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

Ans:B

 
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

Ans:C

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

Ans:C

  
Exp:We know the output wave form of the half wave rectifierFor 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

Answer : D

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

Ans:C

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

Ans:B

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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Thanks for reading.