AC Fundamentals – Exercise – 2

61. A sinusoidal current has an rms value of 14 mA. The peak-to-peak value is ________.

(a) 45.12 mA
(b) 16 mA
(c) 39.6 mA
(d) 22.6 mA

Answer
Answer : (c)
Explanation
Explanation :

Ipeak = √2 * Irms = 1.414 * Irms

Since it is a sine wave, +14 and -14.  Hence, Irms = 28 mA.

Ipeak = 28 * 1.414 = 39.59 mA = 39.6 mA.

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62. A sine wave with an rms value of 12 V is riding on a dc level of 18 V. The maximum value of the resulting waveform is ________.

(a) 6 V
(b) 30 V
(c) 35 V
(d) 0 V

Answer
Answer : (c)
Explanation
Explanation :

Vmax = √2 * Vrms = 1.414 * 12 V =16.9 V.

Vmax + Vdc = 16.9 v+ 18 = 34.9 = 35 V.

One cycle of sine wave has two half waves.

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63. A square wave has a period of 60 μs. The first odd harmonic is ________.

(a) 5 kHz
(b) 50 kHz
(c) 500 kHz
(d) 33.33 kHz

Answer
Answer : (b)
Explanation
Explanation :

f = frequency
t = time period

t = 60 μs

f = 1 / t = 16.66 Hz.

Harmonics are multiples of the fundamental frequency and can therefore be expressed as: 2ƒ, 3ƒ, 4ƒ …. and so on.

First odd harmonic = 3 * f = 3 * 16.66 = 49.98~ = 50 kHz.

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