RC Circuit - Rate energy is dissipated in the resistor

[bornofflame]

Homework Statement

1. A 2.01 uFcapacitor that is initially uncharged is connected in series with a 6.51 kΩ resistor and an emf source with 74.6 V and negligible internal resistance. The circuit is completed at t = 0.

a) Just after the circuit is completed, what is the rate at which electrical energy is being dissipated in the resistor?

Given/Known:
$C = 2.01\cdot10^{-6} ~F$
$Q_0 = 0 ~C$
$R_1 = 6.51\cdot10^3 ~\Omega$
$\mathcal {E} = 74.6 ~V$ (The source can be treated as ideal.)

Homework Equations

$P = \frac {V^2_{ab}} {R_{eq}}$

The Attempt at a Solution

I believe that the question is simply asking for the power across the resistor which is why that is the only equation that I have included. That said:

$P = \frac {V^2_{ab}} {R_{eq}} = \frac {74.6~V} {6.51\cdot10^3 ~\Omega}$
$=\frac {5565~V^2} {6.51\cdot10^3 ~\frac V A} = 854\cdot10^{-3} \frac V A = 854 \frac {mJ} {s}$

is my solution.

It seems pretty simple but I'm not as confident in my knowledge of circuits as I would like to be so I want to make sure that I'm understanding this before I move on to the next step.

 

[gneill]

Yep. Looks good.

You could also express the results in Watts: 854 mW.

 

[bornofflame]

Ok, thank you! mJ/s seemed weired but I guess it felt more fitting as a rate than simply W, which I still don't immediately associate with rate.

 

[gneill]

Watts are Joules per second. So it's a rate of energy exchange. Hope that helps.

 

[safaabrg]

I think that the way to solve this is by differential equations since the difference of potential V is a variable so start by establishing the relation between Vcapacitor ,Vresistor and E (unsing Kirchhoff's law ) and then derive the whole relation and find the differential equation for Vresistor .then You'll find the rate of change of Vre over time (an exponential function ) then replace it in your relevant equation. If I'm wrong then, please correct my answer