AC Fundamentals – Exercise – 1

41. If the rms voltage drop across a 15 kΩ resistor is 16 V, the peak current through the resistor is ________.

(a) 15 mA
(b) 1.5 mA
(c) 10 mA
(d) 1 mA

Answer
Answer : (b)
Explanation
Explanation :

V = 16 V
R = 15 kΩ = 15,000 Ω
Irms = V/R = 16/15,000 = 0.001067 A = 1.067 mA

Imax = 2^1/2 * Irms
Imax = 1.414 * 1.067mA = 1.51 mA

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42. The conductive loop on the rotor of a simple two-pole, single-phase generator rotates at a rate of 400 rps. The frequency of the induced output voltage is ________.

(a) 40 Hz
(b) 100 Hz
(c) 400 Hz
(d) indeterminable

Answer
Answer : (c)
Explanation
Explanation :

N = 400 rps = 400 * 60 rpm,
p = 2

N = 120F/P
F = PN/120 = 400 * 60 * 2 / 120 = 400 Hz

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43. To produce an 800 Hz sine wave, a four-pole generator must be operated at ________.

(a) 200 rps
(b) 400 rps
(c) 800 rps
(d) 1,600 rps

Answer
Answer : (b)
Explanation
Explanation :

F = 800 Hz

N = 120F/P = 120 * 800 / 4 = 24,000 rpm = 24,000 / 60 rps = 400 rps

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44. How many degrees are there in /3 rad?

(a)
(b) 60°
(c) 180°
(d) 27°

Answer
Answer : (b)
Explanation

METHOD-1

1 rad = 57.3°
π/3 rad = (π/3) x 57.3° = (3.142/3) x 57.3° = 60°

METHOD-2

1 degree = π/180 radians,
1 radians =180/π degree
So,
π/3 radians = (180/π) * (π/3) degree = 180/3 degree = 60°

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45. A sine wave of 15 kHz is changing at a faster rate than a sine wave with a frequency of ________.

(a) 25 kHz
(b) 12 kHz
(c) 18 kHz
(d) 1.3 MHz

Answer
Answer : (b)
Explanation
Explanation :

t = 1/f

Time is indirectly proportional to frequency. So, the lesser value of frequency from giving value will be of faster rate. So, 12 kHz is slower than the 15 kHz. Other rates are higher than the 15 kHz.

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