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Alday, Karla B. D3A- Elemele
Take
Home Quiz Engr. Mae T. Dimaculangan 1. Calculate the average voltage induced in a coil of 300 turns through which the flux changes from 250,000 to 20,000 maxwells in 0.15 seconds. SOLUTION: dɸ −8 Е = N ( dt ¿ x 10 = 300 (
volts
250,000−20,000 ¿ x 10−8 0.15 seconds
volts
E = 4.6 volts 2. How many turns of wire are there in a coil in which 35.7 volts are induced when the flux through it increases uniformly at the rate of 3 x 106
maxwells per second?
SOLUTION: dɸ −8 Е = N ( dt ¿ x 10
volts
Е N=
=
(
dɸ ) x 10−8 dt
35.7 volts 3 x 106 x 10−8
N = 1190 turns of wire
3. A voltage wave has the equation e = 170 sin ∝ . Calculate the instantaneous values of voltage for the following angles: 30 ° , 60 ° , 75 ° , 105 ° , 135 ° , 180 ° , 270 ° , and 330 ° .
SOLUTION: e = 170 sin 30 °
e = 170 sin 60 °
e = 170 sin 75°
e = 170 sin 105°
e = 85 V
e = 147.22 V
e = 164.21 V
e = 164.21 V
e = 170 sin 135°
e = 170 sin 180°
e = 170 sin 270 ° e = 170 sin 330°
e = 120.21 V e = 0
e = -170 V
e = -85 V
4. An alternator has 8 poles. (a) At what speed must it be driven to develop 60 cycles? 25 cycles? 50 cycles? 40 cycles? (b) What frequency will be developed if the speed is 825 rpm? 1,350 rpm? 1,800 rpm? 450 rpm? SOLUTION: A. f =
Pxrpm 120
60 cycles =
120 x 60 8
= 900 rpm
25 cycles =
120 x 25 8
= 375 rpm
50 cycles =
120 x 5 0 8
= 750 rpm
40 cycles = B. f825 =
120 x 4 0 8
8 x 825 120
= 600 rpm
= 55
f1350 =
8 x 1350 120
= 90
f1800 =
8 x 1800 120
= 120
f450 =
8 x 450 120
= 30
5. A 117-volt 60-cycle source is connected to a series circuit consisting of three resistors. If the ohmic values of the latter are 20, 30, 40 ohms, respectively, calculate the current through the circuit and the voltage drop across each resistor. SOLUTION: I=
117 20+ 30+40
= 1.3 amperes
E20 = 20 x 1.3 = 26 Volts E30 = 30 x 1.3 = 39 Volts E40 = 40 x 1.3 = 52 Volts 6. Three incandescent lamps (resistors) are connected in parallel, and to a 115-volt 60-cycle source. If the lamp ratings are 75, 100, 150 watts, calculate the rms value of the resultant current. SOLUTION: I75 = 75/115 = 0.65 amp I100 = 100/115 = 0.87 amp I150 = 150/115 = 1.30
