Test: Single Phase HW AC-DC - 2
20 Questions MCQ Test Power Electronics | Test: Single Phase HW AC-DC - 2
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Choose the correct statement with respect to the below given circuit.
The voltage cannot be negative due to the FD (freewheeling diode or commutating diode so connected).
In the below given circuit, the FD (Freewheeling diode) is forward biased at ωt =
It is forward biased at π by the conducting SCR & the current starts to through the FD & Load.
In the below given circuit, when the commutating diode or FD is conducting than the
When the FD is forward biased at π by the conducting SCR & the current starts to through the FD & load.
The output voltage waveform of the below given circuit would be the same that obtained from a
The wave from will be like a half wave diode rectifier circuit. Which is the same as that obtained from a half-wave R load circuit and the above given circuit.
In a single-phase half-wave circuit with RL load and a freewheeling diode, the load voltage during the freewheeling period will be
The FD short circuits the load and voltage across a short circuit would be = 0.
In a single-phase half-wave circuit with RL load and a freewheeling diode, the freewheeling period is
Freewheeling period is the one in which the FD diode conducts.
A single-phase half wave rectifier with a FD is supplied by Vs = 240 V, AC with a load R = 10 Ω, L = 0.5 mH and a firing angle α = 30°. Find the average value of the load voltage.
Vo = (Vm/2π) x (1+cosα)
Vm = √2Vs.
A single-phase HW rectifier with a FD is supplied by Vs = 240 V, 50 Hz, AC with a load R = 10 Ω, L = 0.5 mH and a firing angle α = 30°. Find the average value of the load current.
Vo = (Vm/2π) x (1+cosα)
Vm = √2Vs
Io = Vm/R (Due to the FD).
A single phase half-wave controlled rectifier has 400 sin314t as the input voltage and R as the load. For a firing angle of 60°,the average output voltage is
Vo = (Vm/2π) x (1+cosα) = 400/2π x (1+cos60) = 300/π.
Choose the incorrect statement with respect to the use of FD in half-wave circuits.
PIV is unaffected with the use of FD (freewheeling diodes).
A single-phase half-wave rectifier with a FD is supplied by Vs = 240 V, 50 Hz, AC source with a load R = 10 Ω, L = 0.8 mH. The firing angle is so adjusted such that the output voltage obtained is 100 V. Find the firing angle.
Vo = Vm/2π cos(firing angle).
For the below shown circuit, a motor load (RLE) is connected to the SCR through supply Vs. The minimum value of the firing angle to turn on the SCR would be
The firing angle should be such that the value of V exceeds E. So SCR turns on at excatlly Vm sinωt = E
Therefore ωt = min. angle = Sin-1(E/Vm).
Choose the correct statement with respect to the below given circuit.
Even if no current is flowing through the load when the SCR is off, voltage = E will exists at the load terminals at all times.
By using a freewheeling diode (FD) in a rectifier with RL load, the power consumed by the load
The FD feeds inductor current again to the load.
A 230 V, 50 Hz, one-pulse SCR controlled converter has extinction angle β = 210°. Find the circuit turn-off time
t = 2π-β/ω
Where, π = 90°
ω = 2xπx50
β = 210°.
A 230V, 50Hz, single-pulse SCR is feeding a RL load with α = 40° and β = 210°. Find the value of average output voltage
Vo = Vm/2π x (cosα-cosβ)
Where Vm = √2Vs.
A single-pulse transformer with secondary voltage of 230 V, 50 Hz, delivers power to bulb of R = 10 Ω through a half-wave controlled rectifier circuit. For α = 60°, find the average current in the bulb
First find the average voltage, than Io = Vo/R
Vo = (Vm/2π) x (1+cosα).
A single-pulse transformer with secondary voltage of 230 V, 50 Hz, delivers power to bulb of R = 10 Ω through a half-wave controlled rectifier circuit. For α = 60° and output AC power of 2127 Watts, find the rectification efficiency
Vo = (Vm/2π) x (1+cosα) = 77.64 V
Pdc = Vo2xR = 602.8 W
Rectification efficiency = Pdc/Pac = 28.32 %.
The maximum negative voltage faced by the SCR is –Vm which is nothing but √2Vs.
An SCR rectifier circuit is designed such that the average output voltage is 77.64 V & RMS value of output voltage is 145.873 V. Find the voltage ripple factor.
FF = 145.873/77.64 = 1.879
VRF = √(FF2-1) = 1.5908.
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