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## 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.