GATE 2025 EE – Question 27
The input voltage $v(t)$ and current $i(t)$ of a converter are given by,
$$v(t)=300\sin(\omega t)\ \text{V}$$
$$i(t)=10\sin\left(\omega t-\frac\pi6\right)+2\sin\left(3\omega t+\frac\pi6\right)+\sin\left(5\omega t+\frac\pi2\right)\ \text{A}$$
where, $\omega=2\pi\times50$ rad/s. The input power factor of the converter is closest to
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Correct answer: (A) 0.845
Explanation
Only the fundamental current component produces real power: $P=\dfrac{300}{\sqrt2}\cdot\dfrac{10}{\sqrt2}\cos30^\circ=1500\times0.866=1299$ W. $V_{rms}=212.13$ V and $I_{rms}=\sqrt{\dfrac{10^2+2^2+1^2}{2}}=7.246$ A, so $S=1537$ VA. The power factor is $1299/1537=0.845$.