Question:
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
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The Correct Option is B
Solution and Explanation
Step 1: Identify why only the fundamental current matters for power.
The source voltage is a pure sinusoid at 50 Hz, \(v_s=325\sin(2\pi 50t)\). Average power is the average of the product \(v_s(t)\,i(t)\) over one cycle. When a sinusoid at one frequency is multiplied by a sinusoid at a different frequency and averaged over a full period, the result is zero, because the two waves are orthogonal. So the 150 Hz component of the current, \(9\sin(2\pi 150t)\), contributes nothing to the real power. Only the 50 Hz part of the current, \(12\sin(2\pi 50t)\), interacts with the 50 Hz voltage.
Step 2: Compute the real power.
Both the voltage and the matching current component are pure sine terms with no phase shift between them, so they are in phase.
\[ P=\frac{V_{peak}\,I_{peak,50Hz}}{2}\cos(0^{\circ})=\frac{325\times12}{2}=1950\text{ W} \]
Step 3: Compute the rms voltage.
\[ V_{rms}=\frac{325}{\sqrt{2}}\approx229.81\text{ V} \]
Step 4: Compute the rms current, including both frequencies.
For a waveform made of several sinusoids at different frequencies, the total rms value combines each term's rms value in quadrature.
\[ I_{rms}=\sqrt{\left(\frac{12}{\sqrt{2}}\right)^2+\left(\frac{9}{\sqrt{2}}\right)^2}=\sqrt{\frac{144}{2}+\frac{81}{2}}=\sqrt{\frac{225}{2}}\approx10.6066\text{ A} \]
Step 5: Compute the apparent power.
\[ S=V_{rms}\times I_{rms}=\frac{325}{\sqrt{2}}\times\sqrt{\frac{225}{2}}=325\times\frac{15}{2}=2437.5\text{ VA} \]
Step 6: Compute the load power factor.
The true power factor accounts for the full rms current, including the harmonic content, not just the fundamental.
\[ \text{PF}=\frac{P}{S}=\frac{1950}{2437.5}=0.80 \]
Step 7: Final conclusion.
\[ \boxed{0.80} \]
Hence the correct option is (B).
The source voltage is a pure sinusoid at 50 Hz, \(v_s=325\sin(2\pi 50t)\). Average power is the average of the product \(v_s(t)\,i(t)\) over one cycle. When a sinusoid at one frequency is multiplied by a sinusoid at a different frequency and averaged over a full period, the result is zero, because the two waves are orthogonal. So the 150 Hz component of the current, \(9\sin(2\pi 150t)\), contributes nothing to the real power. Only the 50 Hz part of the current, \(12\sin(2\pi 50t)\), interacts with the 50 Hz voltage.
Step 2: Compute the real power.
Both the voltage and the matching current component are pure sine terms with no phase shift between them, so they are in phase.
\[ P=\frac{V_{peak}\,I_{peak,50Hz}}{2}\cos(0^{\circ})=\frac{325\times12}{2}=1950\text{ W} \]
Step 3: Compute the rms voltage.
\[ V_{rms}=\frac{325}{\sqrt{2}}\approx229.81\text{ V} \]
Step 4: Compute the rms current, including both frequencies.
For a waveform made of several sinusoids at different frequencies, the total rms value combines each term's rms value in quadrature.
\[ I_{rms}=\sqrt{\left(\frac{12}{\sqrt{2}}\right)^2+\left(\frac{9}{\sqrt{2}}\right)^2}=\sqrt{\frac{144}{2}+\frac{81}{2}}=\sqrt{\frac{225}{2}}\approx10.6066\text{ A} \]
Step 5: Compute the apparent power.
\[ S=V_{rms}\times I_{rms}=\frac{325}{\sqrt{2}}\times\sqrt{\frac{225}{2}}=325\times\frac{15}{2}=2437.5\text{ VA} \]
Step 6: Compute the load power factor.
The true power factor accounts for the full rms current, including the harmonic content, not just the fundamental.
\[ \text{PF}=\frac{P}{S}=\frac{1950}{2437.5}=0.80 \]
Step 7: Final conclusion.
\[ \boxed{0.80} \]
Hence the correct option is (B).
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