B Reactive Power Ignore the sign Unit W Please include detai

B) Reactive Power (Ignore the sign; Unit: W)

*Please include detailed steps, thank you*

Find the average power and the reactive power absorbed by the load in the circuit shown above if ig = 20*cos(1250*t) mA

Solution

Current through Load=i_g(Z_s/(Z_s+Z_l )

w=1250

where Z_s=jw9=j11250

Z_l=9+(1/jw*65n)=9-j12308

i_l=(20*10^-3)*(j11250)/(j11250+9-j12308)= -0.2127 +j 0.0018

Power=i_l²Z_l=(-0.2127 +j 0.0018)²(9-j12308)=9.062-j5565.3 W

Avg power=Real(9.062-j5565.3 )W

for reactive power we need to consider only capacitance

Reactive power=(-0.2127 +j 0.0018)²(-j12308)=-9.4245 - j5.56

if we omit the - sign=

Reactive power=9.4245 +j5.56