Investigate the value of form factor of a voltage v = 250sin(2π × 50t)
The question asks us to find the form factor for a given voltage waveform, $v = 250\sin(2\pi \times 50t)$. The form factor of an alternating voltage or current is defined as the ratio of its RMS (Root Mean Square) value to its average value.
The formula for the form factor (FF) is:
$$ FF = \frac{V_{RMS}}{V_{avg}} $$
Let's break down the calculation for the given sinusoidal voltage.
The given voltage equation is $v = 250\sin(2\pi \times 50t)$. This represents a standard sinusoidal waveform of the form $v(t) = V_m \sin(\omega t)$.
For any sinusoidal waveform, the RMS value is related to the peak voltage by the following formula:
$$ V_{RMS} = \frac{V_m}{\sqrt{2}} $$
Substituting the peak voltage ($V_m = 250$ V):
$$ V_{RMS} = \frac{250}{\sqrt{2}} \text{ V} $$
When calculating the form factor, we use the average of the rectified waveform over one half-cycle. For a sinusoidal waveform, the average value (or rectified average) is given by:
$$ V_{avg} = \frac{2V_m}{\pi} $$
Substituting the peak voltage ($V_m = 250$ V):
$$ V_{avg} = \frac{2 \times 250}{\pi} = \frac{500}{\pi} \text{ V} $$
Now, we can calculate the form factor using the RMS and average values we found:
$$ FF = \frac{V_{RMS}}{V_{avg}} = \frac{\frac{250}{\sqrt{2}}}{\frac{500}{\pi}} $$
Simplify the expression:
$$ FF = \frac{250}{\sqrt{2}} \times \frac{\pi}{500} $$
$$ FF = \frac{250 \times \pi}{500 \times \sqrt{2}} $$
$$ FF = \frac{\pi}{2 \sqrt{2}} $$
Now, let's calculate the numerical value:
Using $\pi \approx 3.14159$ and $\sqrt{2} \approx 1.41421$:
$$ FF \approx \frac{3.14159}{2 \times 1.41421} \approx \frac{3.14159}{2.82842} \approx 1.1107 $$
Rounding to two decimal places, the form factor is approximately 1.11.
The calculated form factor for the given sinusoidal voltage waveform $v = 250\sin(2\pi \times 50t)$ is approximately 1.11. This value is characteristic of all pure sinusoidal waveforms.
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