AC Fundamentals – Exercise – 2

56. If the rms current through a 4.7 kΩ resistor is 4 mA, the peak voltage drop across the resistor is ________.

(a) 4 V
(b) 18.8 V
(c) 26.6 V
(d) 2.66 V

Answer
Answer : (c)
Explanation
Explanation :

R = 4.7 kΩ = 4700 Ω
Irms = 4 mA = 0.004 A

Vrms = Irms * R
Vrms = 0.004 * 4700 = 18.8 V.

Vm = √2 * Vrms
Vm = Vrms * 1.414 = 18.8*1.414 = 26.6 V.

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57. A waveform has a baseline of 3 V, a duty cycle of 20%, and an amplitude of 8 V. The average voltage value is ________.

(a) 4 V
(b) 4.6 V
(c) 1.6 V
(d) 11 V

Answer
Answer : (a)
Explanation
Explanation :

Average voltage = baseline + (duty cycle * p-p amplitude)

baseline = 3 V
duty cycle = 0.2
p-p amplitude = amplitude-baseline = 8V – 3V = 5V.

Average voltage = 3 + (0.2 * 5) = 3+1 = 4V.

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58. A sawtooth wave has a period of 10 ms. Its frequency is ________.

(a) 10 Hz
(b) 50 Hz
(c) 100 Hz
(d) 1,000 Hz

Answer
Answer : (c)
Explanation
Explanation :

F = 1 / T  = 1/10 * 1000^-3 = 1000/10 = 100 Hz.

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59. If a sine wave goes through 10 cycles in 20 μs, the period is ________.

(a) 20 s
(b) 4 s
(c) 2 s
(d) 100 s

Answer
Answer : (c)
Explanation
Explanation :

f = frequency
t = time period
p = period

f = 1 / t
Frequency = cycles per second

p = t / cycles
Period = time per cycle

Period = 20 μs / 10 = 2 μs.

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60. A signal with a 400 μs period has a frequency of ________.

(a) 250 Hz
(b) 2,500 Hz
(c) 25,000 Hz
(d) 400 Hz

Answer
Answer : (b)
Explanation
Explanation :

f = frequency
t = time period

Frequency = cycles per second

f = 1 / t = 1 / 400 * 10^-6 = 10000 / 4 = 2,500 Hz.

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