# In a delta connection, if the magnitude of phase voltage is Vp and the magnitude of phase current is Ip, then the expression for the complex power is:Assume load to be capacitive

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## Options:

1. $${S^*} = \sqrt 3 {V_P}{I_P}\angle \phi \;VA$$

2. $${S^*} = \sqrt 3 {V_P}{I_P}\angle - \phi \;VA$$

3. $${S^*} = {V_P}{I_P}\;\angle \phi \;VA$$

4. $${S^*} = 3{V_P}{I_{P\;}}\angle - \phi \;VA$$

### Correct Answer: Option 4 (Solution Below)

This question was previously asked in

UPPCL AE EE Previous Paper 3 (Held On: 4 November 2019 Shift 1)

## Solution:

Total apparent power (S) of 3 ϕ circuit is given by

$$S = \sqrt 3 \;{V_L}{I_L}$$

Total complex power is given by

$${S^*} = \sqrt 3 \;{V_L}I_L^*$$

Where

VL = Line voltage

IL = Line current

IP = Phase current

VP = Phase voltage

For star connection:

VL = √3 Vp

IL = IP

For delta connection:

VL = Vp

IL = √3 IP

Application:

The load is capacitive. So that current is leading the voltage by an angle ϕ.

Generally, we take voltage as a reference. So that ‘ϕ’ is positive.

Now total complex power is expressed as

$${S^*} = \sqrt 3 \;{V_L}I_L^* = \sqrt 3 \;|{V_L}\left| {\angle 0^\circ \;|{I_L}} \right|\angle {\phi ^*}$$

$${S^*} = \sqrt 3 \;|{V_L}\left| {|{I_L}} \right|\angle {\phi ^*}\;\;VA$$

$${S^*} = 3\;{V_P}{I_P}\;\angle - \phi \;VA$$

The above expression is of polar form

Note:

The rectangular form of complex power is

$${S^*} = 3\;{V_P}{I_P}\;cos\phi - j\;3{V_P}{I_P}\;sin\phi$$

Where,

P = 3 VP IP cosϕ = Active power

Q = 3 VP IP sinϕ = Reactive power
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