GATE 2019 EE – Question 53
The voltage across and the current through a load are expressed as follows
$$v(t)=-170\sin\left(377t-\frac\pi6\right)\text{ V},\qquad i(t)=8\cos\left(377t+\frac\pi6\right)\text{ A}.$$
The average power in watts (round off to one decimal place) consumed by the load is ________.
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Correct answer: 585.96 to 591.84
Explanation
Write both as sines: $v=170\sin\left(377t-\frac\pi6+\pi\right)=170\sin\left(377t+\frac{5\pi}6\right)$ and $i=8\sin\left(377t+\frac\pi6+\frac\pi2\right)=8\sin\left(377t+\frac{2\pi}3\right)$. The phase difference is $\frac{5\pi}6-\frac{2\pi}3=\frac\pi6=30^\circ$, so $P=\frac12(170)(8)\cos30^\circ=588.9$ W.