SATHEE: alternating-currents Question 29

alternating-currents Question 29

Question: Q. 4. In the following circuit, calculate (i) the capacitance of the capacitor, if the power factor of the circuit is unity, (ii) the Q-factor of this circuit. What is the significance of the Q-factor in ac circuit? Given the angular frequency of the $a c$ source to be $100$ $rad / s$ . Calculate the average power dissipated in the circuit.

A [O.D. Comptt I, II, III 2017]

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Solution:

Ans. (i) Calculation of Capacitance

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

(i) As power factor is unity,

$\begin{aligned} ∴ X_{L} = X_{C} 1 / 2 & \Rightarrow ω = \frac{1}{\sqrt{L C}} \end{aligned}$

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

(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,

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

[CBSE Marking Scheme 2017]

[A] Q. 5. An $a c$ voltage $V = V_{m} \sin ω t$ 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

(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 $(V_{C} + V_{L})$ having magnitude $| V_{C m} - V_{L m} |$ . Since voltage $V$ is shown as hypotenuse of right angled triangle with sides as $V_{R}$ and $(V_{C} + V_{L})$ , 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 :

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