capacitors?

Question

if a capacitor was charged to 345v and has 1.4 farads, how many amps can be drawn at peak?
i'm not sure if that makes sense, but i want to know how many amps i can draw at most, when it first starts to discharge.

Answer 1

WARNING! This is definitely a lethal amount of energy! It will blow joints out of sockets!

I take it this is actually an array? At any rate, a charged cap represents a specific amount of energy, which is equal in joules to CV^2/2, or, in your case,

345 * 1.4/2 = 83.3kilojoules, aka 83.3 kilowatt-seconds.

Now, if this were an ideal cap with zero internal resistance (impossible), and your load were zero ohms (also impossible), you could draw infinite amps, but for zero time.

But the fact is, neither the cap(s) nor the load are zero ohms resistance, so there will be a finite limit to the current. If the cap is an array of inverter-grade electrolytics, internal resistance might peg source to say 0.05 ohms. Now, what is the impedance of your load? If it is 0.95 ohms, then the peak current will be 345A, rapidly decreasing according to the time constant of the RC, which is

1.4F * 1 ohm = 1.4 seconds

After one time constant of 1.4 seconds, the voltage across the cap will have fallen to 63% of it's initial value, likewise the current.

So it all depends on
1- the internal resistance of the cap(s)
2- the resistance/impedance of the load

email me if you need more help.

BW,
GH

Answer 2

Theoretically, infinite amps but then to get that you need perfect zero impedance in the leads and the rest of the discharge circuit.

Answer 3

345v divided by the internal resistance of the capacitor will give you the initial peak current if you short cirucit the leads.

Answer 4

At that charge and that capacitance you do not want to play with it. You will be toast if u touch that cap