Q: Q. 7.7. A charged $30 \mu \mathrm{~F}$ capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

Capacitance, $\mathrm{C}=30 \mu \mathrm{~F}=30 \times 10^{-6} \mathrm{~F}$ Inductance, $\mathrm{L}=27 \mathrm{mH}=27 \times 10^{-3} \mathrm{H}$ Angular frequency is given as: $$ \begin{aligned} & \omega_{r}=\frac{1}{\sqrt{L C}} \\ & =\frac{1}{\sqrt{27 \times 10^{-3} \times 30 \times 10^{-6}}}=\frac{1}{9 \times 10^{-4}}=1.11 \times 10^{3} \mathrm{rad} / \mathrm{s} \end{aligned} $$ Hence, the angular frequency of free oscillations of the circuit is $1.11 \times 10^{3} \mathrm{rad} / \mathrm{s}$.

Q: Q. 7.3. A 44 mH inductor is connected to $220 \mathrm{~V}, 50 \mathrm{~Hz}$ ac supply. Determine the rms value of the current in the circuit.

Inductance of inductor, $\mathrm{L}=44 \mathrm{mH}=44 \times 10^{-3} \mathrm{H}$ Supply voltage, $\mathrm{V}=220 \mathrm{~V}$ Frequency, $v=50 \mathrm{~Hz}$ Angular frequency, $\omega=2 \pi \nu$ Inductive reactance, $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \nu \mathrm{~L}=2 \pi \times 50 \times 44 \times 10^{-3} \Omega$ rms value of current is given as: $$ I=\frac{V}{X_{L}}=\frac{220}{2 \pi \times 50 \times 44 \times 10^{-3}}=15.92 \mathrm{~A} $$ Hence, the rms value of current in the circuit is 15.92 A .

Question 1

Q. 7.6. Obtain the resonant frequency $\omega$ r of a series LCR circuit with $\mathrm{L}=2.0 \mathrm{H}$, $\mathrm{C}=32 \mu \mathrm{~F}$ and $\mathrm{R}=10 \Omega$. What is the Q-value of this circuit?

Accepted Answer

Inductance, L = 2.0 H

Capacitance, $\mathrm{C}=32 \mu \mathrm{~F}=32 	imes 10^{-6} \mathrm{~F}$

Resistance, $\mathrm{R}=10 \Omega$

Resonant frequency is given by the relation,

$$

\omega_{r}=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{2 	imes 32 	imes 10^{-6}}}=\frac{1}{8 	imes 10^{-3}}=125 \mathrm{rad} / \mathrm{s}

$$

Now, Q-value of the circuit is given as:

$$

Q=\frac{1}{R} \sqrt{\frac{L}{C}}=\frac{1}{10} \sqrt{\frac{2}{32 	imes 10^{-6}}}=\frac{1}{10 	imes 4 	imes 10^{-3}}=25

$$

Hence, the Q-Value of this circuit is 25.

Question 2

Q. 7.7. A charged $30 \mu \mathrm{~F}$ capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

Accepted Answer

Capacitance, $\mathrm{C}=30 \mu \mathrm{~F}=30 	imes 10^{-6} \mathrm{~F}$ Inductance, $\mathrm{L}=27 \mathrm{mH}=27 	imes 10^{-3} \mathrm{H}$

Angular frequency is given as:

$$

\begin{aligned}

& \omega_{r}=\frac{1}{\sqrt{L C}} \

& =\frac{1}{\sqrt{27 	imes 10^{-3} 	imes 30 	imes 10^{-6}}}=\frac{1}{9 	imes 10^{-4}}=1.11 	imes 10^{3} \mathrm{rad} / \mathrm{s}

\end{aligned}

$$

Hence, the angular frequency of free oscillations of the circuit is $1.11 	imes 10^{3} \mathrm{rad} / \mathrm{s}$.

Question 3

Q. 7.8. Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC . What is the total energy stored in the circuit initially? What is the total energy at later time?

Accepted Answer

Capacitance of the capacitor, $\mathrm{C}=30 \mu \mathrm{~F}=30 	imes 10^{-6} \mathrm{~F}$

Inductance of the inductor, $\mathrm{L}=27 \mathrm{mH}=27 	imes 10^{-3} \mathrm{H}$

Charge on the capacitor, $\mathrm{Q}=6 \mathrm{mC}=6 	imes 10^{-3} \mathrm{C}$

Total energy stored in the capacitor can be calculated as:

$$

E=\frac{1}{2} \frac{Q^{2}}{C}=\frac{1}{2} \frac{\left(6 	imes 10^{-3}ight)^{2}}{30 	imes 10^{-6}}=\frac{6}{10}=0.6 \mathrm{~J}

$$

Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor.

Question 4

Q. 7.5. In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.

Accepted Answer

In the inductive circuit,

Rms value of current, $\mathrm{I}=15.92 \mathrm{~A}$

Rms value of voltage, $\mathrm{V}=220 \mathrm{~V}$

Hence, the net power absorbed can be obtained by the relation,

$$

P=V I \cos \Phi

$$

Where,

$\Phi=$ Phase difference between $V$ and $I$.

For a pure inductive circuit, the phase difference between alternating voltage and current is $90^{\circ}$ i.e., $\Phi=90^{\circ}$.

Hence, $\mathrm{P}=0$ i.e., the net power is zero.

In the capacitive circuit, rms value of current, $\mathrm{I}=2.49 \mathrm{~A}$

rms value of voltage, $\mathrm{V}=110 \mathrm{~V}$

Hence, the net power absorbed can be obtained as:

$$

P=V I \operatorname{Cos} \Phi

$$

For a pure capacitive circuit, the phase difference between alternating voltage and current is $90^{\circ}$ i.e., $\Phi=90^{\circ}$.

Hence, $\mathrm{P}=0$ i.e., the net power is zero.

Question 5

Q. 7.1. A $100 \Omega$ resistor is connected to a $220 \mathrm{~V}, 50 \mathrm{~Hz}$ ac supply.

(a) What is the rms value of current in the circuit?

(b) What is the net power consumed over a full cycle?

Accepted Answer

Resistance of the resistor, R = $100 \Omega$

Supply voltage, V $=220$ V

Frequency, $v=50 \mathrm{~Hz}$

(a) The rms value of current in the circuit is given as

$$

I=\frac{V}{R}=\frac{220}{100}=2.20 \mathrm{~A}

$$

(b) The net power consumed over a full cycle is given as:

$\mathrm{P}=\mathrm{VI}=220 	imes 2.2=484 \mathrm{~W}$

Question 6

Q. 7.2. (a) The peak voltage of an ac supply is 300 V . What is the rms voltage?

(b) The rms value of current in an ac circuit is 10 A . What is the peak current?

Accepted Answer

(a) Peak voltage of the ac supply, $\mathrm{V}_{0}=300 \mathrm{~V}$

rms voltage is given as:

$$

V=\frac{V_{o}}{\sqrt{2}}=\frac{300}{\sqrt{2}}=212.1 \mathrm{~V}

$$

(b) The rms value of current is given as:

$$

\mathrm{I}=10 \mathrm{~A}

$$

Now, peak current is given as:

$$

I_{o}=\sqrt{2} I=\sqrt{2} 	imes 10=14.1 \mathrm{~A}

$$

Question 7

Q. 7.3. A 44 mH inductor is connected to $220 \mathrm{~V}, 50 \mathrm{~Hz}$ ac supply. Determine the rms value of the current in the circuit.

Accepted Answer

Inductance of inductor, $\mathrm{L}=44 \mathrm{mH}=44 	imes 10^{-3} \mathrm{H}$

Supply voltage, $\mathrm{V}=220 \mathrm{~V}$

Frequency, $v=50 \mathrm{~Hz}$

Angular frequency, $\omega=2 \pi 
u$

Inductive reactance, $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi 
u \mathrm{~L}=2 \pi 	imes 50 	imes 44 	imes 10^{-3} \Omega$

rms value of current is given as:

$$

I=\frac{V}{X_{L}}=\frac{220}{2 \pi 	imes 50 	imes 44 	imes 10^{-3}}=15.92 \mathrm{~A}

$$

Hence, the rms value of current in the circuit is 15.92 A .

Question 8

Q. 7.4. A $60 \mu \mathrm{~F}$ capacitor is connected to a $110 \mathrm{~V}, 60 \mathrm{~Hz}$ ac supply. Determine the rms value of the current in the circuit.

Accepted Answer

Capacitance of capacitor, $\mathrm{C}=60 \mu \mathrm{~F}=60 	imes 10^{-6} \mathrm{~F}$

Supply voltage, $\mathrm{V}=110 \mathrm{~V}$

Frequency, $v=60 \mathrm{~Hz}$

Angular frequency, $\omega=2 \pi 
u$

Capacitive reactance,

$$

X_{C}=\frac{1}{\omega C}=\frac{1}{2 \pi 
u C}=\frac{1}{2 \pi 	imes 60 	imes 60 	imes 10^{-6}} \Omega

$$

rms value of current is given as:

$$

I=\frac{V}{X_{C}}=\frac{220}{2 \pi 	imes 60 	imes 60 	imes 10^{-6}}=2.49 \mathrm{~A}

$$

Hence, the rms value of current is 2.49 A .

7. Alternating Current

Exercises - 1