Question 1

In some countries the AC frequency is 50 Hz.What is the time interval for one cycle at this frequency?
A) 0.0050 s
B) 0.020 s
C) 2 s
D) 50 s

Accepted Answer

Frequency is defined as the number of cycles per second. Therefore, the time interval for one cycle can be calculated by taking the reciprocal of the frequency. For a frequency of 50 Hz, the time interval for one cycle would be 1/50 or 0.02 seconds. So, the correct answer is option B.

Question 2

An AC voltage source,with a peak output of 120 V,results in dissipation of energy in a resistor at rate of 50.0 W.What is the value of the resistance?
A) 100 Ω
B) 120 Ω
C) 144 Ω
D) 72.0 Ω

Accepted Answer

The RMS voltage is $ V_{rms} = \frac{V_{peak}}{\sqrt{2}} = \frac{120}{\sqrt{2}} \approx 84.85 \, V $. Using the power formula $ P = \frac{V_{rms}^2}{R} $, we solve for $ R $: $ R = \frac{V_{rms}^2}{P} = \frac{84.85^2}{50} \approx 144 \, \Omega $.

Question 3

When a 50-μF capacitor is attached to an AC source,its capacitive reactance is 60 Ω.If instead a 200-μF capacitor is attached to the same source,what will be its capacitive reactance?
A) 80 Ω
B) 15 Ω
C) 30 Ω
D) 120 Ω

Accepted Answer

The capacitive reactance (Xc) is given by the formula Xc = 1/(2πfC), where f is the frequency of the AC source and C is the capacitance. Since the frequency is not given, we can assume it to be constant for both cases.

For the first case, Xc = 60 Ω, C = 50 μF. Substituting these values in the formula, we get

60 = 1/(2πf*50*10^-6)

Solving for f, we get f = 53.05 Hz

For the second case, C = 200 μF. Substituting this value and the frequency in the formula, we get

Xc = 1/(2π*53.05*200*10^-6) = 14.98 Ω

Therefore, the capacitive reactance of the 200-μF capacitor is 15 Ω (rounded off to two significant figures).

Hence, the best choice is B.

Question 4

In an AC series circuit the inductive reactance is 150 Ω and the frequency is 100 Hz.What is the inductance in the circuit?
A) 740 mH
B) 80 mH
C) 500 mH
D) 240 mH

Accepted Answer

The answer of In an AC series circuit the inductive...

Question 5

An AC circuit consists of a 60-Hz source,a 200-Ω resistor,and a 1.06-H inductor,all in series.What is the impedance of this circuit?
A) 400 Ω
B) 280 Ω
C) 600 Ω
D) 450 Ω

Accepted Answer

The answer of An AC circuit consists of a 60-Hz...

Question 6

The output voltage of an AC generator is of the form V = V₀ sin(2πft).A typical frequency in household circuits is 60 Hz.Using radians for angular measure in this equation,what is the factor 2πf in radians per second for this frequency? A) 2π B) 120 C) 377 D) 60

Accepted Answer

The given equation is V = V₀ sin(2πft). Here, the factor 2πf represents the angular frequency in radians per second. For a frequency of 60 Hz, 2πf can be calculated as: 2πf = 2π x 60 = 120π rad/s ≈ 376.99 rad/s Therefore, the best choice is C, which represents 376.99 rad/s (or approximately 377 rad/s).

Question 7

An AC voltage source,with a peak output of 200 V,is connected to a 25-Ω resistor.What is the effective (or rms)current in the circuit?
A) 2.8 A
B) 5.7 A
C) 4.0 A
D) 2.0 A

Accepted Answer

The RMS voltage is V_peak/√2 = 200/√2 = 141.4 V. The RMS current is V_RMS/R = 141.4/25 = 5.7 A.

Question 8

At what frequency does a 0.32-H inductor have the same reactance as a 0.22-μF capacitor? A) 3.8 × 10³ Hz B) 3.6 × 10⁵ Hz C) 24 Hz D) 6.0 × 10² Hz

Accepted Answer

The reactance of an inductor is given by XL = 2πfL and the reactance of a capacitor is given by XC = 1/(2πfC).
Setting XL = XC, we can solve for the frequency:
2πfL = 1/(2πfC)
f = 1/(2π√(LC))
Plugging in the given values, we get:
f = 1/(2π√(0.32*0.22*10^-6)) = 6.0 × 10^12 Hz

However, the answer choices are not in scientific notation, so we must convert our answer:
6.0 × 10^12 Hz = 6.0 × 10^15 Hz = 6.0 petaHertz

Therefore, the answer is D.

Question 9

In an AC series circuit the capacitive reactance is 253 Ω and frequency is 100 Hz.What is the capacitance?
A) 6.29 μF
B) 8.0 μF
C) 50.0 μF
D) 3.2 μF

Accepted Answer

The capacitive reactance (Xc) is given by the formula Xc = 1/(2πfC). Rearranging for capacitance (C), we get C = 1/(2πfXc). Substituting the given values, C = 1/(2π(100)(253)) = 6.29 μF.

Question 10

The peak voltage of an AC source is 400 V.What is the rms voltage?
A) 100 V
B) 200 V
C) 141 V
D) 283 V

Accepted Answer

The answer of The peak voltage of an AC source...

Question 11

What is the effective (rms)current value for an AC current with an amplitude of 20 A?
A) 28 A
B) 7.1 A
C) 3.1 A
D) 14 A

Accepted Answer

The effective (rms) current value for an AC current is given by the formula I_rms = I_peak / √2. For an amplitude (I_peak) of 20 A, I_rms = 20 / √2 ≈ 14 A.

Question 12

The frequency in an AC series circuit is halved.By what factor does this change the capacitive reactance?
A) 1/2
B) 4
C) 1/4
D) 2

Accepted Answer

Capacitive reactance (Xc) is inversely proportional to the frequency (f), given by the formula Xc = 1/(2πfC). If the frequency is halved, the capacitive reactance doubles.

Question 13

An AC voltage source,with a peak output of 200 V,is connected to a 25-Ω resistor.What is the rate of energy dissipated due to heat in the resistor?
A) 566 W
B) 400 W
C) 800 W
D) 200 W

Accepted Answer

The rate of energy dissipated in a resistor can be calculated using the formula P = V^2/R, where P is power, V is voltage, and R is resistance.

In this case, the peak voltage of the AC source is given as 200 V. However, we need to convert this to the RMS voltage in order to use the formula above. The RMS voltage of an AC wave is given by V_RMS = V_peak/sqrt(2). Therefore, the RMS voltage in this case is:

V_RMS = 200/sqrt(2) = 141.4 V

Using this value and the resistance of 25 Ω in the formula above, we get:

P = (141.4^2)/25 = 800 W

Therefore, the rate of energy dissipated in the resistor is 800 W. The correct answer is C.

Question 14

A 400-Ω resistor and a capacitor are connected in parallel across a 300-Hz source.If the amplitude of the current in the resistor is equal to that in the capacitor,what is the value of the capacitance? A) 1.33 × 10^-6 F B) 0.750 × 10^-6 F C) 10.6 × 10^-6 F D) 0.212 F

Accepted Answer

Using the formula for the impedance of a capacitor in parallel with a resistor:

Z = R || (1/jωC)

Where R is the resistance, C is the capacitance, ω is the angular frequency (2πf), and || means parallel.

Since the resistor and capacitor have equal currents, we can equate the amplitudes of their impedances:

|Z_R| = |Z_C|
|r| = |1/jωC|
|r| = ωC

Substituting the given values:

400 = 2π(300)C

Solving for C:

C = 1.33 × 10^-6 F, which is choice A.

Question 15

An AM radio tuning circuit has a coil with an inductance of 4.00 mH and a capacitor set at 7.50 × 10−⁶ μF.What frequency will it detect? A) 1520 kHz B) 919 kHz C) 550 kHz D) 750 kHz

Accepted Answer

The resonant frequency of an LC circuit can be calculated using the formula: 
f = 1/(2π√(LC))

Substituting the given values: 
f = 1/(2π√(4.00 mH × 7.50 × 10^-16 F)) 
f = 919 kHz

Therefore, the circuit will detect a frequency of 919 kHz. Option B) is correct.

Question 16

A 400-Ω resistor and an inductor are connected in parallel across a 300-Hz source.If the amplitude of the current in the resistor is equal to that in the inductor,what is the value of the inductance? A) 0.212 H B) 0.666 H C) 1.33 × 10^-6 H D) 0.333 H

Accepted Answer

The amplitude of the current in the resistor is equal to that in the inductor, which means the resistive reactance (R) equals the inductive reactance (XL). XL = 2πfL, where f is the frequency and L is the inductance. Given R = 400 Ω and f = 300 Hz, we solve for L: L = R / (2πf) = 400 / (2π*300) ≈ 0.212 H.

Question 17

A series LC circuit in a radio is in resonance with AM 1200 kHz.If the radio station is changed to AM 600 kHz,by what factor must the capacitance be multiplied to again achieve resonance?
A) 4
B) 2
C) 1/2
D) 1/4

Accepted Answer

The resonant frequency of an LC circuit is given by f = 1/(2π√(LC)). To halve the frequency from 1200 kHz to 600 kHz, the capacitance must be increased by a factor of 4, since frequency is inversely proportional to the square root of capacitance.

Question 18

An FM radio tuning circuit has a coil with an inductance of 0.0015 mH.What is the value of the capacitance if the set is tuned to 98 MHz? A) 1.8 × 10−⁶ μF B) 0.88 × 10−⁶ μF C) 0.98 × 10−⁶ μF D) 12 × 10−⁶ μF

Accepted Answer

The resonant frequency of an LC circuit is given by:$$f = \frac{1}{2\pi\sqrt{LC}}$$where:* f is the resonant frequency in Hz* L is the inductance in henrys* C is the capacitance in faradsWe can rearrange this equation to solve for C:$$C = \frac{1}{4\pi^2f^2L}$$Plugging in the given values, we get:$$C = \frac{1}{4\pi^2(98 	imes 10^6 Hz)^2(0.0015 	imes 10^{-3} H)} = 1.8 	imes 10^{-6} \mu F$$Therefore, the correct answer is A.

Question 19

If the rotating coil used in a generator has its radius doubled,what change results in the amplitude of the voltage produced? A) It doubles. B) It quadruples. C) It is increased by a factor 2^3/2. D) It stays the same.

Accepted Answer

The amplitude of the voltage produced by a rotating coil in a generator is directly proportional to the radius of the coil. Therefore, if the radius is doubled, the area of the coil increases by a factor of four (since area is proportional to the square of the radius), leading to a quadrupling of the induced voltage.

Question 20

A 17.0-μF capacitor is connected to an AC source with an rms voltage of 120 V and a frequency of 60.0 Hz.What is the rms current in the capacitor?
A) 1.41 A
B) 0.543 A
C) 0 A
D) 0.769 A

Accepted Answer

The rms current in a capacitor connected to an AC source is given by:
Irms = Vrms/ωC
where Vrms is the rms voltage of the source, ω is the angular frequency (2πf), and C is the capacitance of the capacitor.
Substituting the given values, we get:
Irms = 120 V / (2π x 60 Hz x 17.0 x 10^-6 F)
Irms ≈ 0.769 A
Therefore, the correct answer is (D) 0.769 A.

Quiz 22: Reflection and Refraction of Light: Part A

In some countries the AC frequency is 50 Hz.What is the time interval for one cycle at this frequency?

An AC voltage source,with a peak output of 120 V,results in dissipation of energy in a resistor at rate of 50.0 W.What is the value of the resistance?

When a 50-μF capacitor is attached to an AC source,its capacitive reactance is 60 Ω.If instead a 200-μF capacitor is attached to the same source,what will be its capacitive reactance?

In an AC series circuit the inductive reactance is 150 Ω and the frequency is 100 Hz.What is the inductance in the circuit?

An AC circuit consists of a 60-Hz source,a 200-Ω resistor,and a 1.06-H inductor,all in series.What is the impedance of this circuit?

The output voltage of an AC generator is of the form V = V₀ sin(2πft) .A typical frequency in household circuits is 60 Hz.Using radians for angular measure in this equation,what is the factor 2πf in radians per second for this frequency?

An AC voltage source,with a peak output of 200 V,is connected to a 25-Ω resistor.What is the effective (or rms) current in the circuit?

At what frequency does a 0.32-H inductor have the same reactance as a 0.22-μF capacitor?

In an AC series circuit the capacitive reactance is 253 Ω and frequency is 100 Hz.What is the capacitance?

The peak voltage of an AC source is 400 V.What is the rms voltage?

What is the effective (rms) current value for an AC current with an amplitude of 20 A?

The frequency in an AC series circuit is halved.By what factor does this change the capacitive reactance?

An AC voltage source,with a peak output of 200 V,is connected to a 25-Ω resistor.What is the rate of energy dissipated due to heat in the resistor?

A 400-Ω resistor and a capacitor are connected in parallel across a 300-Hz source.If the amplitude of the current in the resistor is equal to that in the capacitor,what is the value of the capacitance?

An AM radio tuning circuit has a coil with an inductance of 4.00 mH and a capacitor set at 7.50 × 10−⁶ μF.What frequency will it detect?

A 400-Ω resistor and an inductor are connected in parallel across a 300-Hz source.If the amplitude of the current in the resistor is equal to that in the inductor,what is the value of the inductance?

A series LC circuit in a radio is in resonance with AM 1200 kHz.If the radio station is changed to AM 600 kHz,by what factor must the capacitance be multiplied to again achieve resonance?

An FM radio tuning circuit has a coil with an inductance of 0.0015 mH.What is the value of the capacitance if the set is tuned to 98 MHz?

If the rotating coil used in a generator has its radius doubled,what change results in the amplitude of the voltage produced?

A 17.0-μF capacitor is connected to an AC source with an rms voltage of 120 V and a frequency of 60.0 Hz.What is the rms current in the capacitor?