circuit board

In this task,i was provied with 4 diodes for rectification, 2 capacitors, a voltage regulator and 2 LED's. For this circuit to work, i would need to have resistors. to get the right ones i would have to calculate the resistance at some points along the circuit.
Also provided was our lecturer's old digital clock, which would provide an AC voltage supply for the cicuit testing. a digital multi meter for measuring the volts from the input, through more points up to the output

the above block diagram shows the lay out of the board.

the ac current is introduced through the recitfer(the four diodes) which converts it to a d.c current. at the first capacitor, the voltage is about 20.7V dc and on the circuit it would pass through the indicator(LED) next. since this is a to high of a voltage for it, i put a resistor. to work out the resistor  the formula

R= (V-Vled) /current

(20.7-2.3)/0.02 = 0.920 ohms...therefore 1K ohms resistor.
this inturn lights up the led indicator.
the current then goes through the regulator and the next capacitor which i measured to have 5 V.
(5-2.3/0.02) = 0.135 ohms therefore  0.15K ohms resistor to which lights up the LED


With jumper cables clipped to the after of the rectifier,i connected a supply of 12V to test the circuit, and both the LED's should light up. With a multimeter set on the volts range, i put the probes by the first capacitor and the reading should be 12 Volts DC,this being after the voltage goes under rectification and is now DC from AC. The voltage regulator then regulates it to 5V, which is the output indicated by the last LED