1. The document contains 13 questions related to 3-phase circuits with detailed answers. 2. Key concepts covered include sinusoidal waveforms, star and delta connections, line and phase voltages/currents, power calculations, and balanced vs unbalanced loads. 3. Formulas are derived for relationships between line and phase quantities under balanced conditions. Effects of unbalanced loads such as neutral shift are also explained.

1. Unit No. 03 Questions and Solution
Que.
No.
Bloom’s
Level
Questions Relevance
to
CO
1 2 Draw sinusoidal waveform of 3 phase emf and indicate the phase sequence.
COI301.3
Marking Scheme waveform – 1.5 Mark, Phase sequence-0.5 Mark
Answer
Phase sequence is R-Y-B
2 1 State any four advantages of polyphase circuits over single phase circuit.
COI301.3
Marking Scheme 1 mark for each of any four = 4 marks
Answer Advantages of polyphase (3-phase) circuits over Single-phase circuits:
i)Three-phase transmission is more economical than single-phase transmission. It requires
less copper material.

2. ii) Parallel operation of 3-phase alternators is easier than that of single-phase alternators.
iii) Single-phase loads can be connected along with 3-ph loads in a 3-ph system.
iv) Instead of pulsating power of single-phase supply, constant power is obtained in 3-phase
system.
v) Three-phase induction motors are self-starting. They have high efficiency, better power
factor and uniform torque.
vi) The power rating of 3-phase machine is higher than that of 1-phase machine of the same
size.
vii) The size of 3-phase machine is smaller than that of 1-phase machine of the same power
rating.
viii) For same power rating, three-phase motors are cheaper than the single-phase motors.
3 2 Compare star and delta connection.( Any four points) COI301.3
Marking Scheme 1 mark for each of any four points = 4 marks

3. Answer
Parameter Star Connection Delta Connection
Basic definition One terminal of each of the three
branches are connected together to
form a common point. Such a
connection is known as Star
Connection
The three branches of the network are
connected in such a way that it forms a
closed loop. Such a connection is known as
Delta Connection
Connection of
terminals
The similar ends of the three coils
are connected together to form a
common point.
The end of each coil is connected to the
starting point of the other coil that means
the opposite terminals of the coils are
connected together to form a closed loop.
Neutral point Neutral or the star point exists in the
star connection.
Neutral point does not exist in the delta
connection.
Relation between
line and phase
current
Line current is equal to the Phase
current.
Line current is equal to √3 times the Phase
Current.
Relation between
line and phase
voltage
Line voltage is equal to √3 times the
Phase Voltage
Line voltage is equal to the Phase voltage.
Diagram
4 2 Delta connected induction motor is supplied by 3 phase 400V, 50 Hz supply the line current is 43.03 amp
and the total power from the supply is 24 kW. Find the resistance and reactance per phase of the motor.
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Marking Scheme Each step 1 Mark

4. Answer

5. 5 3 In a 3 phase star connected system, derive the relationship VL = √ 3 Vph.
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Marking Scheme 2 marks for phasor diagram, 3 marks for stepwise explanation, 1 mark for final answer
Answer

6. 6 2 Define: Phase sequence and write equations for instantaneous value of 3-ph voltages.
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Marking Scheme Definition - 1 mark, Mathematical Expression- 1 Mark
Answer
Phase sequence:
Phase sequence is defined as the order in which the voltages (or any other alternating quantity) of the three phases
attain their positive maximum values. In the following waveforms, it is seen that the R-phase voltage attains the
positive maximum value first, and after angular distance of 120o, Y-phase voltage attains its positive maximum
and further after 120o, B-phase voltage attains its positive maximum value. So the phase sequence is R-Y-B.
7 2 A star connected 3-ph load is supplied from 3-ph, 415V, 50Hz supply. If the
line current is 20 A and total power taken from supply is 10 kW, then
determine:
(i) Load resistance and reactance per phase.
(ii) Load power factor
(iii) Total 3-phase reactive power
Marking Scheme Each step 1 mark

7. Answer
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8 2,3 Explain ‘Neutral Shift’ in case of 3-phase star-connected unbalanced load.

8. Marking Scheme 1 Mark for phasor diagram, 3 Marks for explanation
Answer
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9. 9 2,3 Three equal impedances having R = 20 Ω in series with C = 50 μF are connected in delta across 415 V, 3-ph,
50 Hz AC supply. Determine:
i) Impedance per phase
ii) Phase and line currents
iii) Total 3-ph power consumed by load
Marking Scheme 1 Mark for each bit = 4 Marks
Answer
COI301.3

10. 10 3 Draw complete phasor diagram of voltages & currents for balanced deltaconnected
load and prove the relation between:
(i) Line current and Phase current (ii) Line voltage and Phase voltage
Marking Scheme 1 Mark for each bit = 6 Marks
Answer
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11. 11 2 Draw delta connected load. State relation between:
i) Line voltage and phase voltage.
ii) Line current and phase current.
Marking Scheme 2 Marks for neat labeled diagram and 2 Marks for relations
Answer
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12 2 A Balanced 3-phase star connected load consist of three resistances each of four Ohm’s connected to 400 V, 3 phase
50 Hz supply, find: (i) Phase voltage, (ii) Phase current, (iii) Line current, (iv) Power consumed.

12. Marking Scheme 1 Mark for each bit = 4 Marks
Answer Given Data:
Load is star connected.
Line voltage=V𝐿 = 400 Volt
Frequency = f = 50 Hz
Resistance per phase = Rph = 4 Ω
1) Phase voltage: Vph =VL/ = 400/ = 230.94 Volt
2) Phase current: Iph = Iph / Rph = 230.94/4 = 𝟓𝟕. 𝟕𝟑 𝐚𝐦𝐩𝐞𝐫𝐞.
3) Line Current: IL = Iph = 𝟓𝟕. 𝟕𝟑 𝐚𝐦𝐩𝐞𝐫𝐞
4) Power consumed:
Power consumed = P = 3 × Vph × Iph × cos∅.
= 3 × 230.94 × 57.73 × 1 = 𝟑𝟗. 𝟗𝟗 𝐊𝐖
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13 State difference between 3-phase balanced and 3-phase unbalanced load.
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Marking Scheme 2 Marks
Answer Balanced load in 3 phase system is a condition where all three phases (lines) carry same magnitude of
current, with evenly spaced phase difference. If the load is star connected, with neutral as return path, this
neutral will carry no current. This is so since vector sum of all three phase currents meeting at common
neutral point is zero.
Unbalanced load makes the lines / phases to carry different current magnitudes, and sum total of these at
neutral point is not zero. Load in each phase is different, carrying ts own current. Neutral in this case carries
the net unbalanced current.