I have two questions I CANNOT figure out. Please help.

patweadock said:

I'll give it a shot;
the transformer reduces V by 1/4, so secondary has 20V. Since it's center tapped, it should put 10V across R. I believe peak V is square root of 2, X Vrms so 1.414 X 10 = 12.4V Pk V across R, So Pk I = 12.4/1K = 12.4 mA. I believe average I is .63 X PK so 12.4mA X .63 = 7.81mA. I think the peak inverse V across each diode would be 20V.
If C3 is 750 ohms and has 4mA thru it, the V drop is 3V, which would also be the V across C2 and VC1=5-3=2V. Using the reactance formula, Fr=1/(2Pi*Xc*C), the Freq = 141,471Hz. Xc2 would be 562.5 ohms and the current thru it would be 5.3 mA. So the current thru C1 is 4mA+IC2=9.3mA and it's dropping 2V. So Xc1= 215 Ohms, so it's value must be .005.2 uF. Pat

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