## Saturday, September 15, 2012

### Power Electronics Solved Objective Problems

[1] A 240V, 50Hz supply feeds a highly inductive load of 50 Ohm resistance through a half controlled thyristor bridge. When the firing angle α=45°, the load power is
a) 418 W
b) 512 W
c) 367 W
d) 128 W

Exp: Vav = (Vm/π) (1 + cosα) = [(√2*240)/(π) ) ( 1 + cos45)] = 184.4 V
Iav = Vav / R  = 184.4/50 =3.69A

= 3.69 * √[(180 -45)/180]
= 3.2A
P = 3.2 * 3.2 *50 = 512 W
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[2] A d.c. to d.c. chopper operates from a 48 V battery source into a resistive load of 24Ohm. The frequency of the chopper is set to 250Hz. When chopper on-time is 1 ms the load power is
a)  6W
b) 12W
c)  24W
d)  48W

Exp: Vavg = V* f * Ton = 48 x 250 x (10 - 3) = 12V
Iav = Vav/R = 12/24 = 0.5A
Vrms= V* Square root (Ton) * f = 48 * Square root (0.25) = 24V
Irms = Vrms/R = 24/24 = 1A
P = Irms * Irms * R = 1 * 24 = 24W
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[3] A thyristor half wave controlled converter has a supply voltage of 240V at 50Hz and a load resistance of 100 Ohm. when the firing delay angle is 30 the average value of load current is
a) 126A
b) 2.4A
c) 126mA
d) 24 A

Exp:We know the output wave form of the half wave rectifier For any delay angle alpha, the average load voltage is given by

solving,

substituting the values in the above equation,
Vav = (√2*240) / (2π) * [ 1 + cos30 ]  = 100.8 V
Iav = Vav/R = 100.8/100 = 126 mA

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[4] A full-wave fully controlled bridge has a highly inductive load with a resistance of 55 Ohm, and a supply of 110V at 50Hz. The value of load power for a firing angle α=75° is
a) 10W
b) 11W
c) 10.5W
d) 10.9W

Exp:
Vav = [2Vm/(π)]cosα]
= [(2 *√(2*110)/ 3.14 ] * cos 75
= 99 cos 75
= 25.6V

Iav = Vav / R
= 25.6/55
= 0.446A = Irms

P = Irms * Irms * R
= 0.446 * 0.446 * 55
=10.9W
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[5] A d.c. to d.c. chopper operates from a 48 V source with a resistive load of 24Ohm. The chopper frequency  is  250Hz. When Ton= 3 ms, the rms current is
a) 1.5A
b) 15mA

c) 1.73A
d) 173mA

Exp: Vav = V *f * Ton = 48 * 250 * 3 *(10 - 3) = 36V
Iav = Vav/R = 36/24 = 1.5 A
Vrms =  V* Square root (Ton) * f   = 48 * √0.75 = 41.6V
Irms = Vrms/R = 41.6/24 = 1.73A
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[6]  A 240V, 50Hz supply feeds a highly inductive load of 50 Ohm resistance through a thyristor full control bridge. when the firing angle α= 45°, load power is

a) 456 W
b) 466 W
c) 732 W
d) 120 W

Exp: Vav = (2Vm/π) * cosα  = [(2 * 339)/3.14] cos 45 = 152.6V
Iav = Vav/R = 152.6 / 50 = 3.05A = Irms
P = Square of Irms * R =  3.04 * 3.04 * 50 = 466W
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