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6 of 20 The input voltage applied to a voltage tripler is 120 V rms . What is th

ID: 2316192 • Letter: 6

Question

6 of 20

The input voltage applied to a voltage tripler is 120 Vrms. What is the ideal dc output voltage?

509 V

360 V

170 V

56.56 V

The dc current flowing through the diode in a half-wave rectifier equals

twice the dc load current

half the dc load current

the dc load current

one-fourth the dc load current

The dc current through each diode in a bridge rectifier equals

twice the dc load current

half the dc load current

the load current

one-forth the dc load current

A half-wave rectifier has a dc load current of 50 mA and a filter capacitance of 470 F. The peak-to-peak ripple voltage equals

17.7 Vp-p

3.6 Vp-p

1.77 Vp-p

0.886 Vp-p

For forward-biased diodes, the voltage where the current increases rapidly is called the

knee voltage

breakdown voltage

zener voltage

turn-off voltage

6 of 20

The input voltage applied to a voltage tripler is 120 Vrms. What is the ideal dc output voltage?

509 V

360 V

170 V

56.56 V

The dc current flowing through the diode in a half-wave rectifier equals

twice the dc load current

half the dc load current

the dc load current

one-fourth the dc load current

The dc current through each diode in a bridge rectifier equals

twice the dc load current

half the dc load current

the load current

one-forth the dc load current

A half-wave rectifier has a dc load current of 50 mA and a filter capacitance of 470 F. The peak-to-peak ripple voltage equals

17.7 Vp-p

3.6 Vp-p

1.77 Vp-p

0.886 Vp-p

For forward-biased diodes, the voltage where the current increases rapidly is called the

knee voltage

breakdown voltage

zener voltage

turn-off voltage

509 V

360 V

170 V

56.56 V

Explanation / Answer

+1) Vdc=3*(Vpeak)=3*120*sqrt(2)=509v

2) option c:Id=I load for half wave rectifier

3) option 2:Id=half the dc load current because each diode carries the load current for half cycle, gives the average current value half of the load current.

4)peak to peak ripple voltage=(Vp*T)/(R*C)=I/(f*C)

=>(50*10^-3)/(60*470*10^-6)=1.77 volts

5) knee voltage

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