N9Advanced

A transformer is 90% efficient. It takes 2 kW from a 200 V supply and delivers its output at 1000 V. Find the primary current, output power and secondary current.

Show answer & solution

Concept: Pout = ηPin.

  1. Ip = 2000/200 = 10 A.
  2. Pout = 0.9 × 2000 = 1800 W.
  3. Is = 1800/1000 = 1.8 A.

Answer: 10 A; 1800 W; 1.8 A

N10Advanced

A 40 Ω resistor and a capacitor of reactance 30 Ω are in series across a 100 V rms supply. Find the current and the voltages across R and C. Why is their sum not 100 V?

Show answer & solution

Concept: phasor addition.

  1. Z = √(40² + 30²) = 50 Ω; I = 100/50 = 2 A.
  2. VR = 2 × 40 = 80 V; VC = 2 × 30 = 60 V.
  3. VR and VC are 90° out of phase, so they add as √(80² + 60²) = 100 V, not 140 V.
  4. The current leads the voltage by tan⁻¹(30/40) ≈ 37°.

Answer: 2 A; 80 V and 60 V (they add as phasors to 100 V)

13. NEET-pattern Questions with Detailed Answers

These questions are original and based on concepts that NEET UG asks again and again.

Q1Easy

The peak value of the 220 V household AC supply is about:

  1. 156 V
  2. 220 V
  3. 311 V
  4. 440 V
Show solution

220 V is the rms value, so V₀ = 220√2 ≈ 311 V. (A) divides by √2 instead of multiplying.

Answer: (C)

Q2Easy

In a purely inductive AC circuit, the current:

  1. Leads the voltage by 90°
  2. Lags the voltage by 90°
  3. Is in phase with the voltage
  4. Lags by 180°
Show solution

An inductor opposes changes in current, so the current reaches its peak a quarter cycle after the voltage: it lags by 90°. (A) describes a capacitor.

Answer: (B)

Q3Medium

At resonance in a series LCR circuit:

  1. Impedance is maximum
  2. Current is minimum
  3. Impedance equals R and current is maximum
  4. Power factor is zero
Show solution

XL = XC, so Z = R (its minimum) and the current is maximum. The power factor is 1, not 0.

Answer: (C)

Q4Medium

If the frequency of the AC supply is doubled, the reactance of a capacitor:

  1. Doubles
  2. Halves
  3. Stays the same
  4. Becomes four times
Show solution

XC = 1/(2πfC) ∝ 1/f, so doubling f halves XC. An inductor's reactance would double instead.

Answer: (B)

Q5Medium

The power factor of a series circuit with R = 30 Ω and net reactance 40 Ω is:

  1. 0.6
  2. 0.75
  3. 0.8
  4. 1.33
Show solution

Z = √(900 + 1600) = 50 Ω; cos φ = 30/50 = 0.6. (C) uses 40/50, which is sin φ, and a power factor can never exceed 1.

Answer: (A)

Q6Medium

A step-up transformer increases:

  1. Power
  2. Current
  3. Voltage
  4. Frequency
Show solution

It raises the voltage and lowers the current. The power is at best unchanged, and the frequency stays the same.

Answer: (C)

Q7Hard

In a series LCR circuit, the voltages across R, L and C are 40 V, 90 V and 60 V. The supply voltage is:

  1. 190 V
  2. 50 V
  3. 70 V
  4. 110 V
Show solution

V = √(VR² + (VL − VC)²) = √(1600 + 900) = 50 V. (A) wrongly adds the voltages as plain numbers.

Answer: (B)

Q8Hard

The average power consumed in an AC circuit containing only an ideal inductor and an ideal capacitor is:

  1. VrmsIrms
  2. Zero
  3. ½V₀I₀
  4. Infinite at resonance
Show solution

With no resistance, the phase difference is 90°, so cos φ = 0 and the average power is zero. Energy only moves back and forth between L and C. The current is wattless.

Answer: (B)

14. Quick Revision Summary

Remember before the exam
  • Vrms = V₀/√2; meters read rms.
  • XL = ωL (I lags); XC = 1/ωC (I leads).
  • Z = √(R² + (XL − XC)²); resonance ω₀ = 1/√LC; Q = ω₀L/R.
  • P = VrmsIrms cos φ; cos φ = R/Z.
  • Transformer: Vs/Vp = Ns/Np; AC only.
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