AEE Electrical Engineering Core · Chapter 2

Transients, Two-Port Networks, Three-Phase Circuits and Power Measurement

What to remember

Inductor current and capacitor voltage cannot change instantly. This gives the initial conditions for every transient problem.
Time constant: τ = L / R for RL and τ = R C for RC. After 5τ the circuit is practically at steady state.
For a balanced three-phase load, P = √3 V_L I_L cosφ. The two-wattmeter method measures this power, and its readings give the power factor.

Transients in first-order circuits

RL circuit, DC step V: i(t) = (V/R) (1 - e^(-t/τ)), τ = L/R. Inductor voltage v_L = V e^(-t/τ).
RC circuit charging: v_C(t) = V (1 - e^(-t/τ)), τ = R C. Charging current i = (V/R) e^(-t/τ).
Discharge: v_C(t) = V0 e^(-t/τ).
At t = τ the response has reached 63.2% of its final value (charging) or has fallen to 36.8% (decay). At 5τ it is above 99%.
Initial conditions at t = 0+: an inductor with zero initial current acts as an open circuit; a capacitor with zero initial voltage acts as a short circuit. At steady state with DC, inductor is a short and capacitor is an open circuit.
Energy stored: inductor ½ L I²; capacitor ½ C V².

Worked example: R = 10 ohm, L = 0.5 H, V = 100 V. τ = 0.5 / 10 = 0.05 s. Final current = 10 A. At t = τ, i = 10 x 0.632 = 6.32 A.

Worked example: R = 1 kilo-ohm and C = 100 microfarad give τ = 0.1 s. A capacitor charging from 0 reaches 63.2% of the supply in 0.1 s.

Second-order (series RLC) transients

Define α = R / (2L) and ω0 = 1 / √(LC).

Case	Condition	Response
Over-damped	α > ω0 (R > 2√(L/C))	Two real roots, slow non-oscillatory decay
Critically damped	α = ω0 (R = 2√(L/C))	Fastest decay with no oscillation
Under-damped	α < ω0 (R < 2√(L/C))	Decaying oscillation at ωd = √(ω0² - α²)

Damping ratio ζ = α / ω0. Zero resistance gives undamped oscillation at ω0.

Worked example: L = 1 H, C = 1 microfarad. Critical resistance = 2 √(L / C) = 2 √(10⁶) = 2000 ohm.

Laplace transform pairs: L{1} = 1/s; L{t} = 1/s²; L{e^(-at)} = 1/(s + a); L{sin ωt} = ω / (s² + ω²); L{cos ωt} = s / (s² + ω²). For circuit elements: resistor R, inductor sL (with initial source L i(0)), capacitor 1/(sC) (with initial source v(0)/s). The initial value theorem is f(0+) = lim s F(s) as s tends to infinity; the final value theorem is f(∞) = lim s F(s) as s tends to 0.

Two-port networks

A two-port has an input port (V1, I1) and an output port (V2, I2), both currents taken into the network.

Parameter set	Equations	Notes
Z (open-circuit)	V1 = Z11 I1 + Z12 I2; V2 = Z21 I1 + Z22 I2	Found with ports open
Y (short-circuit)	I1 = Y11 V1 + Y12 V2; I2 = Y21 V1 + Y22 V2	Found with ports shorted
ABCD (transmission)	V1 = A V2 - B I2; I1 = C V2 - D I2	Used for cascade and transmission lines
h (hybrid)	V1 = h11 I1 + h12 V2; I2 = h21 I1 + h22 V2	Used for transistor models

Reciprocal network: Z12 = Z21, Y12 = Y21, AD - BC = 1, h12 = -h21.
Symmetrical network: Z11 = Z22, Y11 = Y22, A = D, and ΔH = 1 for h.
Connections: series connection adds Z parameters; parallel connection adds Y parameters; cascade connection multiplies ABCD matrices.
Conversions: Z = Y⁻¹; the determinant of ABCD for a reciprocal network is 1.
Open-circuit voltage ratio V2/V1 with output open = 1 / A. Input impedance with output open = A / C.

Worked example: a series impedance Z alone has A = 1, B = Z, C = 0, D = 1. A shunt admittance Y alone has A = 1, B = 0, C = Y, D = 1. Cascading them multiplies these matrices.

Three-phase circuits

Three voltages of equal magnitude, 120° apart in phase. The usual sequence is R-Y-B (positive).
Star connection: V_L = √3 V_ph and I_L = I_ph. The line voltage leads the phase voltage by 30°. A neutral is available.
Delta connection: V_L = V_ph and I_L = √3 I_ph. The line current lags the phase current by 30°.
Power (balanced, either connection): P = √3 V_L I_L cosφ = 3 V_ph I_ph cosφ; Q = √3 V_L I_L sinφ; S = √3 V_L I_L. Here φ is the angle between phase voltage and phase current.
A balanced delta load of impedance Z_ph is equal to a star load of Z_ph / 3.
In a balanced system the neutral current is zero. In a four-wire unbalanced star, neutral current is the phasor sum of the three line currents.
Advantages: constant instantaneous power in a balanced load, smaller conductor for the same power, self-starting motors.

Worked example: balanced delta load, each phase 20 ohm resistive, line voltage 400 V. I_ph = 400 / 20 = 20 A, I_L = 20 √3 = 34.6 A. P = 3 x 20² x 20 = 24 kW.

Worked example: balanced star, V_L = 400 V, each phase 100 ohm resistive. V_ph = 400 / √3 = 231 V; I = 2.31 A; P = 3 x 2.31² x 100 = 1600 W.

Power measurement

Wattmeter: has a current coil (in series) and a pressure (voltage) coil (in parallel). Reading = V I cosφ.
Blondel's theorem: to measure total power in an n-wire system, n - 1 wattmeters are enough, with one line as common return. So two wattmeters measure power in a three-wire three-phase system.
Two-wattmeter method (balanced load): W1 = V_L I_L cos(30° - φ), W2 = V_L I_L cos(30° + φ), total P = W1 + W2. Power factor angle: tanφ = √3 (W1 - W2) / (W1 + W2).

Power factor	Wattmeter readings
Unity (φ = 0)	Equal
0.5 lagging (φ = 60°)	One reads zero, the other reads the full power
Below 0.5	One reading is negative (reverse its connections)
Zero (pure reactance)	Equal in magnitude, opposite in sign

Worked example: W1 = 3000 W and W2 = 0. P = 3000 W. tanφ = √3 x 3000 / 3000 = √3, so φ = 60° and pf = 0.5.

Worked example: W1 = 4000 W, W2 = 2000 W. P = 6000 W. tanφ = √3 x 2000 / 6000 = 0.577, so φ = 30° and pf = 0.866.

Reactive power by one wattmeter on a balanced load: connect the current coil in one line and the pressure coil across the other two lines; Q = √3 x wattmeter reading.
Energy meter: induction type meter measures kWh. The meter constant is revolutions per kWh; a pure-LED check uses pulses per kWh.
Instrument transformers: a current transformer (CT) steps current down for meters (secondary usually 1 A or 5 A and never left open); a potential transformer (PT) steps voltage down (secondary commonly 110 V line). Burden is the load on the secondary, measured in VA.
Moving coil (PMMC) instruments read DC average; moving iron instruments read RMS on both AC and DC; dynamometer type is used in wattmeters.
Measurement of power factor: a power factor meter or computation from the two-wattmeter readings.

Exam traps

Current in an inductor and voltage across a capacitor are continuous; the other quantities can jump.
τ = L/R for RL but τ = RC for RC; do not swap.
Critical damping gives the fastest non-oscillatory response, not the slowest.
Z parameters use open circuits; Y parameters use short circuits.
Cascade multiplies ABCD; parallel adds Y; series adds Z.
In delta, line current is √3 times phase current; in star, line voltage is √3 times phase voltage.
Power formula uses φ between phase voltage and phase current, not line quantities.
A CT secondary must never be open-circuited; a PT secondary must never be short-circuited.

One-liners

1. Time constant of RL is L/R.
2. A capacitor reaches 63.2% of its final voltage in one time constant.
3. Critical resistance of a series RLC is 2√(L/C).
4. Reciprocity in Z-parameters means Z12 = Z21.
5. For reciprocal networks AD - BC = 1.
6. Star line voltage = √3 x phase voltage.
7. Delta line current = √3 x phase current.
8. Three-phase power P = √3 V_L I_L cosφ.
9. n - 1 wattmeters measure power in an n-wire system.
10. At 0.5 pf lagging, one wattmeter reads zero in the two-wattmeter method.
11. tanφ = √3 (W1 - W2) / (W1 + W2).
12. A current transformer secondary must not be left open.

Practice questions

The time constant of a series RL circuit is
1. L / R
2. 1 / (R L)
3. R / L
4. R L
Answer
A. L / R
The inductor current rises with time constant L / R.
The time constant of a series RC circuit is
1. R / C
2. R C
3. 1 / (R C)
4. C / R
Answer
B. R C
The capacitor voltage follows time constant RC.
Which quantity cannot change instantly in a circuit?
1. Inductor voltage
2. Resistor voltage
3. Capacitor current
4. Inductor current
Answer
D. Inductor current
Inductor current and capacitor voltage are continuous.
A capacitor charging through a resistor reaches what share of its final voltage in one time constant?
1. 99 percent
2. 63.2 percent
3. 36.8 percent
4. 50 percent
Answer
B. 63.2 percent
v = V (1 - e^-1) = 0.632 V.
At t = 0+ an uncharged capacitor behaves like
1. A resistor of 1 ohm
2. An open circuit
3. An inductor
4. A short circuit
Answer
D. A short circuit
With zero voltage the capacitor initially passes current freely.
At t = 0+ an inductor with zero initial current behaves like
1. A capacitor
2. A resistor
3. An open circuit
4. A short circuit
Answer
C. An open circuit
The current cannot jump, so the inductor opens at the instant.
In steady state with DC, an inductor behaves like
1. A short circuit
2. An open circuit
3. A capacitor
4. A battery
Answer
A. A short circuit
There is no voltage drop across an ideal inductor in DC steady state.
A two-port is reciprocal when
1. Z12 = Z21
2. Z11 = Z22
3. Z11 = Z12
4. Z12 = -Z21
Answer
A. Z12 = Z21
Equality of the transfer impedances gives reciprocity.
For a reciprocal two-port the ABCD parameters satisfy
1. A = D
2. B = C
3. A + D = 1
4. AD - BC = 1
Answer
D. AD - BC = 1
The determinant of the transmission matrix is unity.
In a balanced star connection the line voltage is
1. Root three times smaller than the phase voltage
2. One-third of the phase voltage
3. Root three times the phase voltage
4. Equal to the phase voltage
Answer
C. Root three times the phase voltage
V_L = sqrt(3) V_ph in star.
In a balanced delta connection the line current is
1. Equal to the phase current
2. Root three times the phase current
3. Three times the phase current
4. Half the phase current
Answer
B. Root three times the phase current
I_L = sqrt(3) I_ph in delta.
The power in a balanced three-phase system is
1. 3 x V_L x I_L
2. Root three x V_ph x I_L
3. Root three x V_L x I_L x cos phi
4. V_L x I_L x cos phi
Answer
C. Root three x V_L x I_L x cos phi
P = sqrt(3) V_L I_L cos(phi).
For a series RL circuit with R = 10 ohm and L = 0.5 H, the time constant is
1. 20 s
2. 0.05 s
3. 0.5 s
4. 5 s
Answer
B. 0.05 s
tau = L / R = 0.5 / 10 = 0.05 s.
A circuit has R = 1 kilo-ohm and C = 100 microfarad. The time constant is
1. 10 s
2. 0.01 s
3. 1 s
4. 0.1 s
Answer
D. 0.1 s
tau = R C = 1000 x 100 x 10^-6 = 0.1 s.
A 100 V DC step is applied to R = 10 ohm, L = 0.5 H. The current at t = one time constant is about
1. 6.32 A
2. 10 A
3. 3.68 A
4. 5 A
Answer
A. 6.32 A
Final 10 A, and i = 10 x 0.632 = 6.32 A.
The critical resistance of a series RLC circuit with L = 1 H and C = 1 microfarad is
1. 500 ohm
2. 2000 ohm
3. 1000 ohm
4. 4000 ohm
Answer
B. 2000 ohm
R = 2 sqrt(L / C) = 2 x 1000 = 2000 ohm.
A series RLC circuit has R = 500 ohm, L = 1 H and C = 1 microfarad. The response is
1. Under-damped
2. Over-damped
3. Critically damped
4. Undamped
Answer
A. Under-damped
Critical resistance is 2000 ohm; R below it gives an under-damped response.
A balanced delta load has 20 ohm resistive per phase on 400 V line voltage. The line current is about
1. 60 A
2. 11.5 A
3. 20 A
4. 34.6 A
Answer
D. 34.6 A
I_ph = 400 / 20 = 20 A; I_L = 20 x 1.732 = 34.6 A.
For the same delta load (20 ohm per phase, 400 V) the total power is
1. 8 kW
2. 48 kW
3. 24 kW
4. 12 kW
Answer
C. 24 kW
P = 3 x I_ph squared x R = 3 x 400 x 20 = 24 kW.
A balanced star load of 100 ohm per phase is on a 400 V line supply. The total power is about
1. 2770 W
2. 1600 W
3. 533 W
4. 4800 W
Answer
B. 1600 W
I = 231 / 100 = 2.31 A; P = 3 x 2.31 squared x 100 = 1600 W.
A balanced star system has a line voltage of 400 V. The phase voltage is about
1. 231 V
2. 400 V
3. 133 V
4. 693 V
Answer
A. 231 V
400 / 1.732 = 231 V.
A balanced load takes 10 A at 400 V line voltage with 0.8 power factor. The power is about
1. 9.6 kW
2. 3.2 kW
3. 4.0 kW
4. 5.54 kW
Answer
D. 5.54 kW
P = 1.732 x 400 x 10 x 0.8 = 5543 W.
A balanced three-phase load draws 5 A at 400 V line. The apparent power is about
1. 2.0 kVA
2. 10.4 kVA
3. 3.46 kVA
4. 6.0 kVA
Answer
C. 3.46 kVA
S = 1.732 x 400 x 5 = 3464 VA.
In the two-wattmeter method the readings are 3000 W and 0 W. The power factor is
1. 0.707
2. 0.866
3. 1.0
4. 0.5
Answer
D. 0.5
tan phi = sqrt(3) x 3000 / 3000 = sqrt(3), so phi = 60 degrees and pf = 0.5.
The two wattmeters read 4000 W and 2000 W. The total power is
1. 6000 W
2. 2000 W
3. 3000 W
4. 8000 W
Answer
A. 6000 W
The total is the sum of the two readings.
For readings of 4000 W and 2000 W the power factor is
1. 0.5
2. 0.866
3. 1.0
4. 0.707
Answer
B. 0.866
tan phi = sqrt(3) x 2000 / 6000 = 0.577, phi = 30 degrees, pf = 0.866.
In the two-wattmeter method both readings are 2 kW. The power factor and total power are
1. 0.5 and 2 kW
2. 0.866 and 4 kW
3. Unity and 4 kW
4. Unity and 2 kW
Answer
C. Unity and 4 kW
Equal readings mean zero phase angle, so pf = 1, total = 4 kW.
At power factor below 0.5, in the two-wattmeter method
1. Both read zero
2. Both read equal
3. One reads exactly half
4. One wattmeter reads negative
Answer
D. One wattmeter reads negative
When phi exceeds 60 degrees, W2 becomes negative.
By Blondel's theorem, the number of wattmeters needed for an n-wire system is
1. n + 1
2. n - 1
3. 2 n
4. n
Answer
B. n - 1
One line is taken as a common return.
A single wattmeter used for reactive power on a balanced load reads 2 kW. The reactive power is about
1. 3.46 kVAR
2. 2 kVAR
3. 1.15 kVAR
4. 6 kVAR
Answer
A. 3.46 kVAR
Q = sqrt(3) x 2 = 3.46 kVAR.
The Laplace transform of e^(-3t) is
1. 1 / (s - 3)
2. 3 / s
3. 1 / (s + 3)
4. s / (s + 3)
Answer
C. 1 / (s + 3)
L{e^(-at)} = 1 / (s + a).
An inductor of 2 H carries 3 A. The stored energy is
1. 9 J
2. 18 J
3. 3 J
4. 6 J
Answer
A. 9 J
W = 0.5 x 2 x 9 = 9 J.
A 100 microfarad capacitor is charged to 10 V. The stored energy is
1. 50 mJ
2. 10 mJ
3. 1 mJ
4. 5 mJ
Answer
D. 5 mJ
W = 0.5 x 100 x 10^-6 x 100 = 5 mJ.
A reciprocal two-port has A = 2, B = 3 and C = 1. The parameter D is
1. 1
2. 2
3. 3
4. 0.5
Answer
B. 2
AD - BC = 1 gives 2D - 3 = 1, so D = 2.
A current transformer's secondary should never be
1. Short-circuited
2. Connected to an ammeter
3. Left open-circuited
4. Earthed
Answer
C. Left open-circuited
An open secondary develops a dangerously high voltage.
Consider: 1. Z parameters are found with ports open-circuited. 2. Y parameters are found with ports short-circuited.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both statements are correct.
Consider: 1. Cascaded two-ports multiply ABCD matrices. 2. Series-connected two-ports add Y parameters.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Series connection adds Z parameters; parallel adds Y.
Consider: 1. In star, line current equals phase current. 2. In delta, line voltage equals phase voltage.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both statements are correct.
Consider: 1. The time constant of an RC circuit is R / C. 2. After about 5 time constants a circuit is nearly at steady state.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
The RC time constant is the product R C.
Consider: 1. A critically damped circuit gives the fastest non-oscillatory response. 2. An under-damped circuit has oscillation in the response.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both statements describe the damping cases correctly.
Consider: 1. In a balanced system the neutral current is zero. 2. Three-phase balanced power is pulsating.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Instantaneous power of a balanced load is constant.
Match in order: Z parameters, Y parameters, ABCD, h parameters
1. Hybrid, transmission, short-circuit admittance, open-circuit impedance
2. Transmission, hybrid, open-circuit impedance, short-circuit admittance
3. Short-circuit admittance, open-circuit impedance, hybrid, transmission
4. Open-circuit impedance, short-circuit admittance, transmission, hybrid
Answer
D. Open-circuit impedance, short-circuit admittance, transmission, hybrid
These are the standard names of the parameter sets.
Match in order: Over-damped, Critically damped, Under-damped, Undamped
1. alpha = w0, alpha > w0, R = 0, alpha < w0
2. R = 0, alpha < w0, alpha = w0, alpha > w0
3. alpha < w0, alpha = w0, alpha > w0, R = 0
4. alpha > w0, alpha = w0, alpha < w0, R = 0
Answer
D. alpha > w0, alpha = w0, alpha < w0, R = 0
The damping condition decides the response type.
Three-phase power in a balanced load is constant in time. A key advantage of this is
1. Lower voltage
2. Smooth torque in motors
3. Zero neutral wire needed always
4. Higher line frequency
Answer
B. Smooth torque in motors
Constant instantaneous power gives steady motor torque.
A balanced delta load of 30 ohm per phase is replaced by an equivalent star load. Each star arm is
1. 17.3 ohm
2. 90 ohm
3. 30 ohm
4. 10 ohm
Answer
D. 10 ohm
Z(star) = Z(delta) / 3 = 10 ohm.

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1.The boiling and freezing points of water are exactly _____ degrees apart on the Fahrenheit scale.
2.The center of gravity of a semi-circle lies at a distance of __________ from its base measured along the vertical radius.
3.The crest of a wave is the point where the disturbance is at:
4.The critical angle for diamond is approximately _______.
5.The coordination number of atoms in a body-centered cubic (BCC) lattice is:
6.The atomic radius of a hydrogen atom is:
7.The atomic number of sulphur is-
8.The atomic number of phosphorus atom is:
9.The average temperature of human body is ______.
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