Solution:

Ans. (i) Calculation of Capacitance

(ii) $Q$-factor of circuit and its importance

(i) As power factor is unity,

$$ \begin{aligned} & \therefore \quad X_{L}=X_{C} \quad 1 / 2 \ & \Rightarrow \quad \omega=\frac{1}{\sqrt{L C}} \end{aligned} $$

$$ \begin{align*} C & =\frac{1}{2 \times 10^{3}} \mathrm{~F} \ & =0.5 \times 10^{-3} \mathrm{~F} \ & =0.5 \mathrm{mF} \end{align*} $$

(ii) Quality factor,

$$ Q=\frac{1}{R} \sqrt{\frac{L}{C}} $$

$$ \begin{aligned} & =\frac{1}{10} \sqrt{\frac{200 \times 10^{-3}}{0 \cdot 5 \times 10^{-3}}} \ & =\frac{1}{10} \times 20=2 \end{aligned} $$

Significance : It measures the sharpness of resonance.

Average Power dissipated,

$$ \begin{aligned} P & =V_{r m s} I_{r m s} \cos \phi \ & =50 \times \frac{50}{10} \times 1 \mathrm{~W} \ & =250 \text { watts } \end{aligned} $$

[CBSE Marking Scheme 2017]

[A] Q. 5. An $a c$ voltage $V=V_{m} \sin \omega t$ is applied to a series LCR circuit. Obtain an expression for the current in the circuit and the phase angle between the current and voltage. What is resonance frequency.

A [SQP I 2017]

Ans. (i) Expression for Current and Phase angle 4

(ii) Resonance Frequency

1

(i) In a series $L C R$ circuit shown,

From the phasor relation, voltages $V_{L}+V_{R}+V_{C}=V$, as $V_{C}$ and $V_{L}$ are along the same line and in opposite directions, so they will combine in single phasor $\left(V_{C}+V_{L}\right)$ having magnitude $\left|V_{C m}-V_{L m}\right|$. Since voltage $V$ is shown as hypotenuse of right angled triangle with sides as $V_{R}$ and $\left(V_{C}+V_{L}\right)$, so the Pythagoras Theorem results as :

$$ \begin{aligned} & V_{m}^{2}=V_{R}^{2}+\left(V_{C m}-V_{L m}\right)^{2} \ & V_{m}^{2}=\left(\mathrm{I}{\mathrm{m}} R\right)^{2}+\left(\mathrm{I}{\mathrm{m}} X_{C}-\mathrm{I}{\mathrm{m}} X{L}\right)^{2} \ & V_{m}^{2}=\mathrm{I}{\mathrm{m}}{ }^{2}\left(R^{2}+\left(X{C}-X_{L}\right)^{2}\right) \end{aligned} $$

Now current in the circuit :

$$ \begin{aligned} & I_{m}=\frac{V_{m}}{\sqrt{\left(R^{2}+\left(X_{C}-X_{L}\right)^{2}\right.}} \ & I_{m}=\frac{V_{m}}{Z} \text { as } Z=\sqrt{\left[R^{2}+\left(X_{C}-X_{L}\right)^{2}\right] \mathbf{1}} \end{aligned} $$

As phasor $I$ is always parallel to phasor $V_{R}$, the phase angle $\phi$ is the angle between $V_{R}$ and $V$ and can be determined from figure.

(ii) Resonance Frequency

Frequencies at which the response amplitude is relative maximum are known as system’s resonant frequencies. It is shown as :

$$ \begin{aligned} V_{C m} & =V_{L m} \ f_{0} & =\frac{1}{2 \pi \sqrt{L C}} \end{aligned} $$

[CBSE Marking Scheme 2017]