Hello,

A constant current source is in series with a capacitor ( 10volts,

0.5uF) and a resistor ( load) of 100kilo ohms.

What happens if

1. A constant current source charges up the capacitor upto its maximum

voltage and keep charging it for a long time. The constant current

source has following specs

a. compliance voltage +/- 15 volts

b. current range 0 to +/- 600uA (ac)

c. biploar

d. dc leakage current of 100nA

2. The capacitor will show high impedance to the current source

No, it will initially show a short!

That is why it starts at zero volt across the capacitor.

..6mA in 100kOhm makes 60V, you only have 15V, you example makes no sense.

but

get charged by the DC leakage current constantly. Will the AC flow

through the capacitor anymore?

If you omit the resistor for a moment, then the voltage across the capacitor is

The charge Q = C * U = I * t

So U = (T * t) / C

C in Farad, I in ampere, and t in seconds.

The voltage across the capacitor will then rise LINEAR with time, until it reaches

the 15V (theoretically).

However, with the resistor in parallel, the max voltage will be limited by I * Rp, or

the 15V, whatever is lower.

In this case 15V is lower then 60, so 15V

And the charge current in the capacitor is no longer linear, as

as teh voltatge rises, the current through the resistor is increasing,

substract that from the current in the capacitor.

The capacitor will charge like this:

http://en.wikipedia.org/wiki/RC_circuit
see under 'Time domain considerations'.

There is no longer a linear rise in voltage.