I think you must explain your problem more clearly.

If I understood you you have a source composed of supercapacitors connected in series. The open circuit avolated of the one capacitor is eqaul to its rated voltage of 2.7V. You have 12 capacitors connected in series with a series resistance of one capacitor= 8.7 10^-3 Ohm.So the total resistance will be 12x8.7 10^-3.

About .08 ohm. So, there is about 4 V drop on this resistance and the current amounts to approximatively to 4/.08= 50 A.

So, you have to measure the current withdrawn from the capacitor bank to check that the drop is the voltage is due to the current flowing in the series resistances.

Please give us feedback!

Best wishes

Dear Santosh,

what is the value of the voltage of the input voltage source?. You can not charge the input capacitor beyond the input voltage irrespective of the number of the capacitors in series or in parallel. In order to increase the voltage across the capacitor you have to increase its charging voltage.

Best wishes.

Abdelhalim abdelnaby Zekry thank you for your response,

Sir, I'm not quite sure I understood that completely. I believe you mean there's a maximum voltage for a single cell. I've run a 2nd simulation after your feedback and It'd be helpful to get more insights.

From the datasheet, I've found the Capacitance to be 1200, rated voltage to be 2.7v and the series resistance to be 8.7 milliohms for Electric Vehicle application. And I guess by the value of input voltage you mean the initial voltage of the capacitor, which I've specified as 32.4v.(12 series capacitors = 12 x 2.7). However, the voltage displayed is only 28.13v while it should be 32.4v. why is there a voltage difference?

Please find the attached images below for reference

At first sight, You have losses due to capacitors's internal resistance.

Abdelhalim abdelnaby Zekry you are right