View single post by bobquincy
 Posted: Wed Mar 31st, 2021 01:27 pm
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bobquincy



Joined: Sun Jan 27th, 2013
Location: Florida USA
Posts: 324
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I stand corrected:

T (seconds) = R*C

In that time the voltage of the capacitor,
will drop to ~ 37% of the initial charge voltage. 


If the capacitor is at 2.3 V and the motor draws 0.2 A (about 11 Ohms),
the T is 11 * 2.7 or about 30 seconds to 0.85 V.


Your calculation of 11 seconds run time appears to be good.  :)

boB




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