Q41: The star-delta transformer shown above is excited on the star side with balanced, 4-wire, 3-phase, sinusoidal voltage supply of rated magnitude. The transformer is under no load condition.
With S2 closed and S1 open, the current waveform in the delta winding will be (2009)
(a) a sinusoid at fundamental frequency
(b) flat-topped with third harmonic
(c) only third-harmonic
(d) none of these
Ans: (c)
Sol: As S1 is open, third-harmonic current can not flow in star side, but S2 is closed, therefore delta winding provides path for third harmonic current to flow.
Q42: The star-delta transformer shown above is excited on the star side with balanced, 4-wire, 3-phase, sinusoidal voltage supply of rated magnitude. The transformer is under no load condition.
With both S1 and S2 open, the core flux waveform will be (2009)
(a) a sinusoid at fundamental frequency
(b) flat-topped with third harmonic
(c) peaky with third-harmonic
(d) none of these
Ans: (b)
Sol: As S2 is open, third harmonic current can not flow in delta winding. S1 is open, neutral gets isolated, therefore no third harmonic current in star side. Since third harmonic current cannot flow in either side, the magnetizing current Im is almost sinusoidal. T0 satisfy B-H curve, the core, the core flux must then be non-sinusoidal, it is a flat-topped wave. This can be verified by assuming a sinusoidal im and then finding out the ϕ wave shape from ϕ − im relationship.
Q43: The single phase, 50 Hz iron core transformer in the circuit has both the vertical arms of cross sectional area 20 cm2 and both the horizontal arms of cross sectional area 10 cm2. If the two windings shown were wound instead on opposite horizontal arms, the mutual inductance will (2009)
(a) double
(b) remain same
(c) be halved
(d) become one quarter
Ans: (b)
Sol: Mutual inductance is where the flux of two or more coils are linked so that voltage is induced in one coil proportional to the rate of change of current in another.
The mutual inductance that exist between the coils can be greatly increased by positioning them on a common soft iron core or by increasing the number of turns of either coil.
Consider:
The changing flux on coil 1 is proportional to changing current in coil 2.
Similarly time rate of change of magnetic flux ϕ21 in coil 2 is proportional to time rate of change of current in coil 1
We can observe that none of the parameters changes due to placing of coils on horizontal arms. Hence, mutual inductance will remain unchanged.
Q44: The core of a two-winding transformer is subjected to a magnetic flux variation as indicated in the figure.
The induced emf (ers) in the secondary winding as a function of time will be of the form (2008)
(a) (b) (c)
(d) Ans: (a)
Q45: It is desired to measure parameters of 230 V/115 V, 2 kVA, single-phase transformer. The following wattmeters are available in a laboratory:
- W1 : 250 V, 10 A, Low Power Factor
- W2 : 250 V, 5 A, Low Power Factor
- W3 : 150 V, 10 A, High Power Factor
- W4 : 150 V, 5 A, High Power Factor
The Wattmeters used in open circuit test and short circuit test of the transformer will respectively be (2008)
(a) W1 and W2
(b) W2 and W4
(c) W1 and W4
(d) W2 and W3
Ans: (d)
Sol: Open-circuit Test: In open-circuit test, no-load current I0 is very small (it is usually 2-6% of the rated current).
No-load current I0 lags E by slightly less than 90°, so power factor is very low.
Therefore, wattmeter W2 is suitable for the open-circuit test.
Short-circuit Test: In short-circuit test Vsc needed to circulate the full-load current is very low.
Under these conditions, I0 is only about 0.1 to 0.5% of full load current . As I0 is highly lagging but it is very small as compared to sull load current, therefore the power factor is high.
So, wattmeter W3 is suitable for the short-circuit test.
Q46: Three single-phase transformer are connected to form a 3-phase transformer bank. The transformers are connected in the following manner :
The transformer connecting will be represented by (2008)
(a) Yd0
(b) Yd1
(c) Yd6
(d) Yd11
Ans: (b)
Sol: Phasor diagram of primary and secondary voltage.
It is observed from the phasor diagram that phase to a neutral voltage (equivalent star basis) on the delta side lags by −30° to the phase to neutral voltage on the star side. This connection is known as −30° connection or Yd1 connection.
Q47: A single-phase, 50-kVA, 250 V/500 V two winding transformer has an efficiency of 95% at full load, unity power factor. If it is re-configured as a 500 V/750 V auto-transformer, its efficiency at its new rated load at unity power factor will be (2007)
(a) 95.57%
(b) 97.85%
(c) 98.28%
(d) 99.24%
Ans: (c)
Sol: In two winding transformer
Full load
When two winding transformer is connected as step up autotransformer
I2 = 200, V2 = 750
kVA rating of auto transformer
As current through windings and voltage across winding are equal in two-winding transformer and autotransformer. Losses remain same at full load efficiency at upf and full load
Q48: In a transformer, zero voltage regulation at full load is (2007)
(a) not possible
(b) possible at unity power factor load
(c) possible at leading power factor load
(d) possible at lagging power factor load
Ans: (c)
Sol: So, zero voltage regulation is possible for leading power factor.
When, tanϕ = R/X
Q49: A 300 kVA transformer has 95% efficiency at full load 0.8 pf lagging and 96% efficiency at half load, unity pf.
What is maximum efficiency(in %) at unity pf load? (2006)
(a) 95.1
(b) 96.2
(c) 96.4
(d) 98.1
Ans: (b)
Sol: For maximum efficiency at upf
Q50: A 300 kVA transformer has 95% efficiency at full load 0.8 pf lagging and 96% efficiency at half load, unity pf.
The iron loss (Pi) and copper loss (Pc) in kW, under full load operation are (2006)
(a) Pc = 4.12, Pi = 8.51
(b) Pc = 6.59, Pi = 9.21
(c) Pc = 8.51, Pi = 4.12
(d) Pc = 12.72, Pi = 3.07
Ans: (c)
Sol: Iron loss = Pi, it does not depends on the load.
Full load copper loass Pc, fl, it depends on the load
Pcu = x2Pc,fl
where,
x = Fraction of the full load
Rated kVA = S = 300kVA
η = % efficeincy of a transformer
Q51: Two transformers are to be operated in parallel such that they share load in proportion to their kVA ratings. The rating of the first transformer is 500 kVA and its pu leakage impedance is 0.05 pu. If the rating of second transformer is 250 kVA, its pu leakage impedance is (2006)
(a) 0.2
(b) 0.1
(c) 0.05
(d) 0.025
Ans: (c)
Sol: The curernt carried by two transformer are proportional to their ratings if their per-unit impedances on their own rating are equal.
Z2(p.u.) = Z1(p.u.) = 0.05p.u.
Q52: In transformers, which of the following statements is valid ? (2006)
(a) In an open circuit test, copper losses are obtained while in short circuit test, core losses are obtained
(b) In an open circuit test, current is drawn at high power factor
(c) In a short circuit test, current is drawn at zero power factor
(d) In an open circuit test, current is drawn at low power factor
Ans: (d)
Sol: Circuit model in open-circuit test:
In open-circuit test, the transformer draws only exciting current. The exciting current is only magnetizing in nature and is proportional to the sinusoidal flux and in phase with it. This is represnted by Im lagging the induced emf by 90°. However, the presence of eddy-current, and hysteresis, both demad the flow of active power into the system and as a consequence the exciting current I0 has another component Ii in phase with E1. Thus the exciting current lags the induced emf by an angle slightly less than 90° making power factor very low.
Q53: The three limbed non ideal core shown in the figure has three windings with nominal inductances L each when measured individually with a single phase AC source. The inductance of the windings as connected will be (2006)
(a) very low
(b) L/3
(c) 3L
(d) very high
Ans: (a)
Q54: Which three-phase connection can be used in a transformer to introduce a phase difference of 30° between its output and corresponding input line voltages (2005)
(a) Star-Star
(b) Star-Delta
(c) Delta-Delta
(d) Delta-Zigzag
Ans: (b)
Sol: By using above connection, ±30° phase difference can be introduced between output and input line voltage.
Q55: The equivalent circuit of a transformer has leakage reactance X1, X'2 and magnetizing reactance XM. Their magnitudes satisfy (2005)
(a)
(b)
(c)
(d)
Ans: (d)
Sol: A major part of the total flux is confined to the core as mutual flux ϕm linking both primary and secondary, a small amount of flux does leak through paths which lie mostly in aor and link separating the individual windings.
Since reactance ∝ amount of the flux linkage
With S2 closed and S1 open, the current waveform in the delta winding will be (2009)
With both S1 and S2 open, the core flux waveform will be (2009)
T0 satisfy B-H curve, the core, the core flux must then be non-sinusoidal, it is a flat-topped wave. This can be verified by assuming a sinusoidal im and then finding out the ϕ wave shape from ϕ − im relationship.
(a) double
Mutual inductance is where the flux of two or more coils are linked so that voltage is induced in one coil proportional to the rate of change of current in another.
Similarly time rate of change of magnetic flux ϕ21 in coil 2 is proportional to time rate of change of current in coil 1
We can observe that none of the parameters changes due to placing of coils on horizontal arms. Hence, mutual inductance will remain unchanged.
The induced emf (ers) in the secondary winding as a function of time will be of the form (2008)
(b) 
(c)
(d) 
Ans: (a)
The transformer connecting will be represented by (2008)
It is observed from the phasor diagram that phase to a neutral voltage (equivalent star basis) on the delta side lags by −30° to the phase to neutral voltage on the star side. This connection is known as −30° connection or Yd1 connection.
In two winding transformer
Full load
When two winding transformer is connected as step up autotransformer
I2 = 200, V2 = 750
As current through windings and voltage across winding are equal in two-winding transformer and autotransformer. Losses remain same at full load efficiency at upf and full load
So, zero voltage regulation is possible for leading power factor.
Circuit model in open-circuit test:
(a) very low
By using above connection, ±30° phase difference can be introduced between output and input line voltage.
A major part of the total flux is confined to the core as mutual flux ϕm linking both primary and secondary, a small amount of flux does leak through paths which lie mostly in aor and link separating the individual windings.
