Electronics > Beginners
Replacing a 50uf cap for a 47uf
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Zero999:
The capacitance can't stay the same, because it would mean the energy being stored would double. The energy stored in a capacitor is proportional to the voltage squared.

E = 1∕2CV2

Suppose the two 100µF capacitors are connected in parallel and charged to 100V. The total capacitance would be 200µF.

V = 100V
C = 200×10-6F
E = 1∕2CV2
E = 0.5*200×10-6*1002 = 100×10-6*10000 = 1J

The capacitors are connected in series and charged to 200V. The total capacitance is now 50µF, but the total amount of energy stored in the capacitors is still 1J.

V = 200V
C = 50×10-6F
E = 1∕2CV2
E = 0.5*50×10-6*2002 = 25×10-6*40000 = 1J

Yes, in practise voltage balancing resistors are a good idea, but when you've got plenty of headroom, it's less critical.

A 68μF 200V capacitor would be fine.
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