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QUESTION: 1

Three equal resistances connected in star take a line current of 10 A when fed from 400 V, 50 Hz source. If the load resistances are reconnected in delta, the line current would be

Solution:

or,

QUESTION: 2

The minimum number of wattmeters required to measure the real power in an n-phase system with unbalanced load is

Solution:

QUESTION: 3

The input to the 3-phase, 50 Hz circuit shown in figure below is 100 V. For a phase sequence of ABC, the wattmeters would read

Solution:

Wattmeter readings are:

W_{1} = V_{L}I_{L} cos (30 + φ)

and W_{2} = \/_{L}I_{L} cos (30 - φ)

Here,

and φ = 30°

Now,

and

QUESTION: 4

Three-phase power can be measured by two-wattmeter method in case of

Solution:

QUESTION: 5

In the case of power measured by two-wattmeter method in a balanced 3-phase system with a pure capacitive load

Solution:

W_{1} = V_{L }I_{L} cos (30° + φ)

and W_{2 }= V_{L }I_{L} cos(30° - φ)

When the load is purely capacitance, then φ = - 90°.

So, W_{1} = V_{L}I _{L }cos(30°- 90°)

and W_{2} = V_{L}I_{L} cos (30° + 90°)

Hence, |W_{1}| = |W_{2}| but W_{1} = - W_{2}

QUESTION: 6

Measurement of power factor of a 3-phase system by two-wattmeter method can be obtained in case of

Solution:

QUESTION: 7

Two wattmeters used to measure power of a 3-phase balanced load reads W_{1}, and W_{2}. The reactive Dower drawn bv the load is

Solution:

We know that:

(W_{1 }- W_{2}) = V_{L}I_{L}sin φ = Q_{1 - φ}

So, Q_{3 - φ} = √3(W_{1} - W_{2})

QUESTION: 8

Two wattmeters used to measure power of a 3-phase balanced load reads W_{1} and W_{2}. The' reactive Dower drawn bv the load is

Solution:

We know that:

(W_{1} - W_{2}) = V_{L }I_{L} sin φ = Q_{1-φ}

so, Q_{3 - φ }= √3(W_{1} - W_{2})

QUESTION: 9

A three-phase 220 V supply is applied to a balanced Δ-connected three phase load. The phase current being I_{ab} =. 10∠-30°A as shown in the given figure, The total power received by the Δ-load is

Solution:

I_{ab} = 10∠-30°A

So, I_{bc }= 10∠-150°A

and I_{ca} = 10∠-270° A

Phase current,

I_{a} = I_{ab }- I_{ca} = (10∠-30° - 10∠-270°) A

= 10[cos 30° - jsin 30° - cos 270° + jsin 270°]

= 10[0.866 - j0.5 - 0 - j1]

= 8.66 - j15 = 17.32 ∠-60° A

∴ I_{a} = 17.32 ∠-60°A

∴ Total power received

= √3 V_{L}I_{L} cos φ_{L}

= 3V_{P}I_{P} cos φ_{ph}

= 3 x 220 x 10 x cos 30°

= 5715W

QUESTION: 10

The magnitude of line current in the circuit shown below is

Solution:

Impedance per phase

= 1 + j1 + 3 + j5

= (4 + j6) Ω

QUESTION: 11

A three-phase, 400 V ac system is connected across a delta connected load having load resistance per phase of 200 Ω. If one of the phases of the delta connected load experiences an open circuit, then power consumed by the load will be

Solution:

Initially, in closed - Δ, power consumed by the load is

= 2400W

Let, P_{2} be the power consumed in open delta.

then,

QUESTION: 12

Assertion (A): Generated voltage or current wave have no odd harmonics.

Reason (R): The field system and armature coils of the generators are all symmetrical and so generates mostly symmetrical voltage wave.

Solution:

QUESTION: 13

Assertion (A): RMS value of line voltage is more than √3 times of the rms value of phase voltage.

Reason (R): In a star-connected system, the line voltage contains no triple frequency harmonic content.

Solution:

QUESTION: 14

Assertion (A): In a balanced 3-phase star connected system, line voltages are 30° ahead of the respective phase voltages.

Reason (R): The line voltages are √3 times of the respective phase voltages,

Solution:

QUESTION: 15

Match the List - I (Power factor) with List - II (Wattmeter readings) for the measurement of power in a three-phase system using two-wattmeter method and select the correct answer using the codes given below the lists:

List-I

A. 0.5 lag

B. Unity

C. Zero

D. 0.866

List-II

1. W_{1} = W_{2}

2. W_{1} = 2W_{2}

3. W_{1} > 0, W_{2} = 0

4. W_{1} > 0, W_{2} < 0

Codes:

Solution:

W_{1} = V_{L}I_{L} cos (30 + φ)

and W_{2} = V_{L}I_{L} cos (30 - φ)

By finding φ from cos φ, we can find the values of W_{1} and W_{2}.

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