Question

A capacitor of unknown value and an inductor of $\text{0}\text{.1H}$ and a resistor of $10\Omega$ are connected in series to a $220V$, $50\text{Hz}$ ac source. It is found that the power factor of circuits is unity. Calculate the capacitance of capacitor and maximum amplitude of current.

Answer 1

Hint: At resonant frequency, the power factor is unity. Students can use the formula for the resonant frequency to determine the capacitance and then use the formula for current to calculate the peak value of the current.

Complete step-by-step answer:
The frequency at which the current attains the peak value is known as the resonant frequency. At resonant frequency the power factor, which is defined as the ratio of the true power to the apparent power is unity, which means that at resonant frequency , the true power is equal to the apparent power.
Now, we can write the formula for the resonant frequency as
${{f}_{r}}=\dfrac{1}{2\pi \sqrt{LC}}$
where, ${{\text{f}}_{\text{r}}}$is the resonant frequency, $\text{L}$is the inductance and $\text{C}$is the capacitance of the system.
Now, let us substitute the values given in the question to the above equation and hence determine the value of the capacitance.
$\begin{aligned} & 50=\dfrac{1}{2\pi \sqrt{0.1\times C}} \\\ & \Rightarrow 2500=\dfrac{1}{4{{\pi }^{2}}(0.1C)} \\\ & \Rightarrow C=101.3\mu F \\\ \end{aligned}$
Again in order to determine the peak value of the current, let us first determine the RMS current, denoted by ${{\text{I}}_{\text{rms}}}$.
$\begin{aligned} & {{I}_{rms}}=\dfrac{V}{Z} \\\ & \Rightarrow {{I}_{rms}}=\dfrac{220}{10} \\\ & \Rightarrow {{I}_{rms}}=22A \\\ \end{aligned}$
where, $\text{Z}$ is the impedance.
Now if the peak current is ${{\text{I}}_{\text{0}}}$, then,
$\begin{aligned} & {{I}_{0}}=\sqrt{2}{{I}_{rms}} \\\ & \Rightarrow {{I}_{0}}=\sqrt{2}\times 22 \\\ & \Rightarrow {{I}_{0}}=31.11A \\\ \end{aligned}$
Therefore from the above calculations, we have come to know that the value of capacitance is $C=101.3\mu F$ and maximum amplitude of current is ${{I}_{0}}=31.11A$

Note: Students must note that in calculating the peak current we need to use the RMS current and that the RMS current does not correspond to the peak current. They must also be careful while using the value of the resistor for impedance.