Question

The power factor for series L-C-R circuit at half power frequency is
(A) $1$
(B) ${\displaystyle \frac{1}{2}}$
(C) ${\displaystyle \frac{1}{\sqrt{2}}}$
(D) $0$

Answer 1

Hint: Here, we are looking for a condition where the dissipating power drops by a factor of 2, that is $P={\displaystyle \frac{P_0}{2}}$, where $P_0$ is the power dissipated when the condition of resonance is satisfied. So, in order to solve the question, what you need to do is use the complex notation. By using the given conditions, you can find the reactance of the given L-C-R circuit and then you can easily find the power factor of the series L-C-R circuit.

Complete step by step answer:
When resonant frequency is applied, the power factor is equal to 1. So, we have the power dissipated given as $P_0={\displaystyle \frac{V^2}{R}}$, where $R$ is the resistance used in the circuit and $V$ is the applied voltage. Now, let us see what will be the power dissipated in general. The net impedance of the circuit using the complex notation is given as $Z=R+jX$, where $j$ is the imaginary number equal to $\sqrt{-1}$. The complex power dissipated will be given as $P=V\bar{I}$, where $\bar{I}$ is the complex conjugate of $I$.

$$\begin{gathered} P=V\left(\overline{{\displaystyle \frac{V}{Z}}}\right) \\ \Rightarrow P={\displaystyle \frac{V^2}{\overline{R+jX}}} \\ \Rightarrow P={\displaystyle \frac{V^2}{R-jX}} \\ \Rightarrow P={\displaystyle \frac{V^2}{R-jX}}\left({\displaystyle \frac{R+jX}{R+jX}}\right) \\ \Rightarrow P={\displaystyle \frac{V^2}{R^2+X^2}}\left(R+jX\right) \end{gathered}$$
Now, in order to find the real power, you will consider the real part of the complex power obtained above. Therefore,$P={\displaystyle \frac{V^{2}R}{R^2+X^2}}$. Now, this power is supposed to be equal to ${\displaystyle \frac{P_0}{2}}$, so we get
$$\begin{gathered} {\displaystyle \frac{V^{2}R}{R^2+X^2}}={\displaystyle \frac{V^2}{2R}} \\ \therefore X=R \end{gathered}$$
Hence, the impedance will be given by $Z=R\left(1+j\right)$. From here, you can say that $\tan \phi =1$ and therefore $\cos \phi ={\displaystyle \frac{1}{\sqrt{2}}}$.Hence, the power factor for a series L-C-R circuit at half power frequency is ${\displaystyle \frac{1}{\sqrt{2}}}$.

Hence,option C is correct.

Note:Here, we have discussed in brief about the power dissipated in an L-C-R circuit, you should keep in mind the approach we used here to solve the question. Also, remember the complex notation of the power. Here, the bar above the current denotes the conjugate of the current, so keep this in mind as well. Also, here we used the imaginary number as $j$ and not $i$ because we did not want to mix up current and the imaginary number.