Archived How Is Energy Distributed in a Series RC Circuit?

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In a series RC circuit with a 0.25μF capacitor charged to 50V and connected to a 25Ω and a 100Ω resistor, the total energy stored in the capacitor is calculated using U = 0.5CV², resulting in 0.625 joules. The total resistance of the circuit is 125Ω. To find the energy dissipated by the 25Ω resistor, the energy can be apportioned based on the resistor's proportion of the total resistance. Since the 25Ω resistor represents 1/5 of the total resistance, it dissipates 1/5 of the total energy, equating to 0.125 joules. This method effectively distributes energy dissipation across the resistors in the circuit.
StephenDoty
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A 0.25mu F capacitor is charged to 50 V. It is then connected in series with a 25ohm resistor and a 100ohm resistor and allowed to discharge completely.How much energy is dissipated by the 25ohm resistor?

I know that Q=CV
and U=.5CV^2=.5 * Q^2/C

Now U is the energy released over both resistors. So the energy that is discharged over the 25 ohm resistor relates to the total energy dissipated over the two resistors equaling 125 ohm. But how do I use the total energy dissipated over both resistors and the total resistance of 125 ohm to find the energy dissipated over the 25ohm resistor?

Thanks.
Stephen
 
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One approach is to imagine that the two resistors are combined into a single resistor (no change to circuit behavior will result from this). Knowing the total energy dissipated and assuming that the energy dissipation is uniform over the whole resistor, apportion the total energy according to the relative portion of the whole resistor.

upload_2016-2-7_13-50-3.png


So if R1 is 100 Ω and R2 is 25 Ω, then R2 represent 25/125 = 1/5 of the whole.
 
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