a centre tapped transformer rated 20Vac serves as a supply to a full wave rectifier circuit.if the rectifier circuit is used to power a 3.3kilo Ohm load, calculate the: (1) maximum & d.c value of the output voltage. (2) maximum and d.c value of the load current (3)peak inverse voltage of the diodes(assume ideal diodes for the rectifier)
This sounds like a homework problem for your class in basic electronics. So, I will give you some help in thinking it through so that you can do the computations for answers to the questions.
1. Can we assume the secondary voltage is 20v ac rms? If so, it means the peak voltage is 1.414 times and the peak to peak is 1.414 times 2 times the 20vac. Since it is center tapped the diodes being used for rectifiers are then each getting a peak voltage of 28.3vac.
However, it is not clear whether the 20vac is total secondary making the diode voltages at 10vac or if the total voltage of the secondary is really 40vac with 20vac on each half.
2. Secondly to consider in determining the voltage output of the rectifiers is if a capacitor is being used to filter out the ac components. If a good capacitor is used in a full wave rectifier it can bring the output dc voltage to nearly the peak value of the input wave. If not, then the dc value is that of a half-wave rectified wave or around .63 of the peak value. See this page:
3. Once you know the voltage being fed to the load resistor of 3.3kohm you can compute the outputs. But, you need to know the rms value or take the dc and add the ac current to get the real totals.
Hope this gets you going in the right direction. Let me know if you need more or have any questions.
Good Luck. Study Hard.