Question 1

Whenever the alternating current frequency in a series RLC circuit is halved,
A) the inductive reactance is doubled and the capacitive reactance is halved.
B) the inductive reactance is doubled and the capacitive reactance is doubled.
C) the inductive reactance is halved and the capacitive reactance is halved.
D) the inductive reactance is halved and the capacitive reactance is doubled.
E) the reactance of the circuit remains the same.

Accepted Answer

In a series RLC circuit, the total impedance at any frequency is given by:

Z = √(R² + (Xl - Xc)²)

Where R is the resistance, Xl is the inductive reactance and Xc is the capacitive reactance.

At resonance, Xl = Xc and the total impedance is purely resistive, i.e. Z = R.

When the frequency is halved, the inductive reactance (Xl) and capacitive reactance (Xc) both change according to the formulas:

Xl = 2πfL

Xc = 1/(2πfC)

Where L is the inductance and C is the capacitance.

Halving the frequency means dividing f by 2, so we get:

Xl' = 2π(f/2)L = πL

Xc' = 1/(2π(f/2)C) = 2/(πC)

Thus, the inductive reactance is halved (Xl' = 0.5Xl) and the capacitive reactance is doubled (Xc' = 2Xc).

Therefore, the total impedance of the circuit is altered since it's no longer purely resistive. This change in impedance affects the behavior of the circuit, causing it to behave differently at different frequencies. So the correct choice is D.

Question 2

An alternating current circuit has resistance R, inductance L and capacitance C in series with a voltage source. Which statement is correct?
A) The voltage across the capacitor leads the voltage across the inductor by 90°.
B) The voltage across the inductor leads the voltage across the capacitor by 90°.
C) The voltage across the inductor leads the voltage across the resistor by 180°.
D) The voltage across the inductor is out of phase with the voltage across the capacitor by 180°.
E) Both voltages lead the voltage across the resistor by 90°.

Accepted Answer

The voltage across the inductor is out of phase with the voltage across the capacitor by 180°.

Question 3

All three circuits below have R = 100 Ω, C = 1.0 mF and emf ε = (5.0 V) sin (377 t). The inductors in (B) and (C) are placed sufficiently far apart so that they do not alter one another's inductance. Such inductors add combine like resistors. Which statement regarding the angular resonance frequencies ω_A, ω_B and ω_C is correct? A) ω_C > ω_A = ω_B B) ω_C < ω_A = ω_B C) ω_A = ω_B = ω_C D) ω_B < ω_A = ω_C E) ω_B > ω_A = ω_C

Accepted Answer

The angular resonance frequency of a series RLC circuit can be calculated using the formula ω = 1/√(LC). Since all three circuits have the same values of R, C, and L (assuming the inductors in (B) and (C) add like resistors), their resonance frequencies are the same. Therefore, ω_A = ω_B = ω_C, and the correct answer is C.

Question 4

A 10-μF capacitor in an LC circuit made entirely of superconducting materials (R = 0 Ω) is charged to 100 μC. Then a superconducting switch is closed. At t = 0 s, plate 1 is positively charged and plate 2 is negatively charged. At a later time, V_ab = +10 V. At that time, V_dc is A) 0 V. B) 3.54 V. C) 5.0 V. D) 7.07 V. E) 10 V.

Accepted Answer

The answer of A 10-μF capacitor in an LC circuit...

Question 5

A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RL circuit with R = 10.0 Ω and L = 10.0 mH. The peak current in the circuit is
A) 0.963 A.
B) 10.3 A.
C) 11.0 A.
D) 11.9 A.
E) 29.2 A.

Accepted Answer

The answer of A 60.0-Hz ac generator with a peak...

Question 6

A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RC circuit with R = 10.0 Ω and C = 300 μF. The peak current in the circuit is
A) 8.24 A.
B) 8.84 A.
C) 11.0 A.
D) 12.4 A.
E) 23.5 A.

Accepted Answer

The peak current in an AC circuit is given by I_peak = V_peak / Z, where Z is the impedance. For a series RC circuit, Z = sqrt(R^2 + (1/(ωC))^2), where ω = 2πf. Substituting the given values, Z = sqrt(10^2 + (1/(2π*60*300*10^-6))^2) ≈ 13.35 Ω. Thus, I_peak = 110 V / 13.35 Ω ≈ 8.24 A.

Question 7

All three circuits shown below have R = 100 Ω, L = 0.1 H and emf ε = (5.0 V) sin (377 t). Which statement regarding the angular resonance frequencies ω_A, ω_B and ω_C is correct? A) ω_C > ω_A = ω_B B) ω_C < ω_A = ω_B C) ω_A = ω_B = ω_C D) ω_B < ω_A = ω_C E) ω_B > ω_A = ω_C

Accepted Answer

All three circuits have the same values for R, L, and emf, therefore they will all have the same angular resonance frequency given by:

ω = √(1/(LC) - (R/2L)^2)

Plugging in the values, we get:

ω = √(1/(0.1*100*10^-6) - (100/2*0.1)^2) = 377 rad/s

Therefore, the angular resonance frequency is the same for all three circuits, and the correct choice is C.

Question 8

The graphs below show the phasors ΔV_max and I_max for five RLC series circuits. The solid arrows represent the voltage phasors, ΔV, and the dashed arrows represent the current phasors, I_max. The graph which represents a circuit where the inductive reactance is greater than the capacitive reactance is A) B) C) D) E)

Accepted Answer

The answer of The graphs below show the phasors ΔV_max...

Question 9

The graphs below represent current and voltage phasors at one instant of time. The solid arrows represent the voltage phasors, ΔV_max, and the dashed arrows represent the current phasors, I_max. The graph which shows the correct relationship between current and voltage phasors for a capacitor in an RC circuit is A) B) C) D) E)

Accepted Answer

The answer of The graphs below represent current and voltage...

Question 10

The power output, P_out, of an ideal step-up transformer that receives power input, P_in, and which has N₁ turns in the primary and N₂ turns in the secondary coil is given by A) P_out = P_in. B) P_out = P_in. C) P_out = P_in. D) P_out = P_in. E) P_out = P_in.

Accepted Answer

The answer of The power output, P_out, of an ideal...

Question 11

The average power input to a series alternating current circuit is minimum when A) there are only a resistor and capacitor in the circuit. B) there are only a resistor and inductor in the circuit. C) there is only a resistor in the circuit. D) X_L = X_C and the circuit contains a resistor, an inductor and a capacitor. E) there is only a capacitor in the circuit.

Accepted Answer

The answer of The average power input to a series...

Question 12

The graphs below show the phasors ΔV_max and I_max for five RLC series circuits. The solid arrows represent the voltage phasors, ΔV, and the dashed arrows represent the current phasors, I_max. The graph which represents a circuit where the capacitive reactance is greater than the inductive reactance is A) B) C) D) E)

Accepted Answer

In a purely capacitive circuit, the phasor for the current leads the phasor for the voltage, while in a purely inductive circuit, the phasor for the voltage leads the phasor for the current. In the given options, only option A shows a phasor where the current phasor is ahead of the voltage phasor, indicating a purely capacitive circuit where Xc>XL.

Question 13

The graphs below show a voltage phasor at different instances of time. The voltage phasor which shows the instantaneous value of the voltage with the largest magnitude is
A)

B)

C)

D)

E)

Accepted Answer

The answer of The graphs below show a voltage phasor...

Question 14

A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RC circuit with R = 10.0 Ω and C = 300 μF. The power factor, cos φ, is
A) −1.00.
B) −0.749.
C) +0.749.
D) +0.834.
E) +1.00.

Accepted Answer

We can calculate the impedance of the circuit using:
Z = √(R^2 + (1/ωC)^2) = √(10^2 + (1/(2π*60*300*10^-6))^2) ≈ 32.4 Ω
The current in the circuit is given by:
I = Vpeak/Z = 110/√2/32.4 ≈ 2.15 A
The power factor can be calculated using:
cos φ = R/Z = 10/32.4 ≈ 0.309
However, since this is a series RC circuit, the phase angle between the voltage and current is not just determined by the resistance, but also by the capacitance. In fact, the capacitive reactance, XC = 1/ωC, leads the current by 90 degrees, while the resistance, R, lags the current by 90 degrees. This means that the phase angle between the voltage and current is given by:
tan φ = XC/R = 1/ωCR ≈ 0.762
φ ≈ 37.1 degrees
cos φ = cos(37.1) ≈ 0.749
Therefore, the power factor is 0.749, which is choice C.

Question 15

In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of line has resistance R. For a power line that supplies power to 10000 households, we can conclude that A) IV = I²R. B) I = V/R. C) IV < I²R. D) IV > I²R. E) I²R = 0.

Accepted Answer

In a transmission line, power loss due to resistance is given by I²R, and the power delivered is IV. Since the current is very small and voltage is very large, IV (power delivered) is greater than I²R (power loss).

Question 16

A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RL circuit with R = 10.0 Ω and L = 10.0 mH. The power factor, cos φ, is
A) −1.00.
B) −0.936.
C) +0.943.
D) +0.936.
E) +1.00.

Accepted Answer

To find the power factor, we need to calculate the impedance of the circuit, which is given by:

$Z = \sqrt{R^2+(2\pi fL)^2}$
where f is the frequency of the generator.

Plugging in the given values, we get:

$Z = \sqrt{10^2 + (2\pi 	imes 60 	imes 0.01)^2} = 12.42$ Ω

The current in the circuit is given by:

$I = \frac{V_{peak}}{Z} = \frac{110}{12.42} = 8.86$ A

The phase angle between the voltage and current is given by:

$\phi = 	an^{-1}\left(\frac{2\pi fL}{R}ight) = 	an^{-1}\left(\frac{2\pi 	imes 60 	imes 0.01}{10}ight) = 1.19$ rad

The power factor is then given by:

$\cos\phi = \cos(1.19) = 0.936$

Therefore, the answer is D.

Question 17

In a parallel RLC circuit, where I_R = IR_{, max} sin(ωt), the current through the capacitor, I_C, is A) I_C = −IC_{, max} sin(ωt). B) I_C = IC_{, max} sin(ωt). C) I_C = −IC_{, max} cos(ωt). D) I_C = IC_{, max} cos(ωt). E) I_C = IC_{, max} tan(ωt).

Accepted Answer

The answer of In a parallel RLC circuit, where I_R...

Question 18

The graphs below represent current and voltage phasors at one instant of time. The solid arrows represent the voltage phasors, ΔV, and the dashed arrows represent the current phasors, I_max.The graph which shows the correct relationship between current and voltage phasors for an inductor in an RL circuit is A) B) C) D) E)

Accepted Answer

In an RL circuit, the current lags behind the voltage due to the inductor's opposition to changes in current flow. This means that the current phasor should be shifted to the right (or counterclockwise) from the voltage phasor. Choice A correctly shows the current phasors lagging behind the voltage phasors, indicating the presence of an inductor in the circuit.

Question 19

A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RL circuit with R = 10.0 Ω and L = 10.0 mH. The impedance is
A) 3.77 Ω.
B) 9.26 Ω.
C) 10.0 Ω.
D) 10.7 Ω.
E) 13.8 Ω.

Accepted Answer

The impedance of a series RL circuit is given by:
$$Z=\sqrt{R^2+(\omega L)^2}$$
where $\omega$ is the angular frequency given by $\omega = 2\pi f$. Plugging in the values:
$$\omega = 2\pi(60.0 	ext{ Hz}) = 377 	ext{ rad/s}$$
$$Z = \sqrt{(10.0 	ext{ }\Omega)^2 + (377 	ext{ }	ext{rad/s} \cdot 0.01 	ext{ H})^2} = 10.7 	ext{ }\Omega$$
Therefore, the correct answer is D.

Question 20

A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RC circuit with R = 10.0 Ω and C = 300 μF. The impedance is
A) 4.68 Ω.
B) 8.84 Ω.
C) 10.0 Ω.
D) 13.4 Ω.
E) 18.8 Ω.

Accepted Answer

The impedance of a series RC circuit can be calculated as follows:

Z = √(R^2 + Xc^2)

where Xc is the capacitive reactance given by:

Xc = 1/(2πfC)

Plugging in the values, we get:

Xc = 1/(2π(60)(300×10^-6)) = 88.4 Ω

Z = √(10^2 + 88.4^2) = 13.4 Ω

Therefore, the impedance of the circuit is 13.4 Ω.

Quiz 33: Alternating-Current Circuits

Whenever the alternating current frequency in a series RLC circuit is halved,

An alternating current circuit has resistance R, inductance L and capacitance C in series with a voltage source. Which statement is correct?

A 10-μF capacitor in an LC circuit made entirely of superconducting materials (R = 0 Ω) is charged to 100 μC. Then a superconducting switch is closed. At t = 0 s, plate 1 is positively charged and plate 2 is negatively charged. At a later time, V_ab = +10 V. At that time, V_dc is

A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RL circuit with R = 10.0 Ω and L = 10.0 mH. The peak current in the circuit is

A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RC circuit with R = 10.0 Ω and C = 300 μF. The peak current in the circuit is

All three circuits shown below have R = 100 Ω, L = 0.1 H and emf ε = (5.0 V) sin (377 t) . Which statement regarding the angular resonance frequencies ω_A, ω_B and ω_C is correct?

The power output, P_out, of an ideal step-up transformer that receives power input, P_in, and which has N₁ turns in the primary and N₂ turns in the secondary coil is given by

The average power input to a series alternating current circuit is minimum when

The graphs below show a voltage phasor at different instances of time. The voltage phasor which shows the instantaneous value of the voltage with the largest magnitude is

A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RC circuit with R = 10.0 Ω and C = 300 μF. The power factor, cos φ, is

In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of line has resistance R. For a power line that supplies power to 10000 households, we can conclude that

A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RL circuit with R = 10.0 Ω and L = 10.0 mH. The power factor, cos φ, is

In a parallel RLC circuit, where I_R = IR_{, max} sin(ωt) , the current through the capacitor, I_C, is

A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RL circuit with R = 10.0 Ω and L = 10.0 mH. The impedance is

A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RC circuit with R = 10.0 Ω and C = 300 μF. The impedance is