A capacitor has a peak current of 320 μA when the peak voltage at 320 kHz...

Question

Question

A capacitor has a peak current of 320 μA when the peak voltage at 320 kHz is 2.9 V .

If the peak voltage is held constant, what is the peak current at 640 kHz ?

Answer 1

Answer #1

Sol:

Given
Peak current(Ip)=320 μA
Peak voltageVp=2.9 V
F1= 320 kHz
F2=640 kHz

Assume the AC resistance of the capacitor is neglible.
The impedance of an capacitor is
Zc= 1/2π*f*C

V = I*Zc
=> 2.9= (320*10^-6)*Zc
Zc = 2.9/(320*10^-6)
= 9.062 kOhms

If the frequency(320 KHz) is doubled (i.e F2=2*F1=640 kHz) ,Zc will become HALF.
Because the impedance is inversely proportional to frequency.

So the current will be DOUBLED.

I=2*Ip
=2*320 μA
= 640 μA

Hope this helps you..
Please VOTE my answer

Thank you

Answer 2

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