- #1
Artlav
- 161
- 1
Hello.
I want to know what current i am getting out of a supercapacitor, but i have nothing to directly measure the currents in the given range with.
So i tried to determine it from the voltage drop on the cap, and i want to know if i did it right.
Given is 3000F capacitor, at initial voltage of V1=1.100V.
After being connected to the load in question for 3.5 seconds, the voltage is V2=1.045V.
Thus, the energy in the capacitor is 0.5*C*V*V.
E1=1815J
E2=1638J
Which gives spent energy of 177J.
W=J/s, so for the 3.5 seconds it was giving out 50.5W of power.
Finally, A=W/V, and average V was 1.0725V, which gives 47.1A of current.
Is that a correct way of current estimation?
I want to know what current i am getting out of a supercapacitor, but i have nothing to directly measure the currents in the given range with.
So i tried to determine it from the voltage drop on the cap, and i want to know if i did it right.
Given is 3000F capacitor, at initial voltage of V1=1.100V.
After being connected to the load in question for 3.5 seconds, the voltage is V2=1.045V.
Thus, the energy in the capacitor is 0.5*C*V*V.
E1=1815J
E2=1638J
Which gives spent energy of 177J.
W=J/s, so for the 3.5 seconds it was giving out 50.5W of power.
Finally, A=W/V, and average V was 1.0725V, which gives 47.1A of current.
Is that a correct way of current estimation?
