This set of Power Electronics Multiple Choice Questions & Answers (MCQs) focuses on “1-Phase-Diode Rectifiers HW-1”.
1. In the process of diode based rectification, the alternating input voltage is converted into
a) an uncontrolled alternating output voltage
b) an uncontrolled direct output voltage
c) a controlled alternating output voltage
d) a controlled direct output voltage
Explanation: Rectification is AC to DC. In DIODE biased rectification, control is not possible.
2. In a half-wave rectifier, the
a) current & voltage both are bi-directional
b) current & voltage both are uni-directional
c) current is always uni-directional but the voltage can be bi-directional or uni-directional
d) current can be bi-directional or uni-directional but the voltage is always uni-directional
Explanation: Current is always in one direction only, but voltage can be bi-directional in case of an L load.
3. For a certain diode based rectifier, the output voltage (average value) is given by the equation
1/2π [ ∫Vm sin ωt d(ωt) ] Where the integral runs from 0 to π
The rectifier configuration must be that of a
a) single phase full wave with R load
b) single phase full wave with RL load
c) single phase half wave with R load
d) single phase half wave with RL load
Explanation: Integration is 0 to π from base period of 1/2π so it is a half wave R load.
4. For a single phase half wave rectifier, with R load, the diode is reversed biased from ωt =
a) 0 to π, 2π to 2π/3
b) π to 2π, 2π/3 to 3π
c) π to 2π, 2π to 2π/3
d) 0 to π, π to 2π
Explanation: Diode will be reversed biased in the negative half cycles.
Explanation: PIV = √2 Vs = Vm.
Explanation: Due to the L nature, load current is maximum when the diode will be com-mutated i.e at π.
Explanation: The above is a HW diode rectifier, the RMS o/p voltage equation is given by
Vor = √ [ (1/2π) ∫π Vm2sin2ωt. d(ωt) ] Solving above equation we get, Vor = Vm/2.
8. In a 1-Phase HW diode rectifier with R load, the average value of load current is given by
Take Input (Vs) = Vm sinωt
Explanation: Vo = √ [(1/2π) ∫π Vm sinωt. d(ωt)] Vo = Vm/π
I = Vo/R = Vm/πR.
9. In the circuit shown below,
The switch (shown in green) is closed at ωt = 0. The load current or capacitor current has the maximum value at ωt =
d) none of the mentioned
Explanation: The instant switch is closed the load current will be zero due to the nature of the capacitor.
10. Find the average value of output current for a 1-phase HW diode rectifier with R load, having RMS output current = 100A.
a) 200R A
b) 100/R√2 A
c) 200/R√2 A
d) 200/Rπ A
Explanation: I(rms) = Vm/2R
Therfore, Vm = 200R
I(avg) = Vm/πR = 200R/πR.
Sanfoundry Global Education & Learning Series – Power Electronics.
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