a) Given
Imax = 310*10^-6 A
Vmax = 2.5 V
But we have Xc = Vmax/Imax = 2.5/(310*10^-6) = 8064.52 ohm
Xc = 1/(2pi*f*C) =======> C = 1/(2pi*f*Xc) = 1/(2pi*340000*8064.52)
C = 58*10^-12 F = 58 pF
(b) Vmax = 2.5 V
f = 680000 Hz
C = 58*10^-12 F
Xc = 1/(2pi*f*C) = 4035.4 ohm
Imax = Vmax/Xc = 2.5/4035.4 = 6.195*10^-4 A = 619.5 uA = 58*10^-12 F
Xc = 1/(2pi*f*C) = 4035.4 ohm
Now Imax = Vmax/Xc = 2.5/4035.4 = 619.5 uA
Problem 32.12 A capacitor has a peak current of 310 μA when the peak voltage at...
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