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QUESTION: 1

A 10 kVA, 400/200 V, 1-phase transformer with 2% resistance and 2% leakage reactance. It draws steady short circuit current at angle of

Solution:

The power factor angle will be atan(x/r)=45°.

QUESTION: 2

While conducting open circuit test and short circuit test on a transformer, status of low-voltage and high-voltage windings will be such that in

Solution:

In conducting short circuit test, l.v. winding is short circuited. In OC test h.v. is open circuited.

QUESTION: 3

While conducting testing on the single phase transformer, one of the student tries to measure the resistance by putting an ammeter across one terminal of primary and other to secondary, the reading obtained will be

Solution:

As the primary and secondary are physically isolated, the impedance will be infinite for not electrically connected circuit.

QUESTION: 4

If the per unit leakage impedance for the primary of a transformer is ‘x’ on the given rated base value. If the voltage and volt-amperes are doubled, then the changed per unit impedance will be

Solution:

pu(new base)=(x)*(MVA(new)/MVA(old))*(kV(old)/kV(new))^2

=x*2*(1/4)

=0.5x.

QUESTION: 5

The voltage regulation for transformer is given by

Solution:

Voltage regulation is the change in secondary voltage with secondary rated voltage.

QUESTION: 6

While estimating voltage regulation of a transformer, keeping

Solution:

V.R. is calculated keeping the primary constant because then the core flux will change and the change of secondary voltage can not be fixed.

QUESTION: 7

A 200/400 V single phase transformer has leakage impedance z= r+jx. Then we can expect zero voltage regulation at power factor of

Solution:

ZVR occurs at the leading pf of load at x/r.

QUESTION: 8

A 200/400 V single phase transformer has leakage impedance z= r+jx. Then we can expect magnitude of load pf of ____ at zero voltage regulation.

Solution:

Cosθ = x/z for the transformer at zero V.R.

QUESTION: 9

If the pu impedance of a single phase transformer is 0.01+j0.05, then its regulation at p.f. of 0.8 lagging will be

Solution:

V.R. = (r(pu)*cosθ+x(pu)*sinθ)*100 % = (0.01*0.8 + 0.05*0.6)*100 = 3.8%

QUESTION: 10

The transformer phasor diagram under the short circuit can be identified as

Solution:

For the short-circuit condition of a transformer, voltage across the secondary will be voltage drop across winding only.

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