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In a circuit shown below me and a friend of mine discussed how we could calculate this.

schematic

simulate this circuit – Schematic created using CircuitLab

He simplified the circuit by stating the resistors and capacitors are in parallel, and hence it can be simplified to a capacitor of 17 farad and a resistor of 142k ohm. So energy stored would be: \$ \tfrac{1}{2} CV^2 = 0.5 \cdot 17 \cdot 36= 306 J\$ I thought this was wrong, and one could simplify the circuit to 3 "parallel" rc-circuits, each consisting of a 47k ohm resistor, 50 F capacitor and a 6/3 = 2 V dc supply. So total energy that could be stored is: \$3\cdot\tfrac{1}{2}CV^2 = 1.5 \cdot 50 \cdot 4 = 300 J\$

What is the correct approach, or should a complete different method be taken?

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  • \$\begingroup\$ What circuit? Add a schematic... \$\endgroup\$
    – Matt Young
    Dec 15, 2013 at 23:51
  • \$\begingroup\$ @MattYoung oops, multi-tabbing stackexchange and forgot it >< (though is it worth a negative vote?) \$\endgroup\$
    – paul23
    Dec 15, 2013 at 23:55
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    \$\begingroup\$ "0.5 * 17 * 36" can't possibly equal 306 J. Multiplying any number of dimensionless quantities can't ever result in a value of Joules. -1 for blatant sloppiness with units. \$\endgroup\$ Dec 16, 2013 at 0:36
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    \$\begingroup\$ @OlinLathrop Do you really wish for a "war" on use of units during a calculation? I've always learned to leave those out -and always use the standard SI unites- so that the magnitude and size of the units can be seen easily and there's no difficulty with "m" vs "M" etc. \$\endgroup\$
    – paul23
    Dec 16, 2013 at 0:41
  • \$\begingroup\$ @Paul23 Yes. Olin is the minister for war and proper-units [tm]. I'd agree that if you give the formula first using relevant symbols, and follow it with a numeric equivalent then it is (usually) clear and adequate - BUT, as that is what I often do I'd have to , wouldn't I ? :-). BUT there IS great difficulty with swapping m and M as they are both formally defined symbols and differ in magnitude by a factor of 10^9 (as I know you know). m/F conflation will not usually cause errors but does bend the path of smooth thinking and should be avoided where possible (ie almost always.) \$\endgroup\$
    – Russell McMahon
    Dec 16, 2013 at 1:12

2 Answers 2

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Rather than arguing about combining resistors and capacitors in parallel and series, recall that, in steady state, the current through a capacitor is zero. Thus, assuming steady state, replace the capacitors with open circuits.

Then, by inspection, there is 2V across each resistor (if the resistor values were different, one would used voltage division to determine the voltage across each).

Thus, each capacitor has 2V across. Since the three capacitor values are identical, the stored energy is simply:

$$W = 3 \cdot \dfrac{1}{2}50F\cdot(2V)^2 = 300J$$

If the capacitor values were different, the calculation would be:

$$W = \dfrac{(2V)^2}{2}(C_1 + C_2 + C_3) $$

If the resistor values were different too, the calculation would be:

$$W = \dfrac{(6V)^2}{2(R_1 + R_2 + R_3)^2}(R_1^2C_1 + R_2^2C_2 + R_3^2C_3) $$

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You have a rounding error in the first calculation - the three 50 F capacitors in series will be 16.666666666666... F, not 17. Using that value, I get the same answer both ways.

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