For an LCR circuit, the power transferred from the driving source… - Vidyadip

MCQ

For an $\text{LCR}$ circuit, the power transferred from the driving source to the driven oscillator is $\text{P}=\text{FZ}\cos\phi$.

A
Here, the power factor $\cos\phi\geq0,\text{P}\geq0$.
B
The driving force can give no energy to the oscillator $(P = 0)$ in some cases.
C
The driving force cannot syphon out $(P < 0)$ the energy out of oscillator.
✓
All of the above

✓

Answer

Correct option: D.

All of the above

Here, the power factor $\cos\phi\geq0,\text{P}\geq0$.
The driving force can give no energy to the oscillator $(P = 0)$ in some cases.
The driving force cannot syphon out $(P < 0)$ the energy out of oscillator.

Solution:
Key Concept: Power Factor.

It may be defined as cosine of the angle of lag of lead $($i.e., $\cos\phi).$
It is also defined as the radio of resistance and impedange $\Big(\text{i.e.,} \frac{\text{R}}{\text{Z}}\Big)$.
$\text{The radio}=\frac{\text{True power}}{\text{Apparent power}}=\frac{\text{W}}{\text{VA}}=\frac{\text{kW}}{\text{kVA}}=\cos\phi$

In the given problem power transferred,
$\text{P}=\text{I}^2\text{Z}\cos\phi$
where I is the current, $Z =$ Impedance and $\cos\phi$ is power factor

As power factor, $\cos\phi=\frac{\text{R}}{\text{Z}}$

where $R > 0$ and $Z > 0$
$\Rightarrow\ \cos\phi>0\Rightarrow\ \text{P}>0$

When $\phi=\frac{\pi}{2} ($in case of $L $ of $C), P = 0.$
From $(a)$, it is clear that $P < 0$ is not possible.

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