A single-phase voltage source vs = 325sin(2π50t) V delivers a current, i = 12sin(2π50t) + 9sin(2π150t) A to a load. The load power factor, correct up to two decimal places, is:

The correct option is 0.80.

To find the load power factor, we need to calculate the ratio of the average active power ($P$) delivered to the load to the apparent power ($S$) supplied by the source.
The power factor ($\text{PF}$) is defined as:
$\text{PF}=\frac{P}{S}=\frac{P}{V_{\text{rms}}{I}_{\text{rms}}}$

Step 1: Analyze the voltage and current equations
The source voltage is given by:
$v_s\left(t\right)=325sin(2\pi ·50t)\text{V}$
This is a purely sinusoidal voltage at the fundamental frequency of 50 Hz.
The peak voltage is $V_m=325\text{V}$.
The RMS value of the voltage is:
$V_{\text{rms}}=\frac{V_m}{\sqrt{2}}=\frac{325}{\sqrt{2}}\text{V}$

The load current contains fundamental and third-harmonic components:
$i\left(t\right)=12sin(2\pi ·50t)+9sin(2\pi ·150t)\text{A}$
The peak currents are $I_{m1}=12\text{A}$ (fundamental at 50 Hz) and $I_{m3}=9\text{A}$ (third harmonic at 150 Hz).

Step 2: Calculate the RMS value of the current
For a non-sinusoidal current containing multiple harmonic components, the overall RMS current is:
$I_{\text{rms}}=\sqrt{{I_{\text{rms}1}}^2+{I_{\text{rms}3}}^2}=\sqrt{{\left(\frac{12}{\sqrt{2}}\right)}^2+{\left(\frac{9}{\sqrt{2}}\right)}^2}$
Simplifying this expression:
$I_{\text{rms}}=\sqrt{\frac{144+81}{2}}=\sqrt{\frac{225}{2}}=\frac{15}{\sqrt{2}}\text{A}$

Step 3: Calculate the average active power
Active power is generated only by the interaction of voltage and current components of the same frequency. Since the voltage has only a fundamental component (50 Hz), only the fundamental component of the current contributes to active power:
$P=V_{\text{rms}}{I}_{\text{rms}1}cos\left({\theta }_1\right)$
Here, both the fundamental voltage and current have a phase angle of 0, so the phase difference ${\theta }_1=0$, and $cos\left({\theta }_1\right)=1$.
$P=\left(\frac{325}{\sqrt{2}}\right)·\left(\frac{12}{\sqrt{2}}\right)=\frac{325·12}{2}=325·6=1950\text{W}$

Step 4: Calculate the apparent power
The apparent power is:
$S=V_{\text{rms}}{I}_{\text{rms}}=\left(\frac{325}{\sqrt{2}}\right)·\left(\frac{15}{\sqrt{2}}\right)=\frac{325·15}{2}=2437.5\text{VA}$

Step 5: Determine the power factor
Now we compute the power factor ratio:
$\text{PF}=\frac{P}{S}=\frac{1950}{2437.5}=0.80$
Thus, the load power factor is exactly 0.80.