Chapter 7: Alternating Current

Q7.1Written answer

A 100 Ω resistor is connected to a 220 V, 50 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?

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Worked solution

Given :

The resistance R of the resistor is 100 Ω

The source voltage V is 220 V

The frequency of the supply is 50 Hz.

a) To determine the value of the current in the connection, we can use the following relation:

\(\begin{array}{l}I = \frac{ V }{ R }\end{array} \)

Substituting values, we get

\(\begin{array}{l}I = \frac{ 220 }{ 100 }\end{array} \)

Therefore, the RMS value of the current in the connection is 2.20 A.

b) The total power consumed over an entire cycle can be calculated using the following formula:

P = V x I

Substituting values in the above equation, we get

= 220 x 2.2 = 484 W

Therefore, the total power consumed is 484 W.

Q7.2Written answer

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?

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Worked solution

a) The peak voltage of the AC supply is V₀ = 300 V.

We know that,

\(\begin{array}{l}V _{RMS }\end{array} \)

\(\begin{array}{l}V = \frac{ V _{0}}{ \sqrt{ 2 }}\end{array} \)

Substituting the values, we get

\(\begin{array}{l}V = \frac{ 300 }{ \sqrt{ 2 }}\end{array} \)

Hence, the RMS voltage of the AC supply is 212.2 V.

b) The RMS value of the current in the circuit is I = 10 A

We can calculate the peak current from the following equation,

\(\begin{array}{l}I _{ 0 } = \sqrt{ 2 } I\end{array} \)

Substituting the values, we get

\(\begin{array}{l} \sqrt{ 2 } \times 10 = 14.1 A\end{array} \)

Therefore, the peak current is 14.1 A.

Q7.3Written answer

A 44 mH inductor is connected to a 220 V, 50 Hz ac supply. Determine the RMS value of the current in the circuit.

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Worked solution

As given :

The inductor connected to the AC supply has an inductance of L = 44 m H = 44 × 10 ^{– 3} H

The magnitude of the source voltage V is 220 V

The frequency of the source is ν = 50 Hz

The angular frequency of the source is given by ω = 2 π ν

The inductive reactance X _L can be calculated as follows:

ω _L = 2 π ν L = 2π × 50 × 44 × 10 ^{– 3} Ω

To find the RMS value of the current, we use the following relation:

\(\begin{array}{l}I = \frac{ V }{ X _{ L }} = \frac{ 220 }{ 2\pi \times 50 \times 44 \times 10 ^{ – 3 }}\end{array} \)

Therefore, the RMS value of the current in the network is 15.92 A.

Q7.4Written answer

A 60 µF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the RMS value of the current in the circuit.

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Worked solution

Given:

The capacitance of the capacitor in the circuit is C = 60 μ F or 60 × 10 ^{– 6} F

The source voltage is V = 110 V

The frequency of the source is ν = 60 Hz

The angular frequency can be calculated using the following relation,

ω = 2πν

The capacitive reactance in the circuit is calculated as follows:

\(\begin{array}{l}X _{ C } = \frac{ 1 }{ \omega C } = \frac{ 1 }{ 2\pi \nu C} = \frac{ 1 }{ 2 \pi \times 60 \times 60 \times 10 ^{ – 6 }}\end{array} \)

Now, the RMS value of the current is determined as follows:

\(\begin{array}{l}I = \frac{ V }{ X_{ C }} = \frac{ 110 }{ 2\pi \times 60 \times 60 \times 10^{ – 6 }} = 2.49 A\end{array} \)

Therefore, the RMS current is 2.49 A.

Q7.5Written answer

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

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Worked solution

a) We know that in the case of the inductive network,

The RMS current value is I = 15.92 A

The RMS voltage value is V = 220 V

Therefore, the total power taken can be derived by the following equation:

P = VI cos Φ

Here,

Φ is the phase difference between V and I

In the case of a purely inductive circuit, the difference in the phase of an alternating voltage and an alternating current is 90°,

i.e. Φ = 90°.

Therefore, P = 0

Thus, the total power absorbed by the circuit is zero.

b) In the case of the capacitive network, we know that

The value of RMS current is given by, I = 2.49 A

The value of RMS voltage is given by, V = 110 V

Thus, the total power absorbed can be derived from the following equation:

P = VI Cos Φ

For a purely capacitive circuit, the phase difference between alternating voltage and alternating current is 90°

i.e. Φ = 90°.

Thus , P = 0

Therefore, the net power absorbed by the circuit is zero.

Q7.6Written answer

Obtain the resonant frequency ω_r of a series LCR circuit with L = 2.0H, C = 32 µF and R = 10 Ω. What is the Q-value of this circuit?

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Worked solution

Given:

The inductance of the inductor is L = 2.0 H

The capacitance of the capacitor, C = 32 μF = 32 × 10 ^{– 6} F

The resistance of the resistor R = 10 Ω.

We know that resonant frequency can be calculated by the following relation,

\(\begin{array}{l}\omega _{ r } = \frac{ 1 }{ \sqrt{ LC }} = \frac{ 1 }{ \sqrt{ 2 \times 32 \times 10 ^{ – 6 }}} = \frac{ 1 }{ 8 \times 10 ^{ – 3 }} = 125 \frac{ rad }{ sec }\end{array} \)

Now, the Q-value of the circuit can be calculated as follows:

\(\begin{array}{l}Q = \frac{ 1 }{ R } \sqrt{\frac{ L }{ C }} = \frac{ 1 }{ 10 }\sqrt{\frac{ 2 }{ 32 \times 10 ^{ – 6 }}} = \frac{ 1 }{ 10 \times 4 \times 10 ^{ – 3 }} = 25\end{array} \)

Thus, the Q-value of the circuit is 25.

Q7.7Written answer

A charged 30 µF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

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Worked solution

Given capacitance value of the capacitor, C = 30 μF = 30 × 10 ^{– 6} F

Given inductance value of the charged inductor, L = 27 mH = 27 × 10 ^{– 3} H

Angular frequency is given as:

\(\begin{array}{l}\omega _{ r } = \frac{ 1 }{ \sqrt{ LC }}\end{array} \)

\(\begin{array}{l}\omega _{ r } = \frac{ 1 }{ \sqrt{ 27 \times 10 ^{ – 3 } \times 30 \times 10 ^{ – 6 }}} = \frac{ 1 }{ 9 \times 10 ^{ – 4 }} = 1.11 \times 10 ^{ 3 }\, rad\,sec^{-1}\end{array} \)

Therefore, the calculated angular frequency of free oscillation of the connection is 1.11 x 10 ³ s^-1.

Q7.8Written answer

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 a later time?

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Worked solution

Given capacitance value of the capacitor, C = 30 μF = 30 × 10 ^{– 6} F

Inductance of the inductor, L = 27 mH = 27 × 10 ^{– 3} H

Charge on the capacitor, Q = 6 mC = 6 × 10 ^{– 3} C

The total energy stored in the capacitor can be calculated as follows:

\(\begin{array}{l}E = \frac{ 1 }{ 2 } \times \frac{ Q ^{ 2 }}{ C } = \frac{ 1 }{ 2 }\frac{\left ( 6 \times 10^{ -3 } \right )^{ 2 }}{ 30 \times 10 ^{ – 6}} = \frac{ 6 }{ 10 } = 0.6 J\end{array} \)

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

Q7.9Written answer

A series LCR circuit with R = 20 Ω, L = 1.5 H, and C = 35 µF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

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Worked solution

The supply frequency and the natural frequency are equal at resonance conditions in the circuit.

The given resistance of the resistor, R = 20 Ω

The given inductance of the inductor, L = 1.5 H

The given capacitance of the capacitor, C = 35 μF = 30 × 10 ^{– 6} F

An AC source with a voltage of V = 200 V is connected to the LCR circuit

We know that the impedance of the above combination can be calculated by the following relation:

\(\begin{array}{l}Z = \sqrt{ R ^{ 2 } + \left ( X _{ L } – X _{ C } \right ) ^{ 2 }}\end{array} \)

At resonant condition in the circuit, X _L = X _C

Therefore , Z = R = 20 Ω

We know that the current in the network is given by the relation:

\(\begin{array}{l}I = \frac{ V }{ Z } = \frac{ 200 }{ 20 } = 10 A\end{array} \)

Therefore, the average power that is being transferred to the circuit in one full cycle:

V I = 200 x 10 = 2000 W

Q7.10Written answer

A radio can tune over the frequency range of a portion of the MW broadcast band: 800 kHz to 1200 kHz. If its LC circuit has an effective inductance of 200 µH, what must be the range of its variable capacitor?

[Hint: The condition for tuning is that the natural frequency, which is the frequency of free oscillations of the LC network, must be of the same value as the radio wave frequency]

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Worked solution

Given the frequency range of (ν) of radio is 800 kHz to 1200 kHz.

Given that the lower tuning frequency of the circuit, ν ₁ = 800 kHz = 800 × 10 ³ Hz

Given that the upper tuning frequency of the circuit, ν ₂ = 1200 kHz = 1200 × 10 ³ Hz

Given that the effective inductance of the inductor in the circuit is L = 200 μH = 200 × 10 ^{– 6} H

We know that the capacitance of variable capacitor for ν ₁ can be calculated as follows :

\(\begin{array}{l}C _{ 1 } = \frac{ 1 }{ \omega _{ 1 }^{ 2 } L }\end{array} \)

Here, the variables are,

ω ₁ = Angular frequency for capacitor C₁

ω ₁= 2 π ν ₁

ω ₁ = 2π × 800 × 10³ rad/s

Therefore,

\(\begin{array}{l}C _{ 1 } = \frac{ 1 }{ \left ( 2\pi \times 800 \times 10 ^{ 3 } \right )^{2} \times 200 \times 10 ^{ – 6 }} = 1.9809 \times 10 ^{ – 10 } F = 198.1 pF\end{array} \)

The variable capacitor for v₂ has a capacitance of

\(\begin{array}{l}C _{ 2 } = \frac{ 1 }{ \omega _{ 2 }^{ 2 } L }\end{array} \)

Here, the variables are,

ω ₂ = Angular frequency for capacitor C₂

ω ₂= 2 π ν ₂

ω ₂ = 2π × 1200 × 10³ rad/s

Therefore,

\(\begin{array}{l}C _{ 2 } = \frac{ 1 }{ \left ( 2\pi \times 1200 \times 10 ^{ 3 } \right )^{2} \times 200 \times 10 ^{ – 6 }} = 0.8804 \times 10 ^{ – 10 } F = 88.04 pF\end{array} \)

Thus, the variable capacitor has a range from 88.04 pF to 198.1 pF.

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