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Source P and Q, Ean = -120∠210° V

Short answer

Simplify the voltage to 120∠30° V, then S = (120∠30°)(10∠-60°) = 1200∠-30° = 1039.2 - j600 VA. The source delivers 1039.2 W and absorbs 600 var (receives reactive power).

A voltage source Ean = -120∠210° V and the current through the source Ina = 10∠60° A. Find the values of real power (P) and reactive power (Q) and state whether the source is delivering or receiving each.

Use S=EI∗S = EI^*. The current InaI_{na} flows from n to a inside the source, so it leaves the positive terminal a. A positive PP or QQ therefore means the source is delivering that power.

Simplify the voltage

Ean=−120∠210∘=120∠(210∘−180∘)=120∠30∘ VE_{an} = -120\angle210^\circ = 120\angle(210^\circ - 180^\circ) = 120\angle30^\circ\ \text{V}

Complex power

S=EanIna∗=(120∠30∘)(10∠−60∘)=1200∠−30∘=1200(cos⁡30∘−jsin⁡30∘)=1039.2−j600 VA\begin{aligned} S &= E_{an}I_{na}^* = (120\angle30^\circ)(10\angle-60^\circ)\\ &= 1200\angle-30^\circ\\ &= 1200(\cos30^\circ - j\sin30^\circ)\\ &= 1039.2 - j600\ \text{VA} \end{aligned}

Interpretation

QuantityValueMeaning
PP1039.2 WPositive, so the source delivers real power
QQ−600 varNegative, so the source receives (absorbs) 600 var

The current leads the voltage by 30°, so the source sees a leading (capacitive) load. That load returns reactive power to the source.

Answer: P=1039.2P = 1039.2 W, delivered by the source. Q=600Q = 600 var, absorbed by the source (delivered Q=−600Q = -600 var).

✓ Answer checked by Claude Opus 5.5 (high effort)

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