Power Electronics

This set of Power Electronics Multiple Choice Questions & Answers (MCQs) focuses on “Diode Circuits-3”.

1. For the circuit shown in the figure below, consider the diode as an ideal diode & rms value of source voltage as Vs. The output voltage waveform at R will have a) zero value in the positive half cycle and a peak value of 1.414Vs in the negative half cycle b) sine-wave nature with a peak value 1.414Vs c) zero value in the negative half cycle and a peak value of 1.414Vs in the positive half cycle d) sine-wave nature with a peak value Vs

2. For the circuit shown in the figure below, Vs is the ac voltage source with peak value Vm. The waveform of the load voltage at the resistor will have a) zero value in the positive half cycle and a peak value of –(Vm) in the negative half cycle b) zero value in the negative half cycle and a peak value of –(Vm-V1) in the positive half cycle c) zero value in the positive half cycle and a peak value of –(Vm+V1) in the negative half cycle d) zero value in the positive half cycle and a peak value of –(Vm-V1) in the negative half

3. For the circuit shown below, Vs=230V, Voltage drop across the diode (Vd) = 2V The peak value of voltage at R in the positive and the negative half cycles are ___ & ___ respectively. a) 323V,0V b) 0V, 323V c) 327V, 0V d) 0V, 327V

4. For the circuit shown in the figure below, V9 is the AC voltage source with peak value Vm. The waveform of the load voltage at the resistor has a a) peak value of (Vm+V1) in the negative half cycle b) peak value of (Vm-V1) in the positive half cycle c) peak value of (Vm+V1) in the positive half cycle d) peak value of (Vm-V1) in the negative half cycle

5. The circuit shown below has the following parameters: V1 = 8 Volts V2 = 6 Volts Vs = 10V/√2 (rms) Voltage drop across D1 & D2 = 0.7 Volts At the load (R), the peak value in the positive half cycle will be a) 8.7 V b) 6.7 V c) 8 V d) 10V/√2

6. The circuit shown below has the following parameters: Voltage drop across D1 & D2 = 0.7 Volts V1 = 8 Volts V2 = 6 Volts Vs = 10V/√2 (RMS) At the load (R), the peak value in the negative half cycle is a) 8.7 V b) 6.7 V c) 8 V d) 10V/√2

7. Consider the diode to be an ideal one and Vo = Vr + Vdc during positive half cycle. Thus, during the negative half cycle a) Vo = Vr b) Vo = 0 c) Vo = Vdc+Vr d) Vo = Vdc

8. When a diode is connected in series with an AC source & R load, the conduction time per cycle is a) 0 b) 2π c) π d) π/2

9. When diode is connected in series to an AC source & RL load, the conduction time for the diode a) is always less than π b) is 0 c) is π d) can be greater than π

10. For the circuit shown below, Vdc = 50 V Cut-in voltage for D1 = 0.2 V Cut-in voltage for D2 = 0.6 V R = 5KΩ Current through D1 & D2 would be, a) 5mA, 5mA b) 10mA, 0 c) 0, 10mA d) 9.98mA, 9.94mA