Power electronics and drives miscellaneous Easy Questions and Answers | Page - 5

Power electronics and drives miscellaneous

A single-phase half wave uncontrolled converter circuit is shown in the figure given below. A 2– winding transformer is used at the input for isolation. Assuming the load current to be constant and v = V_m sin ωt, the current waveform through diode D₂ will be

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During 0 < ωt < π, Diode D₁ is forward-biased and diode D₂ sets reversed-biased.
During π < ωt < 2π , diode D₁ is reversed-biased and diode D₂ is forward-biased.
Thus, in π < ωt < 2π, diode D₂ starts conduct, and in π < ωt < 2π , there is no current flows.

Correct Option: D

During 0 < ωt < π, Diode D₁ is forward-biased and diode D₂ sets reversed-biased.
During π < ωt < 2π , diode D₁ is reversed-biased and diode D₂ is forward-biased.
Thus, in π < ωt < 2π, diode D₂ starts conduct, and in π < ωt < 2π , there is no current flows.

A voltage commutation circuit is shown in the figure given below. If the turn off time of the SCRs is 50 μ sec and a safety margin of 2 is considered, then what will be the approximate minimum value of capacitor required for proper commutation?

1. 2.88 μF
1. 1.44 μF
1. 0.91 μF
1. 0.72 μF

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Firing of TH1 thyristor would commute TH2 and firing of TH2 would commute TH1 and in this way load will be transfered between TH1 and TH2 and vice-versa.
Circuit turn-off time for,
TH1, t₁ = R₁C ln 2
and turn-off time for
TH2, t₂ = R₂C ln 2
Safety margin = 2
Then, t₁ = t₂ = 2 × 50 × 10^{– 6} (given)
⇒ R₁ C ln2 = 2 × 50 × 10^{– 6}
⇒ C = 2 × 50 × 10^{– 6} = 2.88 μF
50 × ln 2

Correct Option: A

Firing of TH1 thyristor would commute TH2 and firing of TH2 would commute TH1 and in this way load will be transfered between TH1 and TH2 and vice-versa.
Circuit turn-off time for,
TH1, t₁ = R₁C ln 2
and turn-off time for
TH2, t₂ = R₂C ln 2
Safety margin = 2
Then, t₁ = t₂ = 2 × 50 × 10^{– 6} (given)
⇒ R₁ C ln2 = 2 × 50 × 10^{– 6}
⇒ C = 2 × 50 × 10^{– 6} = 2.88 μF
50 × ln 2

A 3-phase fully controlled bridge converter with free wheeling diode is fed from 400 V, 50 Hz AC source and is operating at a firing angle of 60°. The load current is assumed constant at 10 A due to high load inductance. The input displacement factor (IDF) and the input power factor (IPF) of the converter will be

1. IDF = 0.867IPF = 0.828
1. IDF = 0.867IPF = 0.552
1. IDF = 0.5 IPF = 0.478
1. IDF = 0.5 IPF = 0.318

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For 3-φ full-wave converter,
RMS value of source current,
RMS value of nth harmonic current,
I_Sh = 4I₀ sin nπ
√2 nπ 3

For fundamental component,
I_S1 = 2 √2 I₀ sin π = √6 I₀
π 3 π

Then, CDF (current displacement factor)

and input displacement factor, (IDF) cos α = cos 60° = 0.5
Input power factor, IPF = IDF × CDF = 0.5 × 0.955 ≅ 0.478

Correct Option: C

For 3-φ full-wave converter,
RMS value of source current,
RMS value of nth harmonic current,
I_Sh = 4I₀ sin nπ
√2 nπ 3

For fundamental component,
I_S1 = 2 √2 I₀ sin π = √6 I₀
π 3 π

Then, CDF (current displacement factor)

and input displacement factor, (IDF) cos α = cos 60° = 0.5
Input power factor, IPF = IDF × CDF = 0.5 × 0.955 ≅ 0.478

A solar cell of 350 V is feeding power t o an ac supply of 440 V, 50 Hz through a 3-phase fully controlled bridge converter. A large inductance is connected in the dc circuit to maintain the dc current at 20 A. If the solar cell resistance is 0.5 Ω, then each thyristor will be reverse biased for a period of

1. 125°
1. 120°
1. 60°
1. 55°

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Voltage across 3-φ fully controlled bridge converter
V₀ = – (E – I_dc R_cell)
= – (350 – 20 × 0.5) = – 340V
i.e. , V₀ = -340 = 3V_me cos α
π

⇒ 3 × 440 √2 cos α = - 340
π

⇒ α = 125°
Thus, each thyristor will be reversed biased for a period of (180° – 125°) = 55°

Correct Option: A

Voltage across 3-φ fully controlled bridge converter
V₀ = – (E – I_dc R_cell)
= – (350 – 20 × 0.5) = – 340V
i.e. , V₀ = -340 = 3V_me cos α
π

⇒ 3 × 440 √2 cos α = - 340
π

⇒ α = 125°
Thus, each thyristor will be reversed biased for a period of (180° – 125°) = 55°

The speed of a 3– phase, 440 V, 50 Hz induction motor is to be controlled over a wide range from zero speed to 1.5 time the rated speed using a 3-phase voltage source inverter. It is desired to keep the flux in the machine constant in the constant torque region by controlling the terminal voltage as the frequency changes. The inverter out put voltage vs frequency characteristic should be

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For speed control by varying frequency method, ratio V must be kept constant,
f

so that maximum flux density may remain constant. i.e.,
⇒ V = 4.44 Nf B_m A
⇒ V = 4.44 N B_m A = constant
f

But for frequency above 50 Hz, voltage can’t be increased alone rated voltage. So, above 50 Hz voltage kept constant.

Correct Option: A

For speed control by varying frequency method, ratio V must be kept constant,
f

so that maximum flux density may remain constant. i.e.,
⇒ V = 4.44 Nf B_m A
⇒ V = 4.44 N B_m A = constant
f

But for frequency above 50 Hz, voltage can’t be increased alone rated voltage. So, above 50 Hz voltage kept constant.