Alternating Current: NEET notes.

01 · NCERT §7.1, §7.2

Alternating voltage and a resistor

• The mains supply is not steady: its voltage swings like a sine function of time. Such a voltage is an alternating (ac) voltage, and the current it drives is an alternating current.
• AC is preferred for supply mainly because transformers can raise or lower ac voltages easily and efficiently, which also makes sending energy over long distances economical.
• Take the source voltage as v = vm sin ωt, with amplitude vm and angular frequency ω.
• Across a resistor R, Kirchhoff's loop rule gives vm sin ωt = iR, so i = im sin ωt with im = vm/R. Ohm's law holds for ac just as for dc.
• Voltage and current in a resistor are in phase: they pass through zero, their maxima and their minima at the same moments.
• Over a full cycle the current is positive for half the time and negative for the other half, so its average is zero. That does not make the heating zero, because the heating depends on i², which is never negative.

02 · NCERT §7.2

RMS values and average power

• The power in the resistor at any instant is p = i²R = im²R sin²ωt. Since sin²ωt = ½(1 − cos 2ωt) and cos 2ωt averages to zero over a cycle, the average power is ½im²R.
• To write this like the dc formula P = I²R, define the root mean square (rms) or effective current I = im/√2 = 0.707 im. Then P = I²R.
• Likewise the rms voltage is V = vm/√2 = 0.707 vm, and V = IR, so P = V²/R = IV, exactly as for dc.
• The rms current is the steady current that would heat the same resistor at the same average rate.
• AC values are quoted as rms values unless stated otherwise. The household 220 V is an rms value; its peak is √2 × 220 = 311 V.
• Example 7.1: a bulb rated 100 W for 220 V has R = V²/P = 484 Ω; the source's peak voltage is 311 V; the rms current is P/V = 0.454 A.

03 · NCERT §7.3

Phasors

• In an inductor or a capacitor the current and voltage are not in phase. Phasors make these phase relations easy to see.
• A phasor is an arrow that rotates anticlockwise about the origin with angular speed ω. Its length is the peak value (vm or im) of the quantity it stands for.
• The projection of the phasor on the vertical axis gives the value at that instant, vm sin ωt or im sin ωt. As the phasors turn, the projections trace out the sine curves.
• For a resistor, the V and I phasors point the same way at every instant: the phase angle between them is zero.
• Voltages and currents are scalars. Phasors are only a device: the amplitudes and phases of sinusoidal quantities add in the same way as rotating arrows do.

04 · NCERT §7.4

AC through an inductor

• For a pure inductor (negligible resistance), the loop rule gives v − L di/dt = 0, so di/dt = (vm/L) sin ωt.
• Integrating, i = −(vm/ωL) cos ωt = im sin(ωt − π/2), with im = vm/ωL. There is no constant term, because the source swings evenly about zero.
• The quantity ωL plays the part of resistance. It is the inductive reactance XL = ωL, measured in ohm, and im = vm/XL.
• XL grows in proportion to both the inductance and the frequency.
• The current lags the voltage by π/2, a quarter of a cycle: it reaches each maximum a quarter period after the voltage does.
• The average power into a pure inductor over a cycle is zero. Energy stored in its field during one quarter cycle is handed back to the source in the next.
• Example 7.2: a pure 25.0 mH inductor on 220 V, 50 Hz has XL = 2π × 50 × 25 × 10⁻³ = 7.85 Ω and an rms current of 220/7.85 = 28 A.

05 · NCERT §7.5

AC through a capacitor

• On dc a capacitor passes current only while it charges; once full, the current stops. On ac it is charged, discharged and charged the other way every cycle, so current keeps flowing in the leads.
• With v = q/C = vm sin ωt, the current i = dq/dt = ωCvm cos ωt = im sin(ωt + π/2), where im = ωCvm.
• So 1/ωC acts like a resistance. It is the capacitive reactance XC = 1/ωC, in ohm, and im = vm/XC.
• XC falls as the frequency or the capacitance rises: a capacitor blocks dc (zero frequency) and passes high frequencies easily.
• The current leads the voltage by π/2: it reaches each maximum a quarter period before the voltage.
• As with the inductor, the average power into a pure capacitor over a cycle is zero.
• Example 7.3: a lamp in series with a capacitor stays dark on dc but glows on ac. Making C smaller raises XC and dims the lamp on ac.
• Example 7.4: 15.0 μF on 220 V, 50 Hz has XC = 212 Ω, rms current 1.04 A and peak current 1.47 A, leading the voltage by π/2. At double the frequency XC halves and the current doubles.
• Example 7.5: a bulb in series with an inductor dims when an iron rod is pushed into the coil. The rod raises L and so XL, and more of the supply voltage appears across the coil.

06 · NCERT §7.6, §7.6.1

Series LCR circuit and impedance

• In a series LCR circuit the same current i = im sin(ωt + φ) flows through R, L and C at every instant. φ is the phase of the current relative to the source voltage.
• Draw the current phasor first. The resistor's voltage vRm = imR is along it, the inductor's vLm = imXL is π/2 ahead of it, and the capacitor's vCm = imXC is π/2 behind it.
• VL and VC point in opposite directions, so they combine into one phasor of size |vCm − vLm|. The source phasor V is the hypotenuse of a right triangle with sides VR and that difference: vm² = vRm² + (vCm − vLm)².
• So im = vm/√(R² + (XC − XL)²) = vm/Z, where Z = √(R² + (XC − XL)²) is the impedance, in ohm.
• The phase angle is given by tan φ = (XC − XL)/R. R, (XC − XL) and Z form the impedance triangle.
• If XC > XL the circuit is mainly capacitive and the current leads the voltage; if XL > XC it is mainly inductive and the current lags.
• The phasor method gives the steady-state behaviour only. Right after switching on, a transient part is also present; it dies away with time.
• Example 7.6: 200 Ω and 15.0 μF in series on 220 V, 50 Hz. XC = 212.3 Ω, Z = 291.67 Ω, I = 0.755 A, VR = 151 V and VC = 160.3 V.
• Those two add to 311.3 V, more than 220 V. There is no paradox: VR and VC are 90° apart, so they add as √(VR² + VC²) = 220 V.

07 · NCERT §7.6.2

Resonance

• Systems that tend to oscillate at a natural frequency respond strongly when driven near it, like a swing pushed in time with its own motion. This is resonance.
• In a series LCR circuit, XL = ωL grows and XC = 1/ωC shrinks as ω rises. At one frequency ω₀ they are equal, Z falls to its least value R, and the current amplitude reaches its greatest value vm/R.
• Setting ω₀L = 1/ω₀C gives the resonant frequency ω₀ = 1/√(LC).
• NCERT's graph uses L = 1.00 mH, C = 1.00 nF and vm = 100 V, so ω₀ = 1.00 × 10⁶ rad/s. With R = 100 Ω the peak current at resonance is twice what it is with R = 200 Ω, since im = vm/R there.
• At resonance the voltages across L and C are equal and opposite and cancel; the whole source voltage appears across R. Resonance therefore needs both L and C: an RL or RC circuit cannot resonate.
• Radio and TV tuning uses resonance. The aerial picks up many stations at once; turning the tuning capacitor moves the circuit's resonant frequency onto one station's frequency, so that station's current is the largest.
• Example 7.10: an airport metal detector is a many-turn coil with a capacitor, tuned to resonance. Metal carried through the coil changes the impedance, the current changes noticeably, and the change sets off the alarm.

08 · NCERT §7.7

Power factor

• With v = vm sin ωt and i = im sin(ωt + φ), the average power over a cycle is P = (vm im/2) cos φ = VI cos φ. It can also be written P = I²Z cos φ.
• cos φ is the power factor. The power depends on the phase angle as well as on V and I.
• Purely resistive circuit: φ = 0, cos φ = 1, and the power is the largest possible for that V and I.
• Pure L or pure C: φ = π/2, cos φ = 0, and no power is used even though current flows. Such a current is called wattless current.
• In any LCR circuit, power is used up only in the resistor. At resonance φ = 0, cos φ = 1 and P = I²R, the most power for the circuit.
• Example 7.7: for a given power at a given voltage, a low power factor means a bigger current and so bigger I²R loss in the lines. A lagging wattless current can be cancelled by a capacitor connected in parallel, which raises the power factor towards 1.
• Example 7.8: peak 283 V at 50 Hz on R = 3 Ω, L = 25.48 mH, C = 796 μF. XL = 8 Ω, XC = 4 Ω, Z = 5 Ω, φ = −53.1° (current lags), I = 40 A, P = 4800 W and power factor 0.6.
• Example 7.9: the same circuit resonates at ω₀ = 1/√(LC), about 222 rad/s or 35.3 Hz. There Z = R = 3 Ω, the rms current is 66.7 A and the power is 13.35 kW, more than at 50 Hz.

09 · NCERT §7.8

Transformers

• A transformer changes an ac voltage to a larger or smaller one using mutual induction. Two insulated coils, the primary with Np turns and the secondary with Ns turns, are wound on a soft-iron core, one over the other or on separate limbs.
• AC in the primary sets up an alternating flux in the core. The same flux φ per turn threads the secondary, giving es = −Ns dφ/dt, and it induces a back emf ep = −Np dφ/dt in the primary.
• For an ideal transformer (primary resistance negligible, no flux leakage, small secondary current) vp = ep and vs = es, so vs/vp = Ns/Np.
• If no energy is lost, input and output power are equal, ip vp = is vs, so is/ip = Np/Ns = vp/vs. These ratios hold for peak and rms values alike.
• Ns > Np gives a step-up transformer: higher voltage, lower current. Ns < Np gives a step-down transformer: lower voltage, higher current.
• Example from the text: with 100 primary and 200 secondary turns, 220 V at 10 A in becomes 440 V at 5.0 A out.
• Real transformers lose some energy. Flux leakage is reduced by winding one coil over the other; winding resistance by thick wire; eddy currents in the core by laminating it; hysteresis by a core material with a small hysteresis loss. A well-designed transformer can be more than 95% efficient.
• Power is sent over long distances at high voltage so that the current, and with it the I²R loss in the lines, is small. Sub-stations and pole transformers then step the voltage down in stages for homes.

Revise

Must-know facts

1. 01v = vm sin ωt; in a resistor i = (vm/R) sin ωt, in phase with v.
2. 02The average of an ac current over a cycle is zero, but the average of i² is not.
3. 03rms values: I = im/√2 = 0.707 im and V = vm/√2 = 0.707 vm.
4. 04Average power in a resistor: P = I²R = V²/R = IV.
5. 05Mains 220 V is rms; its peak is 311 V.
6. 06A phasor rotates anticlockwise at ω; its length is the peak value and its vertical projection is the instantaneous value.
7. 07Inductor: XL = ωL; current lags voltage by π/2.
8. 08Capacitor: XC = 1/ωC; current leads voltage by π/2.
9. 09Average power into a pure L or pure C is zero.
10. 10Series LCR: Z = √(R² + (XC − XL)²), im = vm/Z, tan φ = (XC − XL)/R.
11. 11Voltages across R, L and C must be added as phasors, not as plain numbers.
12. 12Resonance: XL = XC at ω₀ = 1/√(LC); Z = R and current is greatest.
13. 13Resonance needs both L and C.
14. 14Average power P = VI cos φ; cos φ is the power factor.
15. 15Wattless current: current with cos φ = 0, which uses no power.
16. 16Ideal transformer: Vs/Vp = Ns/Np = Ip/Is.
17. 17Step-up raises voltage and lowers current; step-down does the reverse.
18. 18Transformer losses: flux leakage, winding resistance, eddy currents, hysteresis.

Where marks go

Common traps

Reference

Formulas

AC voltage and current in R

v = vm sin ωt, i = (vm/R) sin ωt

In phase.

RMS values

I = im/√2, V = vm/√2

0.707 times the peak.

Average power in R

P = I²R = V²/R = IV

rms values.

Inductive reactance

XL = ωL = 2πνL

Current lags by π/2.

Capacitive reactance

XC = 1/ωC = 1/2πνC

Current leads by π/2.

Impedance of series LCR

Z = √(R² + (XC − XL)²)

im = vm/Z.

Phase angle

tan φ = (XC − XL)/R

φ > 0: current leads; φ < 0: current lags.

Resonant frequency

ω₀ = 1/√(LC), ν₀ = 1/2π√(LC)

Z = R at resonance.

Average power

P = VI cos φ

cos φ is the power factor.

Ideal transformer

Vs/Vp = Ns/Np = Ip/Is

Input power = output power.

Glossary

Key terms

Alternating voltage

A voltage that varies sinusoidally with time, v = vm sin ωt.

Amplitude (peak value)

The largest value vm or im reached in a cycle.

RMS value

Peak value divided by √2; the steady value giving the same average heating.

Phasor

A rotating arrow whose length is a peak value and whose vertical projection is the instantaneous value.

Phase angle (φ)

The angle by which the current leads (φ > 0) or lags (φ < 0) the source voltage.

Inductive reactance (XL)

ωL, the opposition an inductor offers to ac; in ohm.

Capacitive reactance (XC)

1/ωC, the opposition a capacitor offers to ac; in ohm.

Impedance (Z)

√(R² + (XC − XL)²), the total opposition of a series LCR circuit; in ohm.

Impedance triangle

Right triangle with sides R and XC − XL and hypotenuse Z.

Resonance

The condition XL = XC in a series LCR circuit, where Z = R and the current is largest.

Resonant frequency

ω₀ = 1/√(LC).

Power factor

cos φ, the fraction of VI that is actually used as power.

Wattless current

Current 90° out of phase with the voltage, which on average uses no power.

Transformer

Two coils on an iron core that change an ac voltage by mutual induction.

Step-up / step-down

A transformer with more / fewer secondary turns than primary turns.

Laminated core

A core made of thin insulated sheets, which cuts eddy-current loss.