Energy stored in capacitors

  • Thread starter nate559
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Two capacitors, C1 = 20 µF and C2 = 5.0 µF, are connected in parallel and charged with a 150 V power supply

A.) Energy Stored I found to be is 2.81*10^-1 J

B.)What potential difference would be required across the same two capacitors connected in series in order that the combination store the same energy as in (a)?

Delta V= sqrt of (2*Energy stored/C) This eqn can b used to solve for delta V

I have no idea what im doing wrong! I have used the capcitor in parallel combo of C= C1+C2..... Then I have converted the microFaraday charges of these sums to Faraday but no success

Any Suggestions??
 

Answers and Replies

  • #2
Kurdt
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You know that the effective capacitance of capacitors in series is given by:

[tex]\frac{1}{C}=\frac{1}{C_1} + \frac{1}{C_2}...... [/tex]

You also know that the energy stored is given by:

[tex]E=\frac{1}{2}CV^2[/tex]

So you have rearranged perfectly well but the problem could just be that you are using the parallel combination rather than the series combination to find the effective capacitance.
 

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