Question

A. Find the value of the phase difference between the current and the voltage in the series LCR circuit shown below. Which one leads in phase : current or voltage?
B. Without making any other change, find the value of the additional capacitor to be connected in parallel with the capacitor C, in order to make the power factor of the circuit unity.

Answer 1

A. The value of the phase difference between the current and the voltage in the series LCR circuit is

The image given in the question is attached below.

Given:

$\mathrm{V}=\mathrm{V_o} \sin(1000+\Phi)$

$\omega=1000 \ \mathrm{s}^{-1}$

$\mathrm{L}=100 \ \mathrm{mH}$

$C=2 \ \mu F$

$R=400 \ \Omega$

To find:

The phase difference is given by the formula:

$\Phi=\tan ^{-1}\left(\frac{X_{L}-X_{c}}{R}\right)$

Solution:

$\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}$

On substituting the values, we get,

$\mathrm{X}_{\mathrm{L}}=1000 \times 100 \times 10^{-3}$

$\therefore \mathrm{X}_{\mathrm{L}}=100 \ \Omega$

$X_{c}=\frac{1}{\omega C}$

On substituting the values, we get,

$X_{c}=\frac{1}{1000 \times 2 \times 10^{-6}}$

$\therefore X_{c}=500 \ \Omega$

Now, on substituting the values in the phase difference formula, we get,

$\Phi=\tan ^{-1}\left(\frac{100-500}{400}\right)$

$\Phi=\tan ^{-1}(-1)$

$\therefore \Phi=-45^{\circ}$

B. The value of the additional capacitor to be connected in parallel

From question, to make power factor of the circuit unity,

R = Z

$X_{L}=X_{C}$

$\omega^{2}=\frac{1}{L C}$

Where,

C = resultant capacitance

On substituting the values in above formula, we get,

$10^{6}=\frac{1}{100 \times 10^{-3} \times C'}$

$C'=10^{-5} \ \mathrm{F}$

From question, the capacitors are in parallel connection.

The resultant capacitance is given by the formula,

$C'=C+C 1$

$10^{-5}=0.2 \times 10^{-5}+\mathrm{C}{1}$

$\therefore C{1}=8 \ \mu F$