Question:medium
A single-phase voltage source \(v_s=325\sin(2\pi 50t)\) V delivers a current, \(i=12\sin(2\pi 50t)+9\sin(2\pi 150t)\) A to a load.
The load power factor, correct up to two decimal places, is:
A single-phase voltage source \(v_s=325\sin(2\pi 50t)\) V delivers a current, \(i=12\sin(2\pi 50t)+9\sin(2\pi 150t)\) A to a load.
The load power factor, correct up to two decimal places, is:
The load power factor, correct up to two decimal places, is:
Show Hint
Only the current component at the same frequency as the source contributes to real power, but the harmonic current still adds to the rms current and lowers the power factor.
Updated On: Jul 20, 2026
- 1.00
- 0.80
- 0.65
- 0.57
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Split the current into a useful part and a distortion part.
The current has two pieces: $12\sin(2\pi 50t)$ at the same frequency as the source, and $9\sin(2\pi 150t)$ at three times that frequency. Only same-frequency waves can exchange real power with the source, so the 150 Hz piece is pure distortion as far as power transfer goes.
Step 2: Get the real power from the matching frequency term alone.
Peak voltage is $325$ V and the peak of the matching current term is $12$ A, both in phase, so
\[ P=\frac{325\times12}{2}=1950\text{ W} \]
Step 3: Find the rms of the full current.
Each sine term contributes its own rms in a Pythagorean sum:
\[ I_{rms}=\sqrt{\left(\frac{12}{\sqrt2}\right)^2+\left(\frac{9}{\sqrt2}\right)^2}=\sqrt{112.5}\approx10.607\text{ A} \]
Step 4: Find the rms voltage and the apparent power.
\[ V_{rms}=\frac{325}{\sqrt2}\approx229.81\text{ V},\qquad S=V_{rms}I_{rms}=2437.5\text{ VA} \]
Step 5: Divide to get the true power factor.
The harmonic current inflates $I_{rms}$ without adding any real power, so it drags the power factor below what the 50 Hz terms alone would give.
\[ \text{PF}=\frac{1950}{2437.5}=0.80 \]
\[ \boxed{0.80} \]
The current has two pieces: $12\sin(2\pi 50t)$ at the same frequency as the source, and $9\sin(2\pi 150t)$ at three times that frequency. Only same-frequency waves can exchange real power with the source, so the 150 Hz piece is pure distortion as far as power transfer goes.
Step 2: Get the real power from the matching frequency term alone.
Peak voltage is $325$ V and the peak of the matching current term is $12$ A, both in phase, so
\[ P=\frac{325\times12}{2}=1950\text{ W} \]
Step 3: Find the rms of the full current.
Each sine term contributes its own rms in a Pythagorean sum:
\[ I_{rms}=\sqrt{\left(\frac{12}{\sqrt2}\right)^2+\left(\frac{9}{\sqrt2}\right)^2}=\sqrt{112.5}\approx10.607\text{ A} \]
Step 4: Find the rms voltage and the apparent power.
\[ V_{rms}=\frac{325}{\sqrt2}\approx229.81\text{ V},\qquad S=V_{rms}I_{rms}=2437.5\text{ VA} \]
Step 5: Divide to get the true power factor.
The harmonic current inflates $I_{rms}$ without adding any real power, so it drags the power factor below what the 50 Hz terms alone would give.
\[ \text{PF}=\frac{1950}{2437.5}=0.80 \]
\[ \boxed{0.80} \]
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