Ques 2. In a 1-phase transformer, the copper loss at full load is 600 watts. A half of the full load the copper loss will be

150 watts✔

75 watts

600 watts

300 watts

Copper loss is proportional to the square of load current. At half load, load current becomes half as voltage remains same, so the copper loss will become (1/2)^{2} i.e 1/4 times of full load copper loss.

So (Pcu)_{full load }= 600 watts

(Pcu)_{ half load} = 600/4 = 150 watts

Ques 3. An autotransformer used with sodium vapour lamp should have high

Winding resistance

Leakage Reactance of winding✔

VA rating

Transformation Ratio

A sodium vapour lamp is also called as the cold-cathode low-pressure lamp which gives high luminous output about three times higher than the other lamp.

For starting discharge through the lamp, it is essential that the striking voltage should be higher than the normal working voltage of the lamp. This high voltage is taken from a high reactance transformer or auto-transformer which has poor voltage regulation. Thus when discharge in the lamp takes place, the lamp current increase due to the decrease in the resistance of the gas in the tube and output voltages of the auto-transformer fall. The lamp then continues to operate normally.

Ques 4. In an auto-transformer, the number of turns in primary winding is 210 and in secondary winding is 140. If the input current is 60 A, the current in output and in common winding are respectively.

40 A, 20A

40A, 100 A

90A, 30 A✔

90A, 150A

Auto-transformer ratio is given as

I1/I2 = N2/N1

60/I2 = 140/210

I2 = 90A

Current in common Winding Ic

Ic = (N2 – N1)I1/N2

= 60 x (210 – 140)/140

= 30 A

Ques 5. A 3-phase transformer has its primary connected in delta and secondary in star. Secondary to primary turns ratio per phase is 6. For a primary voltage of 200 V, the secondary voltage would be

2078 V✔

693 V

1200 V

58 V

Transformer ratio is given by

N1/N2 = V1/V2

0r N2/N1 = V2/V1

6/1 = V2/200

V2 = 1200 volt per phase

For star connection line voltage is

V_{L} = √3 x 1200 = 2078.46 V

Ques 6. A resistance and another circuit element are connected in series across a dc voltage V. The voltage and zero after time. The other element is pure

Capacitance

Both (a) and (c)

Resistance

Inductance✔

Why inductor behave as short circuit for DC voltage?

An inductor has a reactance equivalent XL = 2πfL.Since a DC supply will have frequency = 0, the reactance is 0 and the inductor would be short.

Detail explanation

Imagine a circuit of three elements in series – an input voltage source, inductor and resistive load. The input source has been producing some constant (DC) voltage; If an inductor is connected with DC source, it will not act as inductor but acts like as a simple resistor. The current through it will depend on the resistance of the inductor. Now increase the input voltage to a higher value.

The inductor immediately converts the voltage change to an equivalent “antivoltage” and applies it contrary to the input voltage change, the voltage “produced” by the inductor “jumps” with a magnitude equal to the input voltage change. Thus we have two voltage sources (an “original” and “cloned”:) contrary connected in series and neutralizing each other; as a result, the total (effective) voltage and accordingly the current do not change. After that, the current will increase exponentially (1 – e^{-t}) and reach a constant value. Current will be maximum and inductor will behave like a short circuit.

Ques 7. For RLC series resonance the current is

Minimum at leading P.F

Minimum at Lagging P.F

Maximum at unity P.F

Maximum at leading P.F✔

The total impedance of the series LCR circuit is given as

Z = R + j (X_{1} – X_{²})

where X_{1} is inductive reactance

and X_{2} is capacitive reactance.

At a particular frequency (resonant frequency), we find that X_{1}=X_{2 }because resonance of a series RLC circuit occurs when the inductive and capacitive reactances are equal in magnitude but cancel each other because they are 180 degrees apart in phase. Therefore, the phase angle between voltage and current is zero and the power factor is unity.

Thus, at the resonant frequency, the net reactance is zero because X_{1}=X_{2}. The circuit impedance Z becomes minimum and is equal to the resistance R. “Since the impedance is minimum, the current will be maximum”.

Ques 8. A series RLC circuit resonance at 1 MHZ. At frequency 1.1 MHZ the circuit impedance will be

Resistive

Will depend on the relative amplitude of RLC

Capacitive

Inductive✔

For Series RLC circuit if the frequency is greater than the resonance frequency than the circuit behaves as an inductive circuit.

Ques 9. The equivalent resistance between terminal X and Y of the network shown is

8Ω

100/3Ω

40/3Ω✔

20/9Ω

As the given circuit is Wheatstone bridge balance circuit

Therefore according to balanced equation of Wheatstone bridge

Req = 20||40

= (20 x 40)/(20 + 40)

= 40/3Ω

Ques 10. Application of Thevenin’s theorem in a circuit result in

An ideal voltage source

An ideal current source

A current source and an impedance in parallel

A voltage source and an impedance in series✔

The Thevenin equivalent resistance Rs is viewed from the open terminals A and B is given as. As per Thevenin theorem, when resistance RL is connected across terminals A and B, the network behaves as a source of voltage Vs and internal resistance RT and this is called Thevenin equivalent circuit.

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