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P and Q of a 100 V single-phase source

Short answer

Complex power S = VI* = 1000∠-10° VA = 984.8 - j173.65 VA. The source delivers 984.8 W real power and absorbs 173.6 var reactive power.

A single phase voltage source with V = 100∠0° volts delivers a current I = 10∠10° A, which leaves the positive terminal of the source. Calculate the source real and reactive power and state whether the source delivers or absorbs each of these.

Complex power delivered by a source whose current leaves its positive terminal is S=VI∗S = VI^* (generator convention). A positive P or Q then means the source delivers it; a negative value means it absorbs it.

S=VI∗=(100∠0∘)(10∠−10∘)=1000∠−10∘ VA=1000cos⁡10∘−j 1000sin⁡10∘=984.81−j 173.65 VA\begin{aligned} S &= VI^* = (100\angle0^\circ)(10\angle-10^\circ)\\ &= 1000\angle-10^\circ\ \text{VA}\\ &= 1000\cos10^\circ - j\,1000\sin10^\circ\\ &= 984.81 - j\,173.65\ \text{VA} \end{aligned}
  • P=+984.81P = +984.81 W: positive, so the source delivers real power.
  • Q=−173.65Q = -173.65 var: negative, so the source absorbs 173.65 var (equivalently it delivers −173.65-173.65 var).

The current leads the voltage by 10°, so the load connected to the source is capacitive; a capacitive load supplies reactive power, which flows back into the source.

Answer: P = 984.8 W delivered; Q = 173.6 var absorbed by the source.

✓ Answer checked by Claude Opus 5.5 (high effort)

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