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Problem: An impedance 1000(1 + i) Ohms (and note it contains an imaginary part) is connected across an AC voltage source of amplitude 10 V and frequency 60 Hz. What's the power dissipated during one cycle within the impedance?

Relevant equations:

P = Re(V*I), where V* is complex conjugate of voltage I=V/Z

Solution Attempt:

So it's alternating current, so I first take the RMS V: 10/\sqrt{2}. Then I = V/Z, so I get

V*I = V*V/Z = 100/2 1/(1000(1+i)) = (1-i)/(20*2). Then I take the real part of this, which is 1/40. Am I doing this right?

Thanks!

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    \$\begingroup\$ Every time I see words 'impedance' and 'imaginary part' in the same sentence, I feel so dumb.... \$\endgroup\$
    – BarsMonster
    Commented Jan 24, 2011 at 21:04
  • \$\begingroup\$ homework much? :) \$\endgroup\$
    – Mark
    Commented Jan 24, 2011 at 21:24
  • \$\begingroup\$ ok so i'll give you a hint, if the question is being posed exactly as it is presented here, its a trick question(kinda). They are asking for power dissipated in the load, aka true power which is a separate concept from reactive power or apparent power and that should both influence and simplify(a bit) your calculation. \$\endgroup\$
    – Mark
    Commented Jan 24, 2011 at 21:49

1 Answer 1

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I assume:

"source amplitude" means "peak" (as apposed to peak to peak)

The load impedance is actually calculated at 60Hz

Ztotal = 1000 + 1000j (really should be using j as sqrt(-1) not i in electronics)

Ztotal = 1414.21 @ 45deg (just rewritten in a magnitude / phase representation)

|Ztotal| = 1414.21 Ohms

Vrms = 10/sqrt(2) ~= 7.07 Vrms

|I| = |Vrms| / |Ztotal| = 7.07 / 1414.21

|I| ~= 5mA

P(true power) = I^2 * R = (5mA)^2 * 1000 = 25 mW (this is what is dissipated in the load, your answer)

Q(reactive power) = I^2 * X = (5mA)^2 * 1000 = 25 mVAR (this is power bouncing back and forth, not dissipated)

S(apparent power) = I^2 * Z = (5mA)^2 * 1414.21 = 35.35 mVA (this is the vector sum of the true and reactive power)

Power Factor = P/S = True Power / Apparent Power = 25mW / 35mVA = 0.714

Hopefully I didn't flub up the math there, my calculators batteries just died so i did it all in google.

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  • \$\begingroup\$ So OP had the right answer, because 1/40 W == 25 mW. \$\endgroup\$
    – markrages
    Commented Jan 24, 2011 at 22:24
  • \$\begingroup\$ yep OP had the right answer, but given the nature of the question i don't think OP knew why it was right, so i figured I'd show it in a way more connected to the electronics concepts. \$\endgroup\$
    – Mark
    Commented Jan 24, 2011 at 22:37

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