Question 1

As the frequency of the ac voltage across an inductor approaches zero,the inductive reactance of that coil

Accepted Answer

The inductive reactance of a coil is given by the formula XL = 2πfL, where f is the frequency of the ac voltage and L is the inductance of the coil. As the frequency approaches zero, the inductive reactance approaches zero as well. This is because the coil acts as a pure inductor at low frequencies and the opposition to the flow of current decreases as the frequency decreases.

Question 2

If the maximum voltage of an ac signal is 8.0 V,what is the rms value of this voltage?

Accepted Answer

The rms value of an AC signal is given by the formula rms = (maximum voltage)/(sqrt(2)).
Therefore, rms value = 8.0 V/sqrt(2) = 5.7 V (rounded to one decimal place).
So, the best choice is B.

Question 3

What size capacitor must be placed in series with a 40-mH coil so the circuit will oscillate with a frequency of 1000 Hz?

Accepted Answer

The resonant frequency $f$ of an LC circuit is given by $f = \frac{1}{2\pi\sqrt{LC}}$, where $L$ is the inductance in henries and $C$ is the capacitance in farads. Rearranging this formula to solve for $C$ gives $C = \frac{1}{(2\pi f)^2 L}$. Substituting $f = 1000$ Hz and $L = 40 \times 10^{-3}$ H (or 0.04 H) into this formula and solving for $C$ yields $C \approx 0.63 \times 10^{-6}$ F, or 0.63 μF.

Question 4

A simple ac circuit is composed of a capacitor connected across the terminals of an ac power source.If the frequency of the source is doubled,what happens to the reactance of the capacitor?

Accepted Answer

The reactance of a capacitor is given by Xc = 1/(2πfC), where f is the frequency and C is the capacitance. Doubling the frequency will result in halving the reactance, so the correct answer is D) It decreases by a factor of 2.

Question 5

A pure capacitor is connected to an ac power supply.In this circuit,the current

Accepted Answer

In a pure capacitive circuit, the current leads the voltage by 90 degrees because the capacitor charges and discharges before the applied voltage reaches its maximum or minimum value, resulting in a phase difference where the current is ahead of the voltage.

Question 6

A series circuit contains an 80-μF capacitor,a 0.020-H inductor,and a switch.The resistance of the circuit is negligible.Initially,the switch is open,the capacitor voltage is 50 V,and the current in the inductor is zero.At time t = 0 s,the switch is closed.What is the maximum current in the circuit?

Accepted Answer

The answer of A series circuit contains an 80-μF capacitor,a...

Question 7

A simple circuit consists only of of a 1.0-μF capacitor and a 15-mH coil in series.At what frequency will this circuit oscillate?

Accepted Answer

The frequency of oscillation for an LC circuit is given by the formula:

f = 1 / (2π√(LC))

Given that the circuit consists of a 1.0-μF capacitor and a 15-mH coil in series, we can substitute these values into the formula and solve for the frequency:

f = 1 / (2π√(1.0×10^-6 × 15×10^-3))

f ≈ 1.31 kHz

Therefore, the circuit will oscillate at a frequency of approximately 1.3 kHz, which corresponds to choice B.

Question 8

A simple ac circuit consists of an inductor connected across the terminals of an ac power source.If the frequency of the ac source is decreased by a factor of four,what happens to the reactance of the inductor?

Accepted Answer

The reactance of an inductor (XL) is given by XL = 2πfL, where f is the frequency. If the frequency is decreased by a factor of four, the reactance decreases by the same factor, since reactance is directly proportional to frequency.

Question 9

A series circuit contains an 80-μF capacitor,a 0.020-H inductor,and a switch.The resistance of the circuit is negligible.Initially,the switch is open,the capacitor voltage is 50 V,and the current in the inductor is zero.At time t = 0 s,the switch is closed.When the current in the circuit is 1.5 A,what is the charge on the capacitor?

Accepted Answer

The answer of A series circuit contains an 80-μF capacitor,a...

Question 10

Which of the following statements is true for a series RLC ac circuit in resonance? (There could be more than one correct choice).

Accepted Answer

In resonance, the inductive reactance (XL) equals the capacitive reactance (XC), causing them to cancel each other out, leading to a condition where the impedance is minimal and purely resistive. This results in the current in the circuit being at its maximum.

Question 11

What is the phase angle between the voltages of the inductor and capacitor in a RLC series circuit?

Accepted Answer

In a series RLC circuit, the voltage across the inductor leads the current by 90°, and the voltage across the capacitor lags the current by 90°. Therefore, the phase angle between the voltages of the inductor and capacitor is 180°.

Question 12

In an RLC ac circuit,the values of the inductance and capacitance are both doubled.As a result of this change,the resonance frequency of the circuit is

Accepted Answer

The resonance frequency of an RLC circuit is given by $ f_0 = \frac{1}{2\pi\sqrt{LC}} $. If both L and C are doubled, the new resonance frequency becomes $ f_0' = \frac{1}{2\pi\sqrt{2L \cdot 2C}} = \frac{1}{2\pi\sqrt{4LC}} = \frac{1}{2} \times \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2} f_0 $. Thus, the resonance frequency is reduced to one-half the original value.

Question 13

A series circuit contains an 80-μF capacitor,a 0.020-H inductor,and a switch.The resistance of the circuit is negligible.Initially,the switch is open,the capacitor voltage is 50 V,and the current in the inductor is zero.At time t = 0 s,the switch is closed.When the potential across the capacitor is 30 V,what is the rate of change of the current?

Accepted Answer

The energy stored in the capacitor when the potential across it is 30 V can be calculated using the formula $U = \frac{1}{2}CV^2$, where $U$ is the energy, $C$ is the capacitance, and $V$ is the voltage. Initially, the energy stored in the capacitor when the voltage is 50 V is $U_1 = \frac{1}{2} \times 80 \times 10^{-6} \times 50^2$. When the voltage drops to 30 V, the energy is $U_2 = \frac{1}{2} \times 80 \times 10^{-6} \times 30^2$.The difference in energy, $\Delta U$, is transferred to the inductor and is equal to the energy stored in the inductor, which can be expressed as $U_L = \frac{1}{2}LI^2$, where $L$ is the inductance and $I$ is the current. However, we are interested in the rate of change of current, $\frac{dI}{dt}$, not the current itself.Given that the energy in the inductor is also a function of the rate of change of current through the relation $U_L = \frac{1}{2}L\left(\frac{dI}{dt}\right)^2$, we can equate the change in energy in the capacitor to the energy in the inductor to solve for $\frac{dI}{dt}$.The calculation of $\Delta U$ and solving for $\frac{dI}{dt}$ involves several steps, but the key insight is that the energy lost by the capacitor is gained by the inductor. The correct rate of change of current, given the options, is found by applying these principles and matching the energy transfer with the provided options, leading to the correct answer of 1500 A/s. This involves detailed calculations, but the essence is understanding energy conservation between the capacitor and inductor in an LC circuit.

Question 14

What inductance is needed in series with a 4.7-μF capacitor so the circuit will oscillate at a frequency of 10 kHz?

Accepted Answer

To calculate the inductance needed in series with a 4.7-μF capacitor to produce an oscillation frequency of 10 kHz, we can use the following formula:

f = 1 / (2π√(LC))

where:
f = frequency (in Hz)
L = inductance (in Henries)
C = capacitance (in Farads)
π = 3.14159...

Rearranging the formula to solve for L, we get:

L = 1 / (4π²f²C)

Plugging in the given values, we get:

L = 1 / (4π² x 10,000² x 4.7 x 10^-6)
  = 54 x 10^-6 (or 54 μH)

Therefore, the inductance needed in series with the 4.7-μF capacitor to produce an oscillation frequency of 10 kHz is 54 μH.

Question 15

In the figure,which of the phasor diagrams represents a series RLC circuit driven at resonance?

Accepted Answer

At resonance, the reactive components cancel each other out and the circuit behaves purely resistively. This can be seen in phasor diagram 3, where the impedance vector is horizontal (indicating purely resistive) and in phase with the current vector. Phasor diagrams 1, 2, 4, and 5 show non-resonant behavior, where the current and voltage vectors are out of phase and/or the impedance vector has a non-horizontal component.

Question 16

A pure inductor is connected to an ac power supply.In this circuit,the current

Accepted Answer

A pure inductor has a reactance that opposes the flow of current in the circuit, which means that the current lags behind the voltage by 90°. This is because the voltage reaches its maximum value before the current does. Therefore, the correct choice is D, which says that the current lags the voltage by 90°.

Question 17

Consider a capacitor connected across the ac source.As the capacitance is increased,the current through the capacitor will

Accepted Answer

The current through a capacitor in an AC circuit is given by $I = V\omega C$, where $I$ is the current, $V$ is the voltage, $\omega$ is the angular frequency of the AC source, and $C$ is the capacitance. Increasing the capacitance ($C$) directly increases the current ($I$), assuming voltage and frequency remain constant.

Question 18

In a RLC circuit,the values of the inductance and capacitance are both halved.As a result of this change,the resonance frequency of the circuit is

Accepted Answer

The resonance frequency of an RLC circuit is given by $f = \frac{1}{2\pi\sqrt{LC}}$. Halving both L and C results in the square root of a quarter ($\sqrt{\frac{1}{4}} = \frac{1}{2}$) in the denominator, which is equivalent to multiplying the original frequency by 2, thus doubling the resonance frequency.

Question 19

A series circuit contains an 80-μF capacitor,a 0.020-H inductor,and a switch.The resistance of the circuit is negligible.Initially,the switch is open,the capacitor voltage is 50 V,and the current in the inductor is zero.At time t = 0 s,the switch is closed.How long after closing the switch does the magnetic energy in the inductor first reach its maximum value?

Accepted Answer

The magnetic energy in the inductor reaches its maximum value when the capacitor is fully discharged and all its energy has been transferred to the inductor. This occurs at a quarter of the oscillation period of the LC circuit. The oscillation period $T$ is given by $T = 2\pi\sqrt{LC}$, where $L$ is the inductance and $C$ is the capacitance. Substituting the given values, $L = 0.020 \, \text{H}$ and $C = 80 \times 10^{-6} \, \text{F}$, we find $T = 2\pi\sqrt{(0.020)(80 \times 10^{-6})} = 2\pi\sqrt{1.6 \times 10^{-3}} \approx 2\pi\sqrt{1.6} \times 10^{-1.5}$. A quarter of this period, $T/4$, is when the magnetic energy is at its maximum, which is approximately $2.0 \, \text{ms}$.

Question 20

What capacitance is needed in series with a 3.7-mH inductor so the circuit will oscillate at a frequency of 1000 Hz?

Accepted Answer

The resonant frequency of a series RLC circuit is given by:

f = 1/(2π√LC)

Rearranging the formula and solving for C, we get:

C = 1/(4π^2f^2L)

Substituting the given values, we get:

C = 1/(4π^2(1000)^2(3.7×10^-3))

C ≈ 6.80 μF

Therefore, the capacitance needed in series with the 3.7-mH inductor is 6.8 μF.

The closest answer choice is C.

Quiz 24: Alternating-Current Circuits