Alternating Current: theory

CONCEPT

AC Basics: Peak, RMS & Average Values

In one line
AC changes size and direction again and again. Meters and ratings give its rms value, which is 0.707 of the peak.

ALTERNATING CURRENT

24.1.1 What AC looks like

Some forms of alternating current

The magnitude changes continuously with time between zero and a maximum.
The direction reverses periodically.
The sine form is the one used in practice.

EQUATION

24.1.2 Reading the terms

i = i₀ sin ωt · V = V₀ sin ωt

ω = 2πν = 2πT

Symbol	Meaning	Unit
i₀, V₀	peak value	A, V
ω	angular frequency	rad/s
ν	frequency	Hz
T	time period	s

One cycle of sinusoidal AC

Positive for half a cycle, negative for the other half.
Current is zero 2ν times per second.
Direction also changes 2ν times per second.
Zero to peak takes T4

INSTANT AND TIME

24.1.3 Time to reach a value

Put the value in i = i₀ sin ωt and find ωt.
Then t = ωt2π × T.
With a phase term, set the full angle equal to π2 for the first peak.

Change	Time
0 to peak	T4
0 to i₀2	T12
0 to rms	T8
peak to rms	T8
peak to i₀2	T6

RMS VALUE

24.1.4 RMS value

i_rms = i₀√2 = 0.707 i₀

V_rms = V₀√2 = 0.707 V₀

rms means root of the mean of the squares over one cycle.
Mean of i² = i₀²2
Also called the virtual or effective value.
AC ratings and supply values are rms unless stated. House supply 220 V has a peak of 220√2 = 311 V.
A DC equal to the rms value gives the same heat in the same resistance and time.

The rms line sits at 0.707 of the peak

AVERAGE VALUE

24.1.5 Average value

i_av = 2i₀π = 0.637 i₀ (half cycle)

V_av = 2V₀π = 0.637 V₀

Over a complete cycle the average is zero.
Peak to peak value = 2V₀ = 2.828 V_rms.

OTHER FORMS

24.1.6 When the current has two parts

i = i₁ sin ωt + i₂ cos ωt: i_rms = √(i₁² + i₂²2)

i = I_dc + i₀ sin ωt: i_rms = √(I_dc² + i₀²2)

V₀ for half the cycle and 0 for the rest: V_rms = V₀√2
Bulb rated P, V: i_rms = PV and i₀ = √2 i_rms.

MEASURING AC

24.1.7 Hot wire meters

∝ i²012345DC: θ ∝ i

A hot wire meter has a non-linear scale

A DC meter reads the average, which is zero for AC. So AC cannot be read on it.
AC meters use the heating effect and read the rms value.
A full wave rectifier gives output that flows one way but is not steady.

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Corner

Form factor and peak factor

Three other wave shapes

R_f = rms valueaverage value · R_p = peak valuerms value

Wave	rms	average	R_f	R_p
Sine	i₀√2	2i₀π	1.11	1.41
Half wave	i₀2	i₀π	1.57	2
Full wave	i₀√2	2i₀π	1.11	1.41
Square	i₀	i₀	1	1

Skin effect and quick facts

AC flows mainly along the surface of a conductor. Higher frequency gives a stronger skin effect.
A 50 Hz bulb lights up 100 times per second.

Practise PYQs on this topicAC Basics: Peak, RMS & Average Values · 45 questions with solutions

CONCEPT

Phase, Phasors, Pure R & Pure L

In one line
Phase tells where V and i are in their cycles. In a resistor they stay together; in an inductor current lags voltage by 90°.

PHASE

24.2.1 Phase

X = X₀ sin(ωt ± φ₀)

(ωt ± φ₀) is the phase. Unit: radian.
φ₀ is the initial phase.
For V = V₀ sin(ωt + φ₁) and i = i₀ sin(ωt + φ₂), the difference is φ = φ₁ − φ₂.

time difference = T2π × φ

PHASOR

24.2.2 Phasor: a rotating arrow

The projection of a rotating arrow gives i

A phasor has length equal to the peak value.
It rotates anticlockwise with speed ω.
Series circuit: current on the x-axis. It is common.
Parallel circuit: voltage on the x-axis.

PURE RESISTOR

24.2.3 Pure resistor: in phase

V and i reach zero and peak together

i = i₀ sin ωt · i₀ = V₀R

φ = 0, power factor cos φ = 1.
Power P = V_rms i_rms = V₀i₀2

PURE INDUCTOR

24.2.4 Pure inductor: voltage leads

Phasors in a pure inductor

i = i₀ sin(ωt − π2)

i₀ = V₀X_L = V₀ωL

i reaches each peak a quarter cycle after V

V = L didt, so the voltage peaks first.
If i = I₀ sin ωt, then V_L = ωLI₀ cos ωt.
Time difference = T4
Power factor = 0 and average power = 0.
Energy stored U = ½Li². It goes from maximum to zero in T4

INDUCTIVE REACTANCE

24.2.5 Inductive reactance

X_L = ωL = 2πνL

Unit: ohm. X_L ∝ L and X_L ∝ ν.
On DC, ν = 0, so X_L = 0.
Ratio of reactance on AC to DC is infinite.
If ν is halved: X_L is halved, the current doubles.

Practise PYQs on this topicPure R, L and C Circuits · 65 questions with solutions

CONCEPT

Pure Capacitor & Displacement Current

In one line
In a capacitor current leads voltage by 90°. Its reactance falls as the frequency rises, so a capacitor passes AC and blocks DC.

PURE CAPACITOR

24.3.1 Pure capacitor: current leads

Phasors in a pure capacitor

i = i₀ sin(ωt + π2)

i₀ = V₀X_C = V₀ωC

i reaches each peak a quarter cycle before V

Power factor = 0 and average power = 0.
Time difference = T4
V and i are in quadrature: when one is at its peak, the other is zero.

CAPACITIVE REACTANCE

24.3.2 Capacitive reactance

X_C = 1ωC = 12πνC

Unit: ohm.
X_C ∝ 1ν and X_C ∝ 1C
On DC, X_C = ∞, so no steady current flows.
If ν doubles: X_C is halved, the current doubles.
If both ν and C double: X_C becomes X_C4

Reactances against frequency

AMMETER READING

24.3.3 Ammeter reading

i_rms = V_rms ωC

From V = 200√2 sin 100t: V_rms = 200 V, ω = 100 rad/s.
Do not use the peak value for an ammeter reading.
For the peak current, use V₀ωC.

L AND C COMPARED

24.3.4 L and C compared

	Coil	Capacitor
Reactance	rises	falls
Current	falls	rises
Phase of i	lags 90°	leads 90°
Average power	zero	zero

If X_L = X_C at ω, then at 2ω: X_CX_L = 14

DISPLACEMENT CURRENT

24.3.5 Displacement current

I_d = C dVdt = V₀ωC cos ωt

The changing electric field between the plates acts as a current.
Displacement current equals the conduction current in the wires.
A lower frequency gives a smaller displacement current.
For a parallel plate capacitor C = ε₀Ad

Practise PYQs on this topicPure R, L and C Circuits · 65 questions with solutions

CONCEPT

RL & RC Series Circuits

In one line
In series, the current is common. Resistor voltage and reactance voltage are at 90°, so they add as phasors.

RL CIRCUIT

24.4.1 RL circuit: voltage leads

Series RL circuit and its phasors; V_R lies along i

V = √(V_R² + V_L²)

Z = √(R² + X_L²) = √(R² + 4π²ν²L²)

tan φ = X_LR = ωLR

cos φ = RZ = R√(R² + X_L²)

Current: i = i₀ sin(ωt − φ).
X_L = R gives φ = 45° and Z = √2 R.
X_L = √3 R gives φ = 60° and Z = 2R.

RC CIRCUIT

24.4.2 RC circuit: current leads

Series RC circuit and its phasors; V_R lies along i

V = √(V_R² + V_C²)

Z = √(R² + 1ω²C²)

tan φ = X_CR = 1ωCR

Current: i = i₀ sin(ωt + φ).
Frequency changed to ωn: X_C becomes nX_C.
A dielectric raises C, so X_C falls and V_C falls; the current and V_R rise.

LC CIRCUIT

24.4.3 LC circuit: reactances subtract

Series LC circuit: voltages are opposite

V = |V_L − V_C| · Z = |X_L − X_C|

φ = 90° and the power factor is zero.
Current lags if L dominates, and leads if C dominates.

COIL ON DC AND AC

24.4.4 Coil on DC and AC

DC: R = V_dcI_dc · AC: Z = V_acI_ac

X_L = √(Z² − R²) · L = X_L2πν

A real coil is an RL circuit.
Less current on AC than on DC means series LR. A series capacitor would pass no steady DC.

BULB BRIGHTNESS

24.4.5 Bulb brightness

Change	Bulb + coil	Bulb + C
ν rises	dimmer	brighter
ν falls	brighter	dimmer
iron rod / dielectric	dimmer	brighter
L or C reduced	brighter	dimmer

Practise PYQs on this topicRL and RC Series Circuits · 54 questions with solutions

CONCEPT

Series LCR Circuit

In one line
Take the current along the x-axis. V_L and V_C point in opposite directions, so only their difference combines with V_R.

THE CIRCUIT

24.5.1 The circuit

V_R = iR, V_L = iX_L, V_C = iX_C

VOLTAGES

24.5.2 The phasor sum

Phasors when X_L is greater than X_C

V² = V_R² + (V_L − V_C)²

V_L leads the current by 90° and V_C lags it by 90°.
So V_L and V_C are 180° out of phase.
Voltage across L and C together = |V_L − V_C|.
Do not add the three readings directly.
L, C and R each at 60 V: the supply is 60 V and the LC pair reads zero.

IMPEDANCE

24.5.3 Impedance

Impedance triangle

Z = √(R² + (X_L − X_C)²)

Z = √(R² + (ωL − 1ωC)²)

i₀ = V₀Z · i_rms = V_rmsZ

Unit: ohm.
Reciprocal of Z is admittance: Y = 1Z
Series R values add first: a coil with its own resistance adds to R.
R, L and C are all counted in series even if the figure draws them on different sides.
A capacitor added to an LR circuit cancels part of X_L. So Z falls and the current rises.
X_L = R = 2X_C gives Z = √5R2

PHASE ANGLE

24.5.4 Phase angle: who leads

tan φ = X_L − X_CR = V_L − V_CV_R

Net reactance	Behaves as	Current
X_L > X_C	LR	lags
X_L < X_C	CR	leads
X_L = X_C	R only	in phase

45° TYPE

24.5.5 45° questions

Voltage leads by 45°: X_L − X_C = R.
Current leads by 45°: X_C − X_L = R.
Then solve for L or C.

current leads: L = 1 − 2πνCR4π²ν²C

voltage leads: L = 1 + 2πνCR4π²ν²C

REMOVE L, REMOVE C

24.5.6 Remove L, remove C

L removed: tan φ₁ = X_CR
C removed: tan φ₂ = X_LR
Equal angles mean X_L = X_C. The full circuit is at resonance.
So Z = R and the power factor is 1, whatever the angle.

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Corner

Admittance and susceptance

Y = 1Z · S = 1X

S_L = 1X_L · S_C = 1X_C = ωC

Unit of Y and S: mho.

Practise PYQs on this topicSeries LCR Circuit · 64 questions with solutions

CONCEPT

Series Resonance

In one line
At one frequency X_L = X_C. The reactances cancel, so Z = R, current is maximum and V and i are in phase.

CONDITION

24.6.1 Condition for resonance

The two reactance curves cross at ν₀

X_L = X_C

ω₀ = 1√LC · ν₀ = 12π√LC

Resonant frequency does not depend on R.

RATIO QUESTIONS

24.6.2 Ratio questions: use LC alone

ν₀ ∝ 1√(LC)

Change	New frequency
L → 3L, C → 4C	ν₀√12
L → 3L, C → 9C	ν₀√27
L → 2L, C → 8C	ν₀4
C → 3C, keep ν₀	L → L3

"Increased by 3C" means the new value is C + 3C = 4C.
To lower ν₀, raise L or C. A capacitor added in parallel raises C.
Two circuits with the same ν₀ joined in series still resonate at ν₀.

FINDING L OR C

24.6.3 Finding L or C

L = 1ω²C · C = 1ω²L

Current maximum, V and i in phase, power maximum, power factor 1: all of these mean resonance.

AT RESONANCE

24.6.4 At resonance

Quantity	At resonance
Z	minimum = R
i₀	maximum = V₀R
φ	0
cos φ	1
P	maximum = V²_rmsR
V_L, V_C	equal, opposite

V_R = V, the whole applied voltage.
V_L and V_C can each be much larger than V.
Doubling R halves the current at resonance.

VOLTAGE ACROSS L OR C

24.6.5 Voltage across L or C

V_L = V_C = i ωL = iωC

i = V_RR

From V_R, V_L, R and ω: C = V_RV_L ωR

CHARGE AND ENERGY

24.6.6 Energy ratio at maximum current

U_CU_L = CR²L

This is for a series LCR discharging with no source: at maximum current didt = 0, so V_C = iR.

Practise PYQs on this topicResonance, Bandwidth & Q Factor · 113 questions with solutions

CONCEPT

Resonance Curve, Bandwidth & Q Factor

In one line
Sweep the frequency and the current rises to a peak, then falls. Smaller R gives a taller, narrower peak and better tuning.

FREQUENCY RESPONSE

24.7.1 Z and i against ν

Z first falls, reaches R at ν₀, then rises

Resonance curves for three values of R

Below ν₀ the circuit is capacitive; above ν₀ it is inductive.
Current first rises, reaches a maximum at ν₀, then falls.
A bulb in the circuit brightens, then dims.
Resistance changes the height, never the position, of the peak.

BANDWIDTH

24.7.2 Bandwidth

ω₁ and ω₂ are the half-power frequencies

Δω = ω₂ − ω₁ = RL

At ω₁ and ω₂ the current is i₀√2 and the power is half.
Larger R gives a wider band.

ω₁, ω₂ = ω₀ ∓ Δω2

Q FACTOR

24.7.3 Q factor

Q = ω₀LR = 1ω₀CR = 1R√(LC)

Q = ω₀Δω = ν₀Δν

Dimensionless.
Large Q: sharp peak, used for tuning a radio.
To tune better, choose small R, large L and small C.
Q does not change when L and C are both doubled, but ν₀ is halved.
Ratio Qbandwidth = Q LR, unit: second.

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Corner

Half-power frequencies

i = i₀√2 · P = P_max2

Below ω₀ the circuit is capacitive at ω₁.
Above ω₀ the circuit is inductive at ω₂.

ω₁,₂ = ω₀ ∓ R2L

Energy form of Q

Q = 2π × maximum energy storedenergy dissipated per cycle

Q = V_LV_R = V_CV_R

At resonance Q is the voltage amplification.

Practise PYQs on this topicResonance, Bandwidth & Q Factor · 113 questions with solutions

CONCEPT

Power, Power Factor & Wattless Current

In one line
Only the part of the current in phase with V does work. P = V_rms i_rms cos φ, and pure L or C uses none.

AVERAGE POWER

24.8.1 Average (true) power

P = V_rms i_rms cos φ

P = V₀i₀2 cos φ

P = i_rms² R = V²_rms RZ²

Unit: watt.
Only R uses power. L and C store energy and return it.
Read φ between the V and i equations. Convert mA to A first.
If the power formula has only peak values, keep the ½.

Circuit	φ	cos φ	P
R	0	1	V₀i₀2
L or C	90°	0	0
Series LCR	φ	RZ	V_rms i_rms cos φ
Resonance	0	1	maximum

E = E₀ FORM

24.8.2 Power from E₀, R and Z

P = E₀²2 × RZ²

X_L = R: Z² = 2R, so P = E₀²4R
X_L = √3 R: Z² = 4R², so P = E₀²8R
R on its own draws P; with L in series it draws P(RZ)².

APPARENT POWER

24.8.3 Apparent power

P_app = V_rms i_rms = V₀i₀2

cos φ = true powerapparent power = WVA = kWkVA

Unit: volt ampere (VA). Always positive.

POWER FACTOR

24.8.4 Power factor

cos φ = RZ = V_RV

Same R: Z ∝ 1cos φ
cos φ = 1√2 means X = R.
Doubling ν makes X_L = 2R: cos φ = 1√5
Halving ν makes X_C = 2R: cos φ = 1√5
X_L = 3R, then add X_C = R: cos φ changes from 1√10 to 1√5
A capacitor with X_C = X_L makes the power factor 1.

WATTLESS CURRENT

24.8.5 Wattless current

i cos φ does the work; i sin φ does none

i_wattless = i_rms sin φ = i₀√2 sin φ

It flows at 90° to the voltage.
A circuit with R = 0 draws only wattless current.
Its average over a cycle is zero.

CHOKE COIL

24.8.6 Choke coil

Large L, small r

A choke in series with a tube

A coil with high inductance and low resistance. Controls AC with very little power loss.
Thick Cu wire reduces R.
Soft iron core raises L.
Low frequency: iron core. High frequency: air core.
Used with fluorescent tubes: the tube gets a lower voltage.

ARC LAMP

24.8.7 Arc lamp: choosing L

X_L = √(Z² − R²)

Lamp needs I at V: R = VI
On supply V′: Z = V′I
Then L = X_L2πν

Practise PYQs on this topicPower, Power Factor & Wattless Current · 82 questions with solutions

CONCEPT

Parallel AC Circuits

In one line
In parallel, the voltage is common and the branch currents add as phasors. At resonance the line current is minimum.

PARALLEL RLC

24.9.1 Parallel RLC

R, L and C across one source

i_R = VR · i_L = VX_L · i_C = VX_C

i = √(i_R² + (i_C − i_L)²)

i_L and i_C are 180° out of phase, so they subtract.
L and C alone: i = |i_L − i_C|.
Instantaneous currents: write each branch from the common e.

TWO BRANCHES

24.9.2 Two branches: RC beside RL

Find Z and φ of each branch separately.
One branch leads, the other lags.
If they are 90° apart: I = √(i₁² + i₂²).

PARALLEL RESONANCE

24.9.3 Parallel resonance: rejector

Ideal LC across a source: line current is zero

ν₀ = 12π√LC

Quantity	At parallel resonance
Z	maximum
Line current	minimum
Voltage across LC	maximum
cos φ	1

Called a rejector circuit, because it rejects that frequency.
The opposite of series resonance, where Z is least.
Ideal L and C: Z is infinite and the line current is zero.
The i against ν graph has a dip at ν₀.

LIMITS OF FREQUENCY

24.9.4 Very high and very low frequency

Frequency	L acts as	C acts as
very high	open circuit	short circuit
zero (DC)	short circuit	open circuit

Redraw the circuit with these replacements.
Then solve the resistor network.

L AND C COMBINED

24.9.5 Combining L and C

L_s = L₁ + L₂ · C_p = C₁ + C₂

Combine first, then use ν₀ = 12π√(LC)

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Corner

Parallel RLC: phasor method

Voltage is common, so it lies along the x-axis

i = √(i_R² + (i_C − i_L)²)

1Z = √(1R² + (1X_C − 1X_L)²)

tan φ = i_C − i_Li_R

Coil with resistance across C

R and L in one branch, C in the other

ω₀ = √(1LC − R²L²)

Z_max = LCR · i_min = V₀CRL

Resonance is possible only if R < √(LC).

Practise PYQs on this topicParallel AC Circuits · 11 questions with solutions

CONCEPT

LC Oscillations

In one line
A charged capacitor joined to a coil swaps energy with it again and again. The total ½CV² = ½LI² stays the same.

HOW IT OSCILLATES

24.10.1 Energy moves back and forth

Four instants of one cycle

At t = 0 all energy is electric, in the capacitor.
The capacitor discharges and the current builds up in L.
At maximum current all energy is magnetic.
The current then charges C the other way. The current also reverses.

q = Q cos ωt · i = I₀ sin ωt

FREQUENCY

24.10.2 Natural frequency

ω = 1√LC · ν = 12π√LC

ν ∝ 1√C: C → 2C gives ν√2
A dielectric of constant K: C → KC, so ν → ν√K
K = (old frequencynew frequency)².

MAXIMUM CURRENT

24.10.3 Maximum current

I₀ = V√(CL) = Qω = Q√LC

V is the voltage on C when the battery is removed.
Use SI units: μF and mH must be converted.

AT ANY INSTANT

24.10.4 Current at voltage V₂

½CV₁² = ½CV₂² + ½LI²

I = √(C(V₁² − V₂²)L)

ENERGY SHARE

24.10.5 Equal energy

U_E and U_B vary with period T2

U_E = U_B at q = Q√2

t = T8 = π4√LC

Current maximum first at t = T4
25% of the energy moved to L: sin²ωt = 14, so ωt = π6

ANALOGY

24.10.6 Spring analogy

Circuit	Spring
L	m
C	1k
R	b (damping)

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Corner

Damped oscillations in LCR

Q = Q₀ e^{−Rt ÷ 2L}

Time for the charge amplitude to halve:

t = 2L ln 2R

Practise PYQs on this topicLC Oscillations · 25 questions with solutions

CONCEPT

AC Generator & Transformer

In one line
A generator makes AC by turning a coil in a magnetic field. A transformer changes its voltage, and the turns ratio sets the change.

AC GENERATOR

24.11.1 AC generator

A coil rotating in a magnetic field

e = e₀ sin ωt

e₀ = NBAω = NBA 2πν

The induced emf is sinusoidal

Works on electromagnetic induction.
Convert rpm: ω = 2π × rpm60
Current amplitude: i₀ = e₀R
Average power in R: e₀²2R; peak power: e₀²R

TRANSFORMER

24.11.2 Transformer

Two coils on one soft iron core

Works on mutual induction. So it needs a changing flux.
DC gives no output: the secondary reads zero.
Frequency does not change.

V_sV_p = N_sN_p = I_pI_s

	Step up	Step down
Turns	N_s > N_p	N_s < N_p
Voltage	V_s > V_p	V_s < V_p
Current	I_s < I_p	I_s > I_p

In an ideal transformer the power stays the same: V_pI_p = V_sI_s.

USING THE RATIOS

24.11.3 Transformer variations

Z_pZ_s = (N_pN_s)²

e_s = M di_pdt · peak e_s = Mωi₀

Load on the secondary: I_s = V_sR = PV_s
For an amplifier, take the root of the impedance ratio.

EFFICIENCY

24.11.4 Efficiency

η = output powerinput power = V_s I_sV_p I_p

"50% of power lost" means output = 50% of input.
Find I_p from input and V_s or I_s from output.

ENERGY LOSSES

24.11.5 Energy losses and their cures

Loss	Reduced by
Eddy currents	laminated core
Hysteresis	soft iron core
Copper (i²R)	thick wire
Flux leakage	windings over each other

Lamination breaks the eddy current path.
Power is sent at high voltage and low current, so less is lost as i²R.

Practise PYQs on this topicAC Generator & Transformer · 61 questions with solutions

EXAM-DAY CARD

Just see this before the exam

i_rms = i₀√2

rms value

X_L = ωL, X_C = 1ωC

reactance

Z = √(R² + (X_L − X_C)²)

impedance

ν₀ = 12π√LC

resonance

P = V i cos φ

power

V_sV_p = N_sN_p

transformer

FORMULA SHEET

Whole chapter on one page

Basics

i = i₀ sin ωt, ω = 2πT

i_rms = 0.707 i₀

i_av = 0.637 i₀ (half cycle)

peak to rms: T8; 0 to i₀2: T12

Pure elements

R: φ = 0

L: i lags 90°, X_L = ωL

C: i leads 90°, X_C = 1ωC

I_d = V₀ωC cos ωt

Series LCR

V² = V_R² + (V_L − V_C)²

Z² = R² + (X_L − X_C)²

tan φ = X_L − X_CR

cos φ = RZ

Resonance

ν₀ = 12π√(LC)

Z = R, φ = 0, i₀ = V₀R

Δω = RL

Q = ω₀LR = 1R√(LC)

Power

P = V_rms i_rms cos φ

P = i²_rms R

wattless = i_rms sin φ

arc lamp: L = X_L2πν

LC & transformer

I₀ = V√(CL)

equal energy: q = Q√2

V_sV_p = N_sN_p = I_pI_s

e₀ = NBAω

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Using V₀ for an ammeter reading

Adding V_R + V_L + V_C directly

"Increased by 3C" taken as 3C

Using the turns ratio for current when η is below 100%

CORRECT ✓

Meters read rms: V₀ ÷ √2

V² = V_R² + (V_L − V_C)²

New value = C + 3C = 4C

Use power: output = η × input

Alternating Current

AC Basics: Peak, RMS & Average Values

24.1.1 What AC looks like

24.1.2 Reading the terms

24.1.3 Time to reach a value

24.1.4 RMS value

24.1.5 Average value

24.1.6 When the current has two parts

24.1.7 Hot wire meters

Corner

Phase, Phasors, Pure R & Pure L

24.2.1 Phase

24.2.2 Phasor: a rotating arrow

24.2.3 Pure resistor: in phase

24.2.4 Pure inductor: voltage leads

24.2.5 Inductive reactance

Pure Capacitor & Displacement Current

24.3.1 Pure capacitor: current leads

24.3.2 Capacitive reactance

24.3.3 Ammeter reading

24.3.4 L and C compared

24.3.5 Displacement current

RL & RC Series Circuits

24.4.1 RL circuit: voltage leads

24.4.2 RC circuit: current leads

24.4.3 LC circuit: reactances subtract

24.4.4 Coil on DC and AC

24.4.5 Bulb brightness

Series LCR Circuit

24.5.1 The circuit

24.5.2 The phasor sum

24.5.3 Impedance

24.5.4 Phase angle: who leads

24.5.5 45° questions

24.5.6 Remove L, remove C

Corner

Series Resonance

24.6.1 Condition for resonance

24.6.2 Ratio questions: use LC alone

24.6.3 Finding L or C

24.6.4 At resonance

24.6.5 Voltage across L or C

24.6.6 Energy ratio at maximum current

Resonance Curve, Bandwidth & Q Factor

24.7.1 Z and i against ν

24.7.2 Bandwidth

24.7.3 Q factor

Corner

Power, Power Factor & Wattless Current

24.8.1 Average (true) power

24.8.2 Power from E0, R and Z

24.8.3 Apparent power

24.8.4 Power factor

24.8.5 Wattless current

24.8.6 Choke coil

24.8.7 Arc lamp: choosing L

Parallel AC Circuits

24.9.1 Parallel RLC

24.9.2 Two branches: RC beside RL

24.9.3 Parallel resonance: rejector

24.9.4 Very high and very low frequency

24.9.5 Combining L and C

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LC Oscillations

24.10.1 Energy moves back and forth

24.10.2 Natural frequency

24.10.3 Maximum current

24.10.4 Current at voltage V2

24.10.5 Equal energy

24.10.6 Spring analogy

Corner

AC Generator & Transformer

24.11.1 AC generator

24.11.2 Transformer

24.11.3 Transformer variations

24.11.4 Efficiency

24.11.5 Energy losses and their cures

Just see this before the exam

Whole chapter on one page

Avoid these = easy marks

24.8.2 Power from E₀, R and Z

24.10.4 Current at voltage V₂