Question

A capacitor has a peak current of 240 μA when the peak voltage at 270 kHz is 2.8 V .

What is the capacitance?

If the peak voltage is held constant, what is the peak current at 540 kHz ?

Answer 1

let,

io=240uA =240*10^-6 A

vo=2.8v

frequency f=270kHz =270*10^3 Hz

a)

peak voltage,vo=iO*Xc

vo=iO*(1/w*C)

2.8=240*10^-6*(1/(2pi*270*10^3*C)

==>C=50.525*10^-12

capacitane C=50.525 pF

b)

if f=540*10^3 Hz

vo=io*(1/w*C)

2.8=io*(1/(2pi*540*10^3*50.525*10^-12))

===> io=0.48*10^-3 A

peak current io=0.48 mA

or

peak current io=480 uA