You are watching: What was the total energy dissipated by the resistor during this time?
I know that the energy stored on a capacitor is $E=\\frac CV^22$. And the power loss ~ above the resistor would certainly be $RI^2$ combined over time.
Wouldn\"t $\\frac CV^22$ likewise equal $RI^2$ integrated over time? i am finding that my calculated worths for strength loss across a resistor and energy on a capacitor are not same or even close come one another.
Would it it is in power lost or energy lost throughout the resistor? ns am a little confused around units.
Thank you an extremely much!
electric-circuits electrical-resistance capacitance
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edited Mar 5 \"19 in ~ 19:53
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asked Mar 5 \"19 at 18:50
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Mar 5 \"19 at 19:40
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Your confusion seems to be about the distinction between energy and power. Here $I^2 R$ is strength dissipated, and its integral is an energy. For this reason your opening paragraph need to read \"The power loss in the resistor is $I^2 R$ so the energy lost would be $I^2 R$ integrated over time\".
Here is exactly how it functions out. Ns will an initial treat the situation where a capacitor has actually been charged to some voltage $V_0$ and then beginning at time $t=0$ is discharged through a resistor, v no other components in the circuit (including no battery or voltage resource for example). ~ that i will comment on what happens when charging a capacitor native zero.
For a discharging capacitor the formula for the existing in the circuit have the right to be acquired from circuit laws, the is:$$I = I_0 e^-t / RC$$where $I_0 = V_0/R$ if $V_0$ is the initial voltage ~ above the capacitor, i m sorry is $V_0 = Q/C$ for a stored fee $Q$. The mix $(RC)$ has the dimensions of time and is called the time continuous of the circuit.
The power dissipated in the resistor at any given moment is$$R I^2 = R I_0^2 e^-2 t / RC$$therefore the total energy shed to this dissipation is$$E = \\int_0^\\infty R I_0^2 e^-2t/RC dt = R I_0^2\\left< -(RC/2) e^-2t/RC \\right>_0^\\infty= \\frac12 I_0^2 R^2 C .$$Now, making use of $I_0 = V_0/R$ we can also write this$$E = \\frac12 C V_0^2 $$which we have the right to recognize as the energy initially save on computer in the capacitor.
This illustrates unique the rule of conservation of energy.
Now let\"s act a charging capacitor.
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All the over applies unchanged, due to the fact that the present behaves the same way!Nevertheless, this is a different experiment. Once a capacitor is charged from zero to some last voltage by the use of a voltage source, the above energy loss occurs in the resistive part of the circuit, and also for this reason the voltage source then has actually to provide both the power finally save in the capacitor and also the power lost through dissipation throughout the charging process. Now it is the energy noted by the voltage resource that provides the as whole conservation the energy. The maths reflects this perfect correctly yet one needs to be mindful of this aspect of the 2175forals.com in stimulate to recognize what the maths means.