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I have a capacitor bank with 4 units in parallel and I'm trying to figure out the worst case scenario of current sharing at turn initial turn on. If 3 of the capacitors have the maximum ESR and one is say half the maximum ESR, how do I calculate the maximum current of the low ESR unit?

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For any voltage step, the capacitance doesn't matter and the instantaneous impedance is just the resistance, or ESR.

In any case, when you're applying a fixed voltage to multiple things in parallel, you calculate the current draw of each thing independently, then add them for the total current.

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    \$\begingroup\$ Yes, they can treat them separately as long as they are switching on something that is a close approximation to a voltage source. But if they are being jointly connected to something with non-neglibable source impedance, especially if there is an intentional inrush currently limiting device, then they can no longer be analyzed in isolation. \$\endgroup\$ – Chris Stratton Jan 5 '12 at 18:22
  • \$\begingroup\$ Say the inrush current is limited to 10A. If the low ESR component is approximately 15% of the total ESR, does that mean the inrush current through the low ESR component will be about 85% of the 10A? \$\endgroup\$ – User909 Jan 5 '12 at 18:41
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    \$\begingroup\$ @User909 - no. Resistances don't work that way. The ESR of one can't be lower than the total parallel ESR. Replace the ESR's by actual resistors in parallel, driven by a common current source, and analyze that. \$\endgroup\$ – Chris Stratton Jan 5 '12 at 19:03
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The amount of current flowing through the capacitance is depend on the load current. If the ESR is high in any one of the capacitor, the heat generated by the capacitor is also high. How ever the current through each capacitor remains same.

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    \$\begingroup\$ The OP's capacitors are in parallel, not in series. \$\endgroup\$ – Dave Tweed Aug 26 '13 at 11:26

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