The voltage phasor of a circuit is 10-15-V and the current phasor is 2-45-A- The active and the reactive powers in the circuit are

Complex power=VI^* =10∠15^∘×2∠45^∘=20∠60^∘ =(10+j1 √(3)) =(10+j17.32)=P+jQ ⇒ P=10 W and Q=17.32 VAR ...

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The voltage phasor of a circuit is $10 ∠ 15^{\circ} V$ and the current phasor is $2 ∠ - 45^{\circ} A$ . The active and the reactive powers in the circuit are

10 W a n d 17.32 V A R

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5 W a n d 8.66 V A R

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20 W a n d 60 V A R

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20 \sqrt{2} W a n d 10 \sqrt{2} V A R

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Solution

The correct option is A $10 W a n d 17.32 V A R$
Complex power $= V I^{*}$
$= 10 ∠ 15^{\circ} \times 2 ∠ 45^{\circ} = 20 ∠ 60^{\circ}$
$= (10 + j 1 - \sqrt{3})$
$= (10 + j 17.32) = P + j Q$
$\Rightarrow P = 10 W a n d Q = 17.32 V A R$

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The voltage phasor of a circuit is 10∠15∘V and the current phasor is 2∠−45∘A. The active and the reactive powers in the circuit are

The correct option is A 10W and 17.32 VARComplex power=VI∗ =10∠15∘×2∠45∘=20∠60∘ =(10+j1−√3) =(10+j17.32)=P+jQ ⇒P=10W and Q=17.32VAR

The voltage phasor of a circuit is $10 ∠ 15^{\circ} V$ and the current phasor is $2 ∠ - 45^{\circ} A$ . The active and the reactive powers in the circuit are

The correct option is A $10 W a n d 17.32 V A R$
Complex power $= V I^{*}$
$= 10 ∠ 15^{\circ} \times 2 ∠ 45^{\circ} = 20 ∠ 60^{\circ}$
$= (10 + j 1 - \sqrt{3})$
$= (10 + j 17.32) = P + j Q$
$\Rightarrow P = 10 W a n d Q = 17.32 V A R$