| 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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