Given the following problem:

enter image description here

I'd like to know how exactly one is to find the complex power delivered by the voltage source and the current source, separately, as is shown in the solutions on the right.

I've correctly obtained all the values for the complex power delivered to the resistor, inductor and capacitor. When I attempted to find the power delivered by the voltage source, i ignored the current source, and brought togheter the three impedances to make one, as in:

Z = jwL + R + 1/(jwC)

Then I found the current:

IZ = Voltagesource / Z

And then I found the complex power:

Svoltagesource = 1/2 * Voltagesource * (IZ*)

Which resulted in:

Svoltagesource = 0.295+0.531i

I'd appreciate input on what I might be doing wrong.


1 Answer 1


Generally power cannot be found out separately by using superposition principle, since power is parabolic with reference to current.

When you find the power delivered to an element of impedance Z through which current I flows, the answer is (I^2)*Z;

When you use superposition, then you get current due to both sources individually as I=I1+I2; When you calculate the power individually, due to the individual sources, you would be getting P1=((I1)^2)*Z and P2=((I2)^2)*Z; hence you would calculate the total power to be P=P1+P2=(((I1)^2)+((I2)^2))*Z; which is incorrect since the total power is P=(I^2)Z=(((I1)^2)+((I2)^2)+2(I1)*(I2))*Z;

  • \$\begingroup\$ Then how are you supposed to find the complex power delivered by the voltage source ? \$\endgroup\$
    – user2236
    Mar 2, 2014 at 22:42
  • \$\begingroup\$ If the power through the resistor is 2 Watts, then the current through it should be non complex, but when I do a complete analysis using mesh current systems then I am getting complex currents through the resistor and the inductor. What are you getting>>? \$\endgroup\$ Mar 3, 2014 at 6:48
  • \$\begingroup\$ ya this is the method.But you have to calculate using phasors.The voltage source becomes 5@Angle0 and current source is [email protected] impedance with complex j.Calculate in phasor domain polar form using calculator. \$\endgroup\$
    – Gopi
    Apr 2, 2014 at 7:27

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