Power electronics and drives miscellaneous
- A single - phase fully controlled thyristor bridge ac - dc converter is operating at a firing angle of 25° and an overlap angle 10° with constant dc output current of 20 A. The fundamental power factor (displacement factor) at input ac mains is
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I0 = Vm [cosα - cos(α + μ)] ωLs ∴ 20 = 2 × √2 [cos25° - cos(25 + 10)] 2 × π × 100Ls
Ls = 0.0045 HeneryVo = 2 × √2 × 230 π cos25° - 2 × π × 50 × 0.0045 × 20 = 178.65 π ∴ Displacement factor = Vo Io Vs Is = 178.65 × 20 = 0.78 230 20
Correct Option: A
I0 = Vm [cosα - cos(α + μ)] ωLs ∴ 20 = 2 × √2 [cos25° - cos(25 + 10)] 2 × π × 100Ls
Ls = 0.0045 HeneryVo = 2 × √2 × 230 π cos25° - 2 × π × 50 × 0.0045 × 20 = 178.65 π ∴ Displacement factor = Vo Io Vs Is = 178.65 × 20 = 0.78 230 20
- In the circuit shown in the figure given below, the switch is operated at a duty cycle of 0.5. A large capacitor is connected across the load. The inductor current is assumed to be continuous.
The average voltage across the load and the average current through the diode will respectively be
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The current given is a step-up circuit and output voltage is given as,
Vo = Vs = 20 = 40 Volts 1 - α 1 - 0.5
and, average current through the diode
= IL (1 – α) = 4 × (1 – 0.5) = 2 A
Correct Option: A
The current given is a step-up circuit and output voltage is given as,
Vo = Vs = 20 = 40 Volts 1 - α 1 - 0.5
and, average current through the diode
= IL (1 – α) = 4 × (1 – 0.5) = 2 A
- A three phase fully controlled bridge converter is feeding a load drawing a constant and ripple free load current of 10 A at a firing angle of 30°. The approximate Total Harmonic Distortion (%THD) and the rms value of fundamental component of the input current will respectively be
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Rms value of load current,
Is = Io √2 / 3 = 10 √2 / 3 = 8.165 A
Supply current
Rms value of fundamental current,is1 = 4π0 sin π = 4 × 10 
√3 
= 7.8 A √2π 3 √2π 2
Then, THD (Total harmonic distortion)
= 0.31 = 31%Correct Option: B
Rms value of load current,
Is = Io √2 / 3 = 10 √2 / 3 = 8.165 A
Supply current
Rms value of fundamental current,is1 = 4π0 sin π = 4 × 10 √3 = 7.8 A √2π 3 √2π 2
Then, THD (Total harmonic distortion)
= 0.31 = 31%
- A single phase fully controlled converter bridge is used for electrical braking of a separately excited dc motor. The dc motor load is represented by an equivalent circuit as shown in the figure given below.
Assume that the load inductance is sufficient to ensure continuous and ripple free load current. The firing angle of the bridge for a load current of I0 = 10 A will be
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For continuous conduction Average load current (by KVL),
Vt – 2Ia + 150 = 0∴ I = Vt + 150 = 10 A 2
⇒ Vt + 150 = 20⇒ 2 Vm cos α = -130 π ⇒ 2 × √2 × 230 × cos α = -130 π
⇒ α = 129°Correct Option: C
For continuous conduction Average load current (by KVL),
Vt – 2Ia + 150 = 0∴ I = Vt + 150 = 10 A 2
⇒ Vt + 150 = 20⇒ 2 Vm cos α = -130 π ⇒ 2 × √2 × 230 × cos α = -130 π
⇒ α = 129°
- A single phase voltage source inverter is feeding a purely inductive load as shown in the figure given below.
The inverter is operated at 50 Hz in 180° square wave mode. Assume that the load current does not have any dc component. The peak value of the inductor current i0 will be
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T = π = π = 1 = 0.01 sec ω 2π50 100
And, for 0 < ωt < π,V0 = L di dt ⇒ 200 = 0.1 
IP - (-IP) 
T
⇒ 2IP = 2000T
⇒ IP = 10 Amps
Correct Option: B
T = π = π = 1 = 0.01 sec ω 2π50 100
And, for 0 < ωt < π,V0 = L di dt ⇒ 200 = 0.1 IP - (-IP) T
⇒ 2IP = 2000T
⇒ IP = 10 Amps
