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
or,
The minimum number of wattmeters required to measure the real power in an nphase system with unbalanced load is
The input to the 3phase, 50 Hz circuit shown in figure below is 100 V. For a phase sequence of ABC, the wattmeters would read
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
Threephase power can be measured by twowattmeter method in case of
In the case of power measured by twowattmeter method in a balanced 3phase system with a pure capacitive load
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}
Measurement of power factor of a 3phase system by twowattmeter method can be obtained in case of
Two wattmeters used to measure power of a 3phase balanced load reads W_{1}, and W_{2}. The reactive Dower drawn bv the load is
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})
Two wattmeters used to measure power of a 3phase balanced load reads W_{1} and W_{2}. The' reactive Dower drawn bv the load is
We know that:
(W_{1}  W_{2}) = V_{L }I_{L} sin φ = Q_{1φ}
so, Q_{3  φ }= √3(W_{1}  W_{2})
A threephase 220 V supply is applied to a balanced Δconnected three phase load. The phase current being I_{ab} =. 10∠30°A as shown in the given figure, The total power received by the Δload is
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
The magnitude of line current in the circuit shown below is
Impedance per phase
= 1 + j1 + 3 + j5
= (4 + j6) Ω
A threephase, 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
Initially, in closed  Δ, power consumed by the load is
= 2400W
Let, P_{2} be the power consumed in open delta.
then,
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.
Assertion (A): RMS value of line voltage is more than √3 times of the rms value of phase voltage.
Reason (R): In a starconnected system, the line voltage contains no triple frequency harmonic content.
Assertion (A): In a balanced 3phase 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,
Match the List  I (Power factor) with List  II (Wattmeter readings) for the measurement of power in a threephase system using twowattmeter method and select the correct answer using the codes given below the lists:
ListI
A. 0.5 lag
B. Unity
C. Zero
D. 0.866
ListII
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:
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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