This set of Power Electronics Multiple Choice Questions & Answers (MCQs) focuses on “1-Phase-Diode Rectifiers HW-2”.
1. A 1-phase 230V, 1KW heater is connected across a 1-phase HW rectifier (diode based). The power delivered to the heater is
a) 300 W
b) 400 W
c) 500 W
d) 600 W
Explanation: R = (230 x 230)/1000
V(rms) = (√2 x 230)/2
P = V(rms)2/R = 500W.
2. A 1-phase half wave diode rectifier with R load, has input voltage of 240 V. The input power factor is
b) 0.707 lag
c) 0.56 lag
d) 0.865 lag
Explanation: Input p.f = V(rms)/Vs
Vrms is the RMS value of output voltage. Vrms = (√2 x 230)/2
Vs = 230
pf = 0.707.
3. A 1-phase half wave diode rectifier with R = 1 KΩ, has input voltage of 240 V. The diode peak current is
Explanation: Diode peak current = peak current through the load = Vo/R = Vm/2R.
4. For the below given circuit, after the switch is closed the voltage across the load (shown open) remains constant.
Assuming that all initial conditions are zero. The element across the load would be a/an
d) data not sufficient
Explanation: As the voltage remains constant as soon as the switch is closed, the element is most likely to be a capacitor.
5. For the below given circuit,
After the supply voltage (Vs) is given the
a) diode starts conducting
b) diode starts conducting only when Vs exceeds Vdc
c) diode never conducts
d) diode stops conducting only when Vs exceeds Vdc
Explanation: The diode will be forward biased only when Vs will be greater than Vdc.
Explanation: Due to the FD, we get 1st quadrant operation.
Where, output voltage (Avg) = 1/2π [ ∫Vm sin ωt d(ωt) ], integration runs from 0 to 180 degrees.
Explanation: The output voltage is the voltage across the resister and Vdc. Even if the current falls to zero, the output voltage will be equal to Vdc.
Explanation: Ripple voltage = √(Vrms2 + Vavg2)
Vrms = Vm/2
Vavg = Vm/π
9. For a single phase half wave rectifier, the rectifier efficiency is always constant & it is
Rectifier efficiency = Pdc/Pac
Pdc = (Vm x Im)/π2
Pac = 4/(Vm x Im).
10. For the below given circuit,
The power delivered to the load in Watts is
b) I(avg).Vdc + I(rms)2.R
c) I(avg).Vdc – I(rms)2.R
d) I(avg).Vdc + I(avg)2.R
Explanation: P = power delivered to the resister + power delivered to the emf source.
Sanfoundry Global Education & Learning Series – Power Electronics.
To practice all areas of Power Electronics, here is complete set of 1000+ Multiple Choice Questions and Answers.