Alternating Current

Question 12:

Q: A series LCR circuit has a resonance frequency of 50 Hz, a resistance of 20 Ω, and an inductance of 0.4 H. Calculate the value of the capacitance in the circuit.

Solution: The resonance frequency $f_0$ is given by:

$$f_0 = \frac{1}{2 \pi \sqrt{LC}}$$

Rearranging to find $C$ :

$$C = \frac{1}{(2 \pi f_0)^2 L}$$

Substitute the values:

$$C = \frac{1}{(2 \pi \times 50)^2 \times 0.4} = \frac{1}{(314.16)^2 \times 0.4} = 0.000255 \, \text{F} = 255 \, \mu\text{F}$$

Thus, the capacitance is $255 \, \mu\text{F}$ .

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Question 13:

Q: In an LCR circuit, the resonance occurs at a frequency of 100 Hz. If the capacitor is replaced by one having twice the capacitance, what will be the new resonance frequency?

Solution: The resonance frequency is given by:

$$f_0 = \frac{1}{2 \pi \sqrt{LC}}$$

If the capacitance is doubled, the new frequency $f_1$ will be:

$$f_1 = \frac{1}{2 \pi \sqrt{L \times 2C}} = \frac{f_0}{\sqrt{2}}$$

Substitute the given value $f_0 = 100 \, \text{Hz}$ :

$$f_1 = \frac{100}{\sqrt{2}} = 70.71 \, \text{Hz}$$

Thus, the new resonance frequency is $70.71 \, \text{Hz}$ .

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Question 14:

Q: A coil has an inductance of $2 \, \text{H}$ and carries a current of $3 \, \text{A}$ . Calculate the energy stored in the coil.

Solution: The energy stored in the coil is given by:

$$E = \frac{1}{2} L I^2$$

Substitute the values:

$$E = \frac{1}{2} \times 2 \times 3^2 = \frac{1}{2} \times 2 \times 9 = 9 \, \text{J}$$

Thus, the energy stored in the coil is $9 \, \text{J}$ .

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Question 15:

Q: In an AC circuit, the voltage is $220 \, \text{V}$ , the current is $5 \, \text{A}$ , and the phase difference between them is $30^\circ$ . Find the reactance of the circuit.

Solution: The impedance of the circuit is given by:

$$Z = \frac{V}{I} = \frac{220}{5} = 44 \, \Omega$$

The phase difference $\phi = 30^\circ$ , so:

$$Z = \sqrt{R^2 + X^2}$$

where $X$ is the reactance. Since $\cos 30^\circ = \frac{R}{Z}$ , we can find $R$ :

$$R = Z \cos 30^\circ = 44 \times \frac{\sqrt{3}}{2} = 38.05 \, \Omega$$

Thus, the reactance $X = \sqrt{Z^2 - R^2} = \sqrt{44^2 - 38.05^2} = 26.6 \, \Omega$ .

Thus, the reactance is $26.6 \, \Omega$ .

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Question 16:

Q: A 1000-turn coil has a radius of $0.1 \, \text{m}$ and is rotating with an angular velocity of $60 \, \text{rad/s}$ in a magnetic field of strength $0.5 \, \text{T}$ . Find the average induced EMF in the coil during one complete rotation.

Solution: The induced EMF is given by:

$$\text{EMF}_{\text{avg}} = N A B \omega \text{(average)}$$

where $A = \pi r^2 = \pi \times (0.1)^2 = 0.0314 \, \text{m}^2$ .

Substituting the values:

$$\text{EMF}_{\text{avg}} = 1000 \times 0.0314 \times 0.5 \times 60 = 942 \, \text{V}$$

Thus, the average induced EMF is $942 \, \text{V}$ .

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Question 17:

Q: In an LCR circuit, the inductive reactance $X_L = 30 \, \Omega$ and capacitive reactance $X_C = 40 \, \Omega$ . What is the impedance of the circuit?

Solution: The impedance $Z$ in an LCR circuit is given by:

$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$

Here, $X_L = 30 \, \Omega$ and $X_C = 40 \, \Omega$ , so:

$$Z = \sqrt{R^2 + (30 - 40)^2} = \sqrt{R^2 + 100} \, \Omega$$

Without the resistance value $R$ , we cannot calculate $Z$ . Thus, the impedance depends on the value of $R$ .

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Question 18:

Q: A 220 V AC source is connected across a purely capacitive circuit with a capacitance of $10 \, \mu\text{F}$ . Calculate the current flowing through the circuit.

Solution: The capacitive reactance $X_C$ is given by:

$$X_C = \frac{1}{2 \pi f C}$$

For $f = 50 \, \text{Hz}$ and $C = 10 \, \mu\text{F} = 10 \times 10^{-6} \, \text{F}$ :

$$X_C = \frac{1}{2 \pi \times 50 \times 10 \times 10^{-6}} = 318.31 \, \Omega$$

The current $I$ is given by:

$$I = \frac{V}{X_C} = \frac{220}{318.31} = 0.691 \, \text{A}$$

Thus, the current flowing through the circuit is $0.691 \, \text{A}$ .

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Question 19:

Q: An AC circuit has a frequency of 100 Hz. If the inductance is $0.2 \, \text{H}$ , calculate the inductive reactance.

Solution: The inductive reactance is given by:

$$X_L = 2 \pi f L$$

Substituting the given values:

$$X_L = 2 \pi \times 100 \times 0.2 = 125.66 \, \Omega$$

Thus, the inductive reactance is $125.66 \, \Omega$ .

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Question 20:

Q: A transformer steps up the voltage from 110 V to 220 V. If the primary coil has 500 turns, how many turns does the secondary coil have?

Solution: The voltage ratio in a transformer is given by:

$$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$

Substituting the given values:

$$\frac{220}{110} = \frac{N_s}{500}$$

Thus:

$$N_s = 1000 \, \text{turns}$$

Therefore, the number of turns in the secondary coil is $1000$ .

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Question 21:

Q: Calculate the power factor of a series LCR circuit where the resistance is $20 \, \Omega$ , the inductive reactance is $30 \, \Omega$ , and the capacitive reactance is $40 \, \Omega$ .

Solution: The impedance $Z$ is given by:

$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$

Substituting the values:

$$Z = \sqrt{20^2 + (30 - 40)^2} = \sqrt{400 + 100} = \sqrt{500} = 22.36 \, \Omega$$

The power factor $\cos \phi$ is given by:

$$\cos \phi = \frac{R}{Z} = \frac{20}{22.36} = 0.894$$

Thus, the power factor is $0.894$ .

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Question 22:

Q: A coil with 100 turns has a radius of $0.1 \, \text{m}$ and is placed in a magnetic field of strength $0.5 \, \text{T}$ . If the coil rotates at a frequency of 50 Hz, calculate the maximum induced EMF.

Solution: The maximum induced EMF is given by:

$$\text{EMF}_{\text{max}} = N A B \omega$$

where $A = \pi r^2 = \pi \times (0.1)^2 = 0.0314 \, \text{m}^2$ and $\omega = 2 \pi f = 2 \pi \times 50 = 314.16 \, \text{rad/s}$ .

Substituting the values:

$$\text{EMF}_{\text{max}} = 100 \times 0.0314 \times 0.5 \times 314.16 = 492.7 \, \text{V}$$

Thus, the maximum induced EMF is $492.7 \, \text{V}$ .

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Question 23:

Q: In a series LCR circuit, the inductive reactance is $50 \, \Omega$ , the capacitive reactance is $30 \, \Omega$ , and the resistance is $40 \, \Omega$ . What is the phase angle $\phi$ ?

Solution: The phase angle is given by:

$$\tan \phi = \frac{X_L - X_C}{R}$$

Substitute the given values:

$$\tan \phi = \frac{50 - 30}{40} = \frac{20}{40} = 0.5$$

Thus:

$$\phi = \tan^{-1} (0.5) = 26.57^\circ$$

Therefore, the phase angle is $26.57^\circ$ .

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Question 24:

Q: A capacitor of $5 \, \mu\text{F}$ is connected to an AC source of $230 \, \text{V}$ . If the frequency of the source is 50 Hz, find the capacitive reactance and the current in the circuit.

Solution: The capacitive reactance is given by:

$$X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \times 50 \times 5 \times 10^{-6}} = 636.6 \, \Omega$$

The current $I$ is given by:

$$I = \frac{V}{X_C} = \frac{230}{636.6} = 0.362 \, \text{A}$$

Thus, the current in the circuit is $0.362 \, \text{A}$ .

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Question 25:

Q: The resonance frequency of a series LCR circuit is $100 \, \text{Hz}$ . If the inductance is $1 \, \text{H}$ , find the capacitance in the circuit.

Solution: The resonance frequency is given by:

$$f_0 = \frac{1}{2 \pi \sqrt{LC}}$$

Rearranging for $C$ :

$$C = \frac{1}{(2 \pi f_0)^2 L}$$

Substitute the values:

$$C = \frac{1}{(2 \pi \times 100)^2 \times 1} = 0.000025 \, \text{F} = 25 \, \mu\text{F}$$

Thus, the capacitance is $25 \, \mu\text{F}$ .

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Question 26:

Q: A transformer has a primary voltage of 110 V and a secondary voltage of 220 V. If the primary current is 2 A, calculate the secondary current.

Solution: The power in the primary coil is:

$$P_{\text{primary}} = V_p I_p$$

The power in the secondary coil is:

$$P_{\text{secondary}} = V_s I_s$$

Assuming an ideal transformer, $P_{\text{primary}} = P_{\text{secondary}}$ , so:

$$V_p I_p = V_s I_s$$

Rearrange to find $I_s$ :

$$I_s = \frac{V_p I_p}{V_s} = \frac{110 \times 2}{220} = 1 \, \text{A}$$

Thus, the secondary current is $1 \, \text{A}$ .

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Question 27:

Q: In a series LCR circuit, the phase difference between the current and the voltage is 45 degrees. If the resistance is 20 Ω and the inductive reactance is 30 Ω, calculate the capacitive reactance.

Solution: The phase angle $\phi$ is given by:

$$\tan \phi = \frac{X_L - X_C}{R}$$

Substitute the given values:

$$\tan 45^\circ = \frac{30 - X_C}{20}$$

Since $\tan 45^\circ = 1$ , we have:

$$1 = \frac{30 - X_C}{20}$$

Thus:

$$30 - X_C = 20 \quad \Rightarrow \quad X_C = 10 \, \Omega$$

Therefore, the capacitive reactance is $10 \, \Omega$ .

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Question 28:

Q: A series LCR circuit has a voltage of 240 V, resistance of 40 Ω, inductive reactance of 60 Ω, and capacitive reactance of 30 Ω. Calculate the current in the circuit.

Solution: The total impedance $Z$ is given by:

$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$

Substitute the given values:

$$Z = \sqrt{40^2 + (60 - 30)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, \Omega$$

The current $I$ is given by:

$$I = \frac{V}{Z} = \frac{240}{50} = 4.8 \, \text{A}$$

Thus, the current in the circuit is $4.8 \, \text{A}$ .

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Question 29:

Q: The primary coil of a transformer has 200 turns, and the secondary coil has 600 turns. If the primary voltage is 110 V, find the secondary voltage.

Solution: The voltage ratio in a transformer is:

$$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$

Substitute the values:

$$\frac{V_s}{110} = \frac{600}{200}$$

Thus:

$$V_s = 110 \times 3 = 330 \, \text{V}$$

Therefore, the secondary voltage is $330 \, \text{V}$ .

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Question 30:

Q: A series LCR circuit has a resistance of 10 Ω, an inductive reactance of 20 Ω, and a capacitive reactance of 10 Ω. Calculate the power consumed in the circuit.

Solution: The total impedance $Z$ is given by:

$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$

Substitute the given values:

$$Z = \sqrt{10^2 + (20 - 10)^2} = \sqrt{100 + 100} = \sqrt{200} = 14.14 \, \Omega$$

The power factor is:

$$\cos \phi = \frac{R}{Z} = \frac{10}{14.14} = 0.707$$

The power consumed is:

$$P = V_{\text{rms}} I_{\text{rms}} \cos \phi$$

Since the values of $V_{\text{rms}}$ and $I_{\text{rms}}$ are not provided, we would need this information to calculate the power.