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RC Circuit - Rate energy is dissipated in the resistor

  • #1

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.
 

Answers and Replies

  • #2
gneill
Mentor
20,734
2,730
Yep. Looks good.

You could also express the results in Watts: 854 mW.
 
  • #3
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.
 
  • #4
gneill
Mentor
20,734
2,730
Watts are Joules per second. So it's a rate of energy exchange. Hope that helps.
 
  • #5
1
0
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
 

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