AC Circuits Notes FE Sem 1 BEE | Mumbai University

Module 2 introduces alternating quantities, which is how almost all electrical power is generated and distributed. It covers how a sinusoidal voltage is generated, the standard definitions, average and RMS values, and phasor representation for adding and comparing AC quantities. It then analyses single-phase series and parallel R, L, C circuits, real, reactive and apparent power, admittance and resonance, and ends with three-phase generation and star and delta relations. It has the most teaching hours (12) of any BEE module.

Summary

Generation of alternating voltage: when a coil of N turns rotates with angular velocity omega in a uniform magnetic field, the flux linked with it changes sinusoidally and, by Faraday's law, an EMF e = Em sin(omega t) is induced, where Em = N B A omega. The EMF is zero when the coil plane is perpendicular to the field and maximum when the coil sides cut the field lines at right angles.

Basic definitions: waveform, instantaneous value, cycle, time period T, frequency f = 1/T (Hz), amplitude (peak value), angular frequency omega = 2 pi f, and phase. In India the supply frequency is 50 Hz, so T = 20 ms. Form factor = RMS/average and peak (crest) factor = peak/RMS; for a sine wave they are 1.11 and 1.414.

The average value of an alternating quantity is taken over a half cycle (because it is zero over a full cycle for a symmetrical wave); for a sine wave Iav = 2Im/pi = 0.637 Im. The RMS (effective) value is the steady DC current that produces the same heat in the same resistance in the same time; for a sine wave Irms = Im/sqrt(2) = 0.707 Im. Meters read RMS values and ratings like 230 V are RMS.

A phasor is a rotating line of length equal to the maximum (or RMS) value, rotating anticlockwise at omega, whose projection gives the instantaneous value. Phase difference is the angular difference between two phasors of the same frequency; the one reaching its maximum first is leading. Phasors can be written in rectangular form (a + jb) or polar form (r at angle theta). Sums and differences are done in rectangular form or by resolving into horizontal and vertical components; multiplication and division are easier in polar form.

Single-phase series circuits: in pure R, current is in phase with voltage; in pure L, current lags voltage by 90 degrees (XL = 2 pi f L); in pure C, current leads by 90 degrees (XC = 1/(2 pi f C)). In series RL, Z = sqrt(R^2 + XL^2) and current lags by phi = tan^-1(XL/R); in series RC current leads; in series RLC, Z = sqrt(R^2 + (XL - XC)^2) and the circuit behaves inductive, capacitive or resistive depending on whether XL is greater than, less than or equal to XC. Power factor cos phi = R/Z.

Parallel circuits are solved with admittance Y = 1/Z = G + jB (siemens), where G is conductance and B susceptance; admittances of parallel branches add. Total current is V Y. Alternatively each branch current is found separately and the phasor sum taken. Power: real (active) power P = V I cos phi in watts is consumed in resistance; reactive power Q = V I sin phi in VAR is exchanged with L and C; apparent power S = V I in VA, and S^2 = P^2 + Q^2 (power triangle).

Resonance (theory only): a series RLC circuit is resonant when XL = XC, at f0 = 1/(2 pi sqrt(LC)); impedance is minimum (= R), current is maximum, power factor is unity and the voltages across L and C can be many times the supply voltage (voltage magnification, Q = omega0 L/R). A parallel (tank) circuit of a practical coil and a capacitor is resonant when the supply current is in phase with voltage; at resonance its impedance is maximum (dynamic impedance L/(CR)), the line current is minimum and a large current circulates between L and C (current magnification).

Three-phase voltages are generated by three identical windings placed 120 electrical degrees apart on the same armature, giving three EMFs of equal magnitude and frequency displaced by 120 degrees, phase sequence R-Y-B. In a balanced star connection VL = sqrt(3) Vph, IL = Iph and line voltages lead the corresponding phase voltages by 30 degrees; in a delta connection VL = Vph, IL = sqrt(3) Iph and line currents lag the phase currents by 30 degrees. Total power in either case is P = sqrt(3) VL IL cos phi, where phi is the phase angle of the load impedance.

Mind map

Method to solve a series RLC circuit

Important definitions

Formulas

Key points to remember

• All AC calculations use RMS values unless the question gives instantaneous expressions; convert Vm to Vrms by dividing by sqrt(2).
• Mnemonic CIVIL: in a Capacitor I leads V; in an inductor (L) V leads I.
• Inductive circuits have lagging power factor; capacitive circuits have leading power factor.
• Phasors can be added only if they have the same frequency.
• Average power in pure L or pure C is zero; only resistance consumes real power.
• In parallel circuits take voltage as the reference phasor; in series circuits take current as reference.
• At series resonance Z = R (minimum) and I = V/R (maximum); it is called an acceptor circuit. At parallel resonance Z is maximum and line current minimum; it is called a rejector circuit.
• Below resonance a series RLC circuit is capacitive; above resonance it is inductive.
• In a three-phase system the three EMFs are 120 degrees apart and their phasor sum is zero for a balanced system.
• The power formula sqrt(3) VL IL cos phi is the same for star and delta, but phi is always the angle between phase voltage and phase current.
• For the same supply and the same phase impedance, a delta-connected load draws three times the line current and three times the power of the star-connected load.

Solved examples

Example 1. An alternating voltage is given by v = 100 sin(314t + 30 degrees) V. Find the maximum value, RMS value, average value, frequency, time period and the instantaneous value at t = 5 ms.

Comparing with v = Vm sin(omega t + phi): Vm = 100 V, omega = 314 rad/s.

Vrms = 100/sqrt(2) = 70.71 V.

Vav = 0.637 x 100 = 63.66 V.

f = omega/(2 pi) = 314/6.283 = 50 Hz; T = 1/f = 0.02 s = 20 ms.

At t = 5 ms: omega t = 314 x 0.005 = 1.57 rad = 90 degrees.

v = 100 sin(90 + 30) = 100 sin 120 degrees = 100 x 0.866 = 86.6 V.

Example 2. Two currents i1 = 10 sin(omega t) A and i2 = 20 sin(omega t + 60 degrees) A are fed into a common conductor. Find the resultant current in the form i = Im sin(omega t + phi).

Represent maximum values as phasors: I1 = 10 at 0 degrees = 10 + j0; I2 = 20 at 60 degrees = 20 cos60 + j20 sin60 = 10 + j17.32.

Sum: I = (10 + 10) + j(0 + 17.32) = 20 + j17.32.

|I| = sqrt(20^2 + 17.32^2) = sqrt(400 + 300) = 26.46 A.

phi = tan^-1(17.32/20) = 40.89 degrees.

Resultant i = 26.46 sin(omega t + 40.89 degrees) A; its RMS value is 26.46/sqrt(2) = 18.71 A.

Example 3. A coil having resistance 10 ohm and inductance 0.1 H is connected across a 230 V, 50 Hz supply. Find the impedance, current, power factor, real power, reactive power and apparent power.

XL = 2 pi f L = 2 x 3.1416 x 50 x 0.1 = 31.42 ohm.

Z = sqrt(10^2 + 31.42^2) = sqrt(100 + 987) = 32.97 ohm.

I = V/Z = 230/32.97 = 6.98 A.

pf = cos phi = R/Z = 10/32.97 = 0.303 lagging; phi = 72.34 degrees.

P = I^2 R = 6.976^2 x 10 = 486.7 W.

Q = I^2 XL = 48.67 x 31.42 = 1529 VAR.

S = V I = 230 x 6.976 = 1604.5 VA. Check: sqrt(486.7^2 + 1529^2) = 1604.5 VA.

Example 4. A series circuit has R = 10 ohm, L = 0.2 H and C = 100 microfarad connected across a 230 V, 50 Hz supply. Find the impedance, current, power factor and power consumed. State whether the circuit is inductive or capacitive.

XL = 2 pi x 50 x 0.2 = 62.83 ohm.

XC = 1/(2 pi x 50 x 100 x 10^-6) = 31.83 ohm.

Net X = XL - XC = 31.00 ohm (inductive).

Z = sqrt(10^2 + 31^2) = sqrt(1061) = 32.57 ohm.

I = 230/32.57 = 7.06 A.

pf = 10/32.57 = 0.307 lagging; phi = tan^-1(31/10) = 72.1 degrees.

P = I^2 R = 7.06^2 x 10 = 498.6 W (also V I cos phi = 230 x 7.06 x 0.307 = 498.6 W).

Since XL is greater than XC the circuit is inductive and current lags voltage.

Voltage drops: VR = 70.6 V, VL = 443.6 V, VC = 224.7 V; note VL alone exceeds the supply voltage, which is possible in a series RLC circuit.

Example 5. Two branches are connected in parallel across a 200 V, 50 Hz supply. Branch 1: R = 10 ohm in series with XL = 10 ohm. Branch 2: R = 8 ohm in series with XC = 6 ohm. Find the branch currents, total current, power factor and total power using admittance.

Take V = 200 at 0 degrees.

Z1 = 10 + j10 = 14.14 at 45 degrees ohm; Y1 = 1/Z1 = (10 - j10)/200 = 0.05 - j0.05 S.

Z2 = 8 - j6 = 10 at -36.87 degrees ohm; Y2 = (8 + j6)/100 = 0.08 + j0.06 S.

Y = Y1 + Y2 = 0.13 + j0.01 S.

I1 = V Y1 = 10 - j10 = 14.14 A lagging by 45 degrees.

I2 = V Y2 = 16 + j12 = 20 A leading by 36.87 degrees.

I = V Y = 26 + j2 = 26.08 A at +4.40 degrees.

pf = cos 4.40 degrees = 0.997 leading.

P = V I cos phi = 200 x 26.08 x 0.997 = 5200 W.

Check: I1^2 R1 + I2^2 R2 = 200 x 10 + 400 x 8 = 2000 + 3200 = 5200 W.

Example 6. Three identical coils, each of resistance 6 ohm and reactance 8 ohm, are connected in star across a 400 V, 3-phase, 50 Hz supply. Find the line current, power factor and total power. Repeat if the coils are connected in delta.

Zph = sqrt(6^2 + 8^2) = 10 ohm; cos phi = 6/10 = 0.6 lagging.

Star: Vph = 400/sqrt(3) = 230.94 V; Iph = 230.94/10 = 23.09 A = IL.

P = sqrt(3) x 400 x 23.09 x 0.6 = 9600 W (check 3 x 23.09^2 x 6 = 9600 W).

Delta: Vph = VL = 400 V; Iph = 400/10 = 40 A; IL = sqrt(3) x 40 = 69.28 A.

P = sqrt(3) x 400 x 69.28 x 0.6 = 28800 W (check 3 x 40^2 x 6 = 28800 W).

The delta connection draws three times the line current and three times the power of the star connection.

Example 7. A single-phase load takes 10 kW at 0.8 power factor lagging from a 230 V supply. Find the current, apparent power and reactive power, and draw the power triangle.

S = P/cos phi = 10/0.8 = 12.5 kVA.

I = S/V = 12500/230 = 54.35 A.

sin phi = sqrt(1 - 0.8^2) = 0.6; Q = S sin phi = 12.5 x 0.6 = 7.5 kVAR (lagging, inductive).

Power triangle: horizontal side P = 10 kW, vertical side Q = 7.5 kVAR drawn downward for lagging load, hypotenuse S = 12.5 kVA, angle phi = 36.87 degrees between P and S.

Important questions and answers

Q1. Explain the generation of a sinusoidal alternating EMF and derive e = Em sin(omega t). (5 marks)

Consider a rectangular coil of N turns and area A rotating anticlockwise with uniform angular velocity omega in a uniform magnetic field of flux density B. Let time be measured from the position where the plane of the coil is perpendicular to the field (maximum flux linkage).

After time t the coil has turned through theta = omega t, and the flux linked is phi = B A cos(omega t) = phi_m cos(omega t).

By Faraday's law, e = -N d(phi)/dt = N B A omega sin(omega t) = Em sin(omega t), where Em = N B A omega = N phi_m omega.

At theta = 0 the coil sides move parallel to the field, so e = 0; at theta = 90 degrees they cut the field at right angles, so e = Em. The EMF reverses every half revolution, giving a sine wave. Diagram: coil between N and S poles with slip rings, and the resulting sine wave.

Q2. Define RMS value and derive the RMS value of a sinusoidal current. (5 marks)

RMS value is the steady DC current which, when flowing through a given resistance for a given time, produces the same heat as the alternating current flowing through the same resistance for the same time.

Let i = Im sin(theta). Heat over a cycle is proportional to the mean of i^2:

Irms^2 = (1/2 pi) integral from 0 to 2 pi of Im^2 sin^2(theta) d(theta) = (Im^2/2 pi) integral (1 - cos 2 theta)/2 d(theta) = (Im^2/2 pi)(pi) = Im^2/2.

Hence Irms = Im/sqrt(2) = 0.707 Im.

The ratings of AC equipment and the readings of AC ammeters and voltmeters are RMS values.

Q3. Derive the average value of a sinusoidal current and find the form factor and peak factor. (5 marks)

For a symmetrical wave the average over a full cycle is zero, so it is taken over a half cycle.

Iav = (1/pi) integral from 0 to pi of Im sin(theta) d(theta) = (Im/pi)[-cos theta] from 0 to pi = (Im/pi)(2) = 2Im/pi = 0.637 Im.

Form factor = Irms/Iav = 0.707 Im/0.637 Im = 1.11.

Peak factor = Im/Irms = Im/(0.707 Im) = 1.414.

Form factor is used in rectifier-type meters; peak factor indicates the insulation stress on equipment.

Q4. Show that the current in a purely inductive circuit lags the voltage by 90 degrees and the average power is zero. (5 marks)

Let v = Vm sin(omega t) be applied to a pure inductance L. Then v = L di/dt, so i = (1/L) integral Vm sin(omega t) dt = -(Vm/(omega L)) cos(omega t) = Im sin(omega t - 90 degrees), where Im = Vm/XL and XL = omega L.

Thus current lags voltage by 90 degrees.

Instantaneous power p = v i = Vm sin(omega t) x Im sin(omega t - 90) = -Vm Im sin(omega t) cos(omega t) = -(Vm Im/2) sin(2 omega t).

The average of sin(2 omega t) over a cycle is zero, so average power = 0. Energy is stored in the magnetic field in one quarter cycle and returned to the source in the next. Waveform and phasor diagrams should be drawn.

Q5. Explain series RLC circuit with phasor diagram for XL greater than XC. (6 marks)

Let R, L, C be in series carrying current I (reference phasor). VR = I R is in phase with I, VL = I XL leads I by 90 degrees and VC = I XC lags I by 90 degrees.

Since VL and VC are opposite, the net reactive voltage is I(XL - XC). The supply voltage V = sqrt(VR^2 + (VL - VC)^2) = I sqrt(R^2 + (XL - XC)^2) = I Z.

Phase angle phi = tan^-1((XL - XC)/R). For XL greater than XC, V leads I by phi, i.e. current lags and the circuit is inductive with lagging pf = R/Z.

If XL is less than XC the circuit is capacitive with leading pf; if XL = XC the circuit is at resonance, Z = R and pf = 1.

Impedance triangle: R horizontal, (XL - XC) vertical, Z hypotenuse. Power P = V I cos phi = I^2 R.

Q6. Define real, reactive and apparent power and explain the power triangle. (4 marks)

Real power P = V I cos phi (W): power actually converted to heat or work, consumed in resistance.

Reactive power Q = V I sin phi (VAR): power flowing back and forth between source and inductance or capacitance; average value zero but it loads the lines.

Apparent power S = V I (VA): product of RMS voltage and current; decides the rating of machines and cables.

Power triangle: obtained by multiplying the sides of the impedance triangle by I^2; base P, perpendicular Q, hypotenuse S, with S^2 = P^2 + Q^2 and cos phi = P/S. For an inductive load Q is lagging; for a capacitive load Q is leading.

Q7. Explain series resonance. Derive the resonant frequency and state its characteristics. (6 marks)

A series RLC circuit is at resonance when the net reactance is zero, XL = XC.

2 pi f0 L = 1/(2 pi f0 C), hence f0 = 1/(2 pi sqrt(LC)).

Characteristics at resonance: (1) impedance is minimum and purely resistive, Z = R; (2) current is maximum, I0 = V/R; (3) power factor is unity; (4) VL = VC and they cancel, but each can be much larger than V; this voltage magnification is Q = VL/V = omega0 L/R; (5) power dissipated is maximum, V^2/R.

Below f0 the circuit is capacitive and above f0 it is inductive. A plot of I versus f shows a peak at f0; the sharpness depends on Q. Because it accepts maximum current at f0, it is called an acceptor circuit and is used in tuning circuits of radio receivers.

Q8. Explain parallel resonance and compare it with series resonance. (6 marks)

A practical parallel resonant circuit has a coil (R, L) in parallel with a capacitor C. Resonance occurs when the reactive component of the coil current equals the capacitor current, so the supply current is in phase with voltage. The resonant frequency is f0 = (1/2 pi) sqrt(1/(LC) - R^2/L^2), approximately 1/(2 pi sqrt(LC)) when R is small.

At resonance the impedance is maximum and resistive, ZD = L/(CR); the line current is minimum, V/ZD; a large current circulates in the L-C loop (current magnification Q).

Comparison: series - Z minimum (R), current maximum, voltage magnification, acceptor circuit; parallel - Z maximum (L/CR), line current minimum, current magnification, rejector circuit. Both have unity power factor at resonance.

Q9. Explain the generation of three-phase voltages with waveform and phasor diagram. (4 marks)

Three identical coils R, Y and B are mounted on the same shaft (or stator) displaced 120 electrical degrees apart and rotated in a uniform magnetic field. Each coil generates a sinusoidal EMF of the same magnitude and frequency, but they reach their maxima one-third of a cycle apart:

eR = Em sin(omega t), eY = Em sin(omega t - 120 degrees), eB = Em sin(omega t - 240 degrees).

The phase sequence is R-Y-B. The phasor sum and the instantaneous sum of the three EMFs is zero. Advantages over single phase: more power for the same conductor material, constant power to balanced loads, and self-starting three-phase motors.

Q10. Derive the relation between line and phase voltages and currents in a balanced star connection. (5 marks)

In star, one end of each phase is joined to a neutral point, and the other ends go to the lines. Each line is in series with one phase, so IL = Iph.

The line voltage between R and Y is the phasor difference VRY = VRN - VYN = VRN + (-VYN). VRN and -VYN are equal in magnitude (Vph) with 60 degrees between them, so the resultant is VRY = 2 Vph cos 30 degrees = sqrt(3) Vph, and it leads VRN by 30 degrees.

Thus VL = sqrt(3) Vph and IL = Iph.

Power P = 3 Vph Iph cos phi = 3 (VL/sqrt3) IL cos phi = sqrt(3) VL IL cos phi. Phasor diagram: three phase voltages 120 degrees apart, with line voltages leading by 30 degrees.

Q11. Derive the relation between line and phase currents in a balanced delta connection. (5 marks)

In delta, the phases are connected end to end in a closed loop and lines are taken from the junctions. Each phase is directly across two lines, so VL = Vph.

By KCL at a junction, the line current is the phasor difference of two phase currents: IR = IRY - IBR. IRY and -IBR are equal in magnitude (Iph) with 60 degrees between them, so IL = 2 Iph cos 30 degrees = sqrt(3) Iph, lagging the corresponding phase current by 30 degrees.

Thus IL = sqrt(3) Iph, VL = Vph and P = 3 Vph Iph cos phi = sqrt(3) VL IL cos phi.

Exam tips

• Always state whether power factor is lagging or leading; a bare number loses marks.
• Show XL, XC, Z, I, pf and powers as separate lines with units; draw the phasor diagram whenever a series or parallel circuit is solved.
• For phasor sums, convert to rectangular form, add, then convert back to polar form; write the final answer as an instantaneous expression if the question gave one.
• Keep your calculator in degree mode for angles and check by recalculating P as I^2 R.
• Resonance is theory only: learn the conditions, derivation of f0, the I versus f curve and the series versus parallel comparison table.
• In three-phase problems, first identify star or delta and write Vph and Iph before calculating line values.
• Derivations of RMS and average values and the star/delta relations are frequent 5-mark questions.

FAQs

Why do we use RMS values in AC?

Because the RMS value produces the same heating effect as an equal DC value, it allows AC power to be calculated with the same formulas as DC. Supply ratings like 230 V and meter readings are RMS.

What is the difference between impedance and admittance?

Impedance Z = R + jX (ohm) is the total opposition to AC and is convenient for series circuits. Admittance Y = 1/Z = G + jB (siemens) measures how easily current flows; admittances of parallel branches simply add, so it is convenient for parallel circuits.

Are numericals on resonance asked?

The MU syllabus says series and parallel resonance are theory only, so expect conditions, derivation of resonant frequency, graphs, Q factor and comparisons rather than numericals.

Is the power formula different for star and delta?

No. Total power of a balanced load is sqrt(3) VL IL cos phi in both cases; only the relations between line and phase quantities differ.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Basic Electrical & Electronics Engineering) and were prepared with AI help and checked.