Question 1

The 60-Hz ac source of the series circuit shown in the figure has a voltage amplitude of 120 V. The capacitive reactance is 790 Ω, the inductive reactance is 270 Ω, and the resistance is 500Ω.  
(a) What is the capacitance of the capacitor?
(b) What is the inductance of the inductor?

Accepted Answer

(a) 3.4 µF 
(b) 720 mH

Question 2

An LRC series circuit has R = 15.0 Ω, L = 25.0 mH, and C = 30.0 μF. The circuit is connected to a 120-V (rms) ac source with frequency 200 Hz.
(a) What is the impedance of the circuit?
(b) What is the rms current in the circuit?
(c) What is the rms voltage across the resistor?
(d) What is the rms voltage across the inductor?
(e) What is the rms voltage across the capacitor?

Accepted Answer

(a) 15.8 Ω 
(b) 7.61 A 
(c) 114 V (d) 239 V 
(e) 202 V

Question 3

An alternating current is supplied to an electronic component with a warning that the voltage across it should never exceed 12 V. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit in the warning?

Accepted Answer

The warning specifies a peak voltage limit of 12 V. The RMS voltage is related to the peak voltage by the equation VRMS = Vpeak / √2. Rearranging the equation gives Vpeak = VRMS x √2. So, the highest peak voltage that can be supplied is 12 V, which corresponds to an RMS voltage of 12 V / √2 ≈ 8.5 V. Of the available choices, only A is less than 8.5 V, so A is the correct answer.

Question 4

Which one of the phasor diagrams shown below best represents a series LRC circuit driven at resonance?

Accepted Answer

The answer of Which one of the phasor diagrams shown...

Question 5

The 60-Hz ac source of a series circuit has a voltage amplitude of 120 V. The resistance, capacitive reactance, and inductive reactance are as shown in the figure. What is the rms current in the circuit?

Accepted Answer

The answer of The 60-Hz ac source of a series...

Question 6

A transformer changes the 10,000 V power line to 120 V. If the primary coil contains 750 turns, how many turns are on the secondary?

Accepted Answer

The answer of A transformer changes the 10,000 V power...

Question 7

The primary coil of a transformer has 100 turns and its secondary coil has 400 turns. If the ac voltage applied to the primary coil is 120 V, what voltage is present in its secondary coil?

Accepted Answer

The ratio of the number of turns in the secondary coil to the number of turns in the primary coil is 4:1. Therefore, the voltage in the secondary coil will be four times the voltage in the primary coil.
Voltage in secondary coil = 4 x voltage in primary coil
Voltage in secondary coil = 4 x 120 V
Voltage in secondary coil = 480 V

Question 8

When an LRC series circuit is driven at resonance, which of the following statements about the circuit are correct? (There may be more than one correct choice.)

Accepted Answer

At resonance in an LRC series circuit, the inductive reactance (XL) and capacitive reactance (XC) are equal and cancel each other out, leading to the circuit having its minimum impedance, which is primarily resistive.

Question 9

When an LRC series circuit is at resonance, which one of the following statements about that circuit is accurate? (There may be more than one correct choice.)

Accepted Answer

The answer of When an LRC series circuit is at...

Question 10

An ac series circuit consists of a voltage source, an 880-Ω resistor, and an inductance L. (There is no capacitance in the circuit.) The current amplitude is 0.60 A, and the phase angle between the source voltage and the current has a magnitude of 41°
(a) Does the source voltage lag or lead the current?
(b) What is the voltage amplitude of the source?

Accepted Answer

The answer of An ac series circuit consists of a...

Question 11

In a series LRC circuit, the frequency at which the circuit is at resonance is f0. If you double the resistance, the inductance, the capacitance, and the voltage amplitude of the ac source, what is the new resonance frequency?

Accepted Answer

The answer of In a series LRC circuit, the frequency...

Question 12

When a current of 2.0 A flows in the 100-turn primary of an ideal transformer, this causes 14 A to flow in the secondary. How many turns are in the secondary?

Accepted Answer

The answer of When a current of 2.0 A flows...

Question 13

In a transformer, how many turns are necessary in a 110-V primary if the 24-V secondary has 100 turns?

Accepted Answer

The number of turns in a transformer's primary coil is directly proportional to the voltage across it, following the formula $V_p/V_s = N_p/N_s$, where $V_p$ and $V_s$ are the primary and secondary voltages, and $N_p$ and $N_s$ are the primary and secondary turns, respectively. Substituting the given values, $110/24 = N_p/100$, solving for $N_p$ gives approximately 458 turns.

Question 14

An ideal transformer consists of a 500-turn primary coil and a 2000-turn secondary coil. If the current in the secondary is 3.0 A, what is the current in the primary?

Accepted Answer

The current in the primary coil can be found using the transformer equation $I_1 = \frac{N_2}{N_1}I_2$, where $I_1$ is the current in the primary coil, $N_1$ and $N_2$ are the number of turns in the primary and secondary coils respectively, and $I_2$ is the current in the secondary coil. Substituting the given values, $I_1 = \frac{2000}{500} \times 3.0 = 4 \times 3.0 = 12$ A.

Question 15

A 120-V rms voltage at 1000 Hz is applied to an inductor, a 2.00-μF capacitor and a 100-Ω resistor, all in series. If the rms value of the current in this circuit is 0.680 A, what is the inductance of the inductor?

Accepted Answer

First, we can find the total impedance of the circuit using the formula:

$Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$

where $R$ is the resistance, $\omega$ is the angular frequency (which is $2\pi$ times the frequency), $L$ is the inductance, and $C$ is the capacitance.

Plugging in the given values, we get:

$Z = \sqrt{100^2 + (2\pi \cdot 1000 \cdot L - \frac{1}{2\pi \cdot 1000 \cdot 2	imes 10^{-6}})^2}$

Simplifying and setting $Z$ equal to the ratio of the rms voltage and rms current (which is just the impedance in a series circuit), we get:

$\frac{120 	ext{ V}}{0.680 	ext{ A}} = \sqrt{100^2 + (2\pi \cdot 1000 \cdot L - \frac{1}{2\pi \cdot 1000 \cdot 2	imes 10^{-6}})^2}$

Squaring both sides and solving for $L$, we get:

$L = \frac{1}{(2\pi \cdot 1000)^2} \left[\frac{(120/0.680)^2 - 100^2}{\left(\frac{120}{0.680}ight)^2} + \frac{1}{(2\pi \cdot 1000 \cdot 2	imes 10^{-6})^2}ight] \approx 35.8 	ext{ mH}$

Therefore, the best choice is B.

Question 16

The inductor in a radio receiver carries a current of amplitude 200 mA when a voltage of amplitude 2.40 V is across it at a frequency of 1400 kHz. What is the value of the inductance?

Accepted Answer

The answer of The inductor in a radio receiver carries...

Question 17

A series ac circuit consists of an inductor having a reactance of 80 Ω and an inductance of 190 mH, a 40-Ω resistor, a capacitor whose reactance is 100 Ω, and an ac source. The rms current in the circuit is measured to be 2.2 A. What is the rms voltage of the source?

Accepted Answer

The rms voltage of the source can be calculated using the formula $V = IZ$, where $V$ is the rms voltage, $I$ is the rms current, and $Z$ is the impedance of the circuit. The impedance $Z$ of a series RLC circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$, where $R$ is the resistance, $X_L$ is the inductive reactance, and $X_C$ is the capacitive reactance. Substituting the given values, $Z = \sqrt{40^2 + (80 - (-100))^2} = \sqrt{1600 + 180^2} = \sqrt{1600 + 32400} = \sqrt{34000} = 184.39 \Omega$. Therefore, the rms voltage $V = 2.2 A \times 184.39 \Omega = 405.658 V$. However, this calculation seems to have an error in interpretation of the correct answer options provided, indicating a mistake in the calculation process. Correctly calculating, the impedance should be considered with the correct subtraction for reactances, and the final voltage calculation should align with the provided options. Correcting the calculation: $Z = \sqrt{40^2 + (80 - 100)^2} = \sqrt{1600 + 400} = \sqrt{2000} = 44.72 \Omega$. Then, $V = IZ = 2.2 \times 44.72 = 98.384 V$, which rounds to approximately 98 V, matching option A. The initial calculation provided was incorrect; the correct approach gives us the answer as 98 V, aligning with option A.

Question 18

An ac circuit is shown in the figure. The rms current in the circuit is measured to be 1.8 A. What is the capacitance of the capacitor?

Accepted Answer

The answer of An ac circuit is shown in the...

Question 19

An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and inductor, or a capacitor. At time t = 0 the voltage is zero. At time t = T/4 the current in the unknown element is equal to zero, and at time t = T/2 the current is I = -Imax, where Imax is the current amplitude. What is the unknown element?

Accepted Answer

The current in a capacitor leads the voltage by 90 degrees (T/4 in time), so at t = T/4, the current is zero when the voltage is at its maximum. At t = T/2, the current is at its negative maximum, which matches the behavior of a capacitor.

Question 20

If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the frequency is closest to:

Accepted Answer

We can use the formula for capacitive reactance, which relates the capacitance, frequency, and current through the capacitor:

Xc = 1/(2πfC)

where Xc is the capacitive reactance, f is the frequency, and C is the capacitance.

At the given voltage amplitude and current amplitude, the capacitive reactance is:

Xc = V/I = 12.0 V / 3.33 mA = 3.60 kΩ

We can rearrange the formula to solve for frequency:

f = 1/(2πXcC)

Substituting the given values, we get:

f = 1/(2π × 3.60 kΩ × 8.50 nF) ≈ 5.20 MHz

Therefore, the answer is B, 5.20 MHz.

Quiz 30: Alternating Current

The 60-Hz ac source of the series circuit shown in the figure has a voltage amplitude of 120 V. The capacitive reactance is 790 Ω, the inductive reactance is 270 Ω, and the resistance is 500Ω. (a) What is the capacitance of the capacitor? (b) What is the inductance of the inductor?