RC circuit - Rate o'energy dissipated = rate o'energy stored

[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.

b) At what value of t is the rate at which electrical energy is being dissipated in the resistor equal to the rate at which electrical energy is being stored in the capacitor.

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.)
$P = 854 mW$ (From the prior part of the question.)

Homework Equations

$P = Wt$
$W = \frac 1 2 CU^2$
$U = \frac 1 2 CV^2$

The Attempt at a Solution

Using the relevant equations to solve for W:
$W = \frac 1 2 C U^2 = \frac 1 2 C\cdot (\frac 1 2 C V^2)^2 = \frac 1 8 C^2 V^4$

Plugging in numbers gives me: $W = 31.4\cdot10^{-12}~J$

Then using $W = Pt$ to solve for $t: t = \frac W P$, I get $t=27.2\cdot10^{10}~s = 27.2 ~Gs$

This seems obviously wrong, I don't know exactly how many seconds it would take but this looks to be grossly out of proportion. I'm sure that the reason is because I'm not applying the equation for work properly but I'm not sure exactly what I am missing.

 

[haruspex]

P=854mW

What question did that answer? The power will be varying.

Hint: you need to find the current as a function of time.

 

[Delta2]

First of all the equation $W=Pt$ is valid only when the power P is constant, independent of time t. But in the case of a charging capacitor we know that the power P (which is the rate with which capacitor energy is stored) depends on time t. The correct equation that holds for all cases is $W=\int P(t)dt$.

There is a trick we can do to find this. Suppose $I$ is the current at any time t and $V_C$ is the voltage at the capacitor. Then the current at the resistor is also $I$ (connected in series) and the voltage is $V_R=E-V_C$. The exercise asks for the time at which $P_C=P_R\Rightarrow V_CI=V_RI\Rightarrow V_C=V_R=E-V_C\Rightarrow V_C=\frac{E}{2}$. So the requested time is that time at which the voltage at the capacitor equals $\frac{E}{2}$.

 

[bornofflame]

What question did that answer? The power will be varying.

Hint: you need to find the current as a function of time.

That was part a of this question: Just after the circuit is completed, what is the rate at which electrical energy is being dissipated in the resistor?
The answer was 855 mW.

First of all the equation $W=Pt$ is valid only when the power P is constant, independent of time t. But in the case of a charging capacitor we know that the power P (which is the rate with which capacitor energy is stored) depends on time t. The correct equation that holds for all cases is $W=\int P(t)dt$.

There is a trick we can do to find this. Suppose $I$ is the current at any time t and $V_C$ is the voltage at the capacitor. Then the current at the resistor is also $I$ (connected in series) and the voltage is $V_R=E-V_C$. The exercise asks for the time at which $P_C=P_R\Rightarrow V_CI=V_RI\Rightarrow V_C=V_R=E-V_C\Rightarrow V_C=\frac{E}{2}$. So the requested time is that time at which the voltage at the capacitor equals $\frac{E}{2}$.

Using the trick provided by Delta$^2$ I solved for $V_c = 37.3~V$ then took $I = \frac {dq} {dt} \rightarrow I dt = dq \rightarrow \int I dt = \int dq \rightarrow I t = Q \rightarrow t = \frac Q I$
$t = \frac {CV} I \rightarrow t = \frac {CIR} I \rightarrow t = RC = 2.01\cdot10^{-6}F \text{x} 6.51\cdot10^{-3}\Sigma = 13.1\cdot10^{-3}s$

However, according to the actual answer $t = RC ln2 = 9.07\dot10^{-3}s$
I see that $e$ is incorporated in $q = Q_f (1 - e^{-t / RC})$ but I didn't end up using that in my answer. Is this the actual value for Q that I should have used in my calculations?

 

[Delta2]

Er yes you should have use the equation $q=Q_f(1-e^{-\frac{t}{RC}})$ and plug in $q=Q_f/2$ (since it is $V_C=E/2=V_f/2$)) and solve the equation
$\frac{Q_f}{2}=Q_f(1-e^{-\frac{t}{RC}})$ for t.

 

[Delta2]

Up to the point where you write $\int Idt=\int dq$ you are correct but then you write $It=Q$ which is not correct, this holds only if the current is constant in time. But we know in this case the current is not constant , it goes exponentially with time (negatively exponentially to be more accurate).

 

[bornofflame]

So I should have continued to manipulate the equation algebraically before integrating? Replacing $I$ with something else?

 

[Delta2]

Well yes you should have replaced current I with $I=\frac{E}{R}e^{-\frac{t}{RC}}$ and then do the integration and then all you ll get is the equation $q=EC(1-e^{\frac{-t}{RC}})$ which is essentially the equation for q you write after (and I mention in post #5) where $Q_f=EC$.

 

[bornofflame]

Oh! Thank you! I was struggling to make the connection. I overlooked that entirely. It makes sense now. Thank you very much!