Given data:

The voltage phasor is,

v(t)=169.7sin(377t+45°) V (1)

The current phasor is,

i(t)=5.567sin(377t) A (2)

Formula used:

Write the expression to find the complex power S.

S=Vrms(Irms)* (3)

Here,

Vrms is the root mean square value of voltage, and

Irms is the root mean square value of current.

Write the expression for complex power S.

S=P+jQ (4)

Here,

P is the real power or average power, and

Q is the reactive power.

Calculation:

From equation (1), the rms value of the voltage is

Vrms=169.72∠45°      {∵Vm=169.7, θv=45°}=120∠45° V

From equation (2), the rms value of the current is

Irms=5.6572∠0°      {∵Im=5.657, θi=0°}=4∠0° A

Substitute 220∠45° V for Vrms and 4∠0° A for Irms in equation (3) to find the complex power S in VA.

S=220∠45° V(4∠0° A)*=(220∠45° V)(4∠−0° A)

S=480∠45° VA (5)

Convert equation (5) from polar form to rectangular form. Therefore,

S=339.4 W+j339.4 VAR (6)

On comparing equation (4) and (6), the average power P is 339.4 W and the reactive power Q is 339.4 VAR.

Conclusion:

Thus, the complex power is S=339.4 W+j339.4 VAR, average power is P=339.4 W, and reactive power is Q=339.4 VAR.

Given data:

The voltage phasor,

v(t)=339.4sin(377t+90°) V (7)

The current phasor,

i(t)=5.657sin(377t+45°) A (8)

Calculation:

From equation (7), the rms value of the voltage is

Vrms=339.42∠90°      {∵Vm=339.4, θv=90°}=240∠90° V

From equation (8), the rms value of the current is

Irms=5.6572∠45°      {∵Im=5.657, θi=45°}=4∠45° A

Substitute 240∠90° V for Vrms and 4∠45° A for Irms in equation (3) to find the complex power S in VA.

S=240∠90° V(4∠45° A)*=(240∠90° V)(4∠−45° A)

S=960∠45° VA (9)

Convert equation (9) from polar form to rectangular form. Therefore,

S=678.8 W+j678.8 VAR (10)

On comparing equation (4) and (10), the average power P is 678.8 W and the reactive power Q is 678.8 VAR.

Conclusion:

Thus, the complex power is S=678.8 W+j678.8 VAR, average power is P=678.8 W, and reactive power is Q=678.8 VAR.

Given data:

The voltage phasor,

Vrms=900∠90° V (11)

The impedance phasor,

Z=75∠45° Ω (12)

Formula used:

Write the expression for complex power S.

S=Vrms2Z* (13)

Here,

Z* is the complex conjugate of the impedance.

Calculation:

From equation (11),

Vrms= |Vrms| =900 V

Substitute 900 for Vrms and 75∠45° Ω for Z in equation (13) to find the complex power S in VA.

S=(900 V)2(75∠45° Ω)*=810000 V⋅V75∠−45° Ω=(1080 ∠45°)×103×10−3 VA         {∵1 A=1 V1 Ω}

S=10.8∠45° kVA (14)

Convert equation (14) from polar form to rectangular form. Therefore,

S=7.637 kW+j7.637 kVAR (15)

On comparing equation (4) and (15), the average power P is 7.637 kW and the reactive power Q is 7.637 kVAR.

Conclusion:

Thus, the complex power is S=7.637 kW+j7.637 kVAR, average power is P=7.637 kW, and reactive power is Q=7.637 kVAR.

Given data:

The current phasor,

Irms=100∠60° A (16)

The impedance phasor,

Z=50∠60° Ω (17)

Write the expression to find the complex power S.

S=Irms2Z (18)

Calculation:

From equation (16),

Irms= |Irms| =100 A

Substitute 50∠60° Ω for Z and 100 A for Irms in equation (18) to find the complex power S in VA.

S=(100 A)2(50∠60° Ω)=(500000∠60°)×103×10−3 A2⋅Ω

S=500∠60° kVA              {∵ 1 V=1 A×1 Ω} (19)

Convert equation (19) from polar form to rectangular form. Therefore,

S=250 kW+j433 kVAR (20)

On comparing equation (4) and (20), the average power P is 250 kW and the reactive power Q is 433 VAR.

Conclusion:

Thus, the complex power is S=250 kW+j433 kVAR, average power is P=250 kW, and reactive power is Q=433 kVAR.