# AC Fundamentals – Exercise – 2

51. A 20 kHz pulse waveform consists of pulses that are 15 μs wide. The duty cycle ________.

(a) is 1%
(b) is 30%
(c) is 100%
(d) cannot be determined

Explanation
Explanation :

Duty Cycle = (time of signal active / total period of signal) * 100
Total period = 1 / frequency

Therefore,
D = t / p * 100% = f * p * 100% = 15 * 10^-6 * 20 * 10^3 * 100 = 30%.

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52. A pulse waveform has a high time of 8 ms and a pulse width of 32 ms. The duty cycle is ________.

(a) 25%
(b) 50%
(c) 1%
(d) 100%

Explanation
Explanation :

Duty cycle = (high time / pulse width) * 100 = (8 ms / 32 ms) * 100 = 25%

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53. The average value of a 12 V peak sine wave over one complete cycle is ________.

(a) 0 V
(b) 1.27 V
(c) 7.64 V
(d) 6.37 V

Explanation
Explanation :

The sine wave has the same values for both positive and negative cycles. Hence, the average is equal to zero. Average value over complete sine wave cycle is always zero.

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54. The average half-cycle value of a sine wave with a 40 V peak is ________.

(a) 25.48 V
(b) 6.37 V
(c) 14.14 V
(d) 50.96 V

Explanation
Explanation :

METHOD-1

Vavg = 2Vp / π = 2 * 40 / 3.14 = 25.48 V

METHOD-2

Vavg = Vp * 0.637 = 40 * 0.637 = 25.48 V

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55. If the rms current through a 6.8 kΩ resistor is 8 mA, the rms voltage drop across the resistor is ________.

(a) 5.44 V
(b) 54.4 V
(c) 7.07 V
(d) 8 V