Test: 1P Diode Rectifiers HW - Electrical Engineering (EE) MCQ

For a single phase half wave rectifier, with R load, the diode is reversed biased from ωt =

For the circuit shown below,

The secondary transformer voltage Vs is given by the expression
Vs = Vm sin ωt
Find the PIV of the diode.

 For the circuit shown below,

The peak value of the load current occurs at ωt = ?

Find the rms value of the output voltage for the circuit shown below.

Voltage across the secondary is given by Vm sinωt.

 In a 1-Phase HW diode rectifier with R load, the average value of load current is given byTake Input (Vs) = Vm sinωt

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 =

Find the average value of output current for a 1-phase HW diode rectifier with R load, having RMS output current = 100A.

A 1-phase 230V, 1KW heater is connected across a 1-phase HW rectifier (diode based). The power delivered to the heater is

A 1-phase half wave diode rectifier with R load, has input voltage of 240 V. The input power factor is

A 1-phase half wave diode rectifier with R = 1 KΩ, has input voltage of 240 V. The diode peak current is

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

For the below given circuit,

After the supply voltage (Vs) is given the

 For the below given circuit,

With Vs = Vm sin ωt (secondary side). The expression for the average voltage is

 For the below given circuit, the

For the below given circuit,

Vs = 325 sin ωt (secondary side)
The ripple voltage is

For a single phase half wave rectifier, the rectifier efficiency is always constant & it is

For the below given circuit,

The power delivered to the load in Watts is