Question

In the circuit shown, R1 = 10 Ω, R2 = 4 Ω, R3 = 10 Ω, ε = 10 V, and capacitor has a capacitance C = 2 μF and the switch S has been closed for a long time. After switch S is opened, how much total energy is dissipated through R3?
http://tinypic.com/r/24v678w/5<-picture

Answer 1

When \(S\) is closed for a long time, the capacitor will begin to charge up with a charge according to \(C=\frac{Q}{V}\), the definition of capacitance. The amount of charge on one plate is thus given by \(Q=CV\). The voltage across the capacitor \(V\) is simply the voltage \(\mathcal E\) provided by the battery minus the voltage drop across resistor \(R_1\). Recall Ohm's law \(V=IR\). Combining what we know so far gives us the following expression for the initial charge \(Q_i\) on one plate of the capacitor. Let \(I_1\) be the current through \(R_1\). \[Q_i=C(\mathcal E - I_1R_1)\]To find \(I_1\), let us consider the resistance of the entire circuit as a whole and then apply Ohm's law again since this wire comes directly from the battery. The resistance of the whole circuit is given by \(R=R_1+\frac{1}{\frac{1}{R_2}+\frac{1}{R_3}} = R_1 + \frac{R_2R_3}{R_2+R_3}\) since \(R_1\) is connected in series and \(R_2\) and \(R_3\) are connected in parallel (once the capacitor is fully charged, it just acts like a break in the wire at that point). By Ohm's law, the current is thus \(I_1 = \frac{\mathcal E}{R_1 + \frac{R_2R_3}{R_2+R_3}}\). We know have the following expression for \(Q_i\).\[Q_i=\mathcal EC\left(1-\frac{R_1}{R_1 + \frac{R_2R_3}{R_2+R_3}}\right)\]After switch \(S\) is opened, we can ignore the entire left portion of the circuit since it is no longer closed. We will now only consider the parallel series \(R_2\), \(R_3\) and \(C\). As we calculated before, the combined resistance of \(R_2\) and \(R_3\) is \(\frac{R_2R_3}{R_2+R_3}\). By Kirchoff's voltage law, the sum of the voltage drops across the entire circuit must be zero. We will consider the parallel resistors as one entity, as well as consider \(I\) as the current through the single wire of the circuit as well as let \(Q\) be the time-variant charge on one of the capacitor plates. Recognize now that \(I=\frac{dQ}{dt}\).
\[\underbrace{\frac{Q}{C} }_{C = \frac{Q}{V}}+\underbrace{\frac{dQ}{dt}\frac{R_2R_3}{R_2+R_3}}_{V=IR}=0\]We will find the general solution of this differential equation via separation of variables. Don't forget that \(\int \frac{dQ}{Q}=\ln|Q|+K\)!\[-\int \frac{dt}{\frac{R_2R_3}{R_2+R_3}C}=\int \frac{dQ}{Q}\]\[-\frac{(R_2+R_3)t}{R_2R_3 C}=\ln|Q|+K\]\[Q(t) = Q_i e^{-\frac{(R_2+R_3)t}{R_2R_3C}}\]Note that when we pull the constant of integration out of the exponential, we know it must be \(Q_i\) because when \(t=0\) the exponential is just 1. Let's substitute now our found value for \(Q_i\).\[Q(t) = \mathcal EC\left(1-\frac{R_1}{R_1 + \frac{R_2R_3}{R_2+R_3}}\right) e^{-\frac{(R_2+R_3)t}{R_2R_3C}}\]Now recall that power is given by \(P=\frac{V^2}{R}\). Even though the resistor is in series, the voltage across the capacitor is also the voltage across our resistor \(R_3\). By the definition of capacitance, the voltage across the capacitor is \(V(t) = \frac{Q(t)}{C}\) and the resistance we will use is \(R_3\). Plugging these values in gives us the following expression for power. Take note that squaring the exponential only brings a factor of 2 into the exponent. \[P = \frac{\left(\mathcal E\left(1-\frac{R_1}{R_1 + \frac{R_2R_3}{R_2+R_3}}\right) e^{-\frac{(R_2+R_3)t}{R_2R_3C}}\right)^2}{R_3}=\frac{\mathcal E^2}{R_3}\left(1-\frac{R_1}{R_1 + \frac{R_2R_3}{R_2+R_3}}\right)^2 e^{-\frac{2(R_2+R_3)t}{R_2R_3C}}\]Now, work done (and thus energy released) is just the integral with respect to time of power, we will use an improper integral to evaluate the energy released from \(t=0\) to \(t=\infty\). While this may seem unweildy at first, you honestly couldn't wish for an easier function to integrate other than the expoential function. \[W = \int\limits_0^\infty \frac{\mathcal E^2}{R_3}\left(1-\frac{R_1}{R_1 + \frac{R_2R_3}{R_2+R_3}}\right)^2 e^{-\frac{2(R_2+R_3)t}{R_2R_3C}}dt =\]\[W= \left. \frac{\frac{\mathcal E^2}{R_3}\left(1-\frac{R_1}{R_1 + \frac{R_2R_3}{R_2+R_3}}\right)^2}{-\frac{2(R_2+R_3)t}{R_2R_3C}} e^{-\frac{2(R_2+R_3)}{R_2R_3C}} \right|_0^\infty \]\[ W=\frac{\frac{\mathcal E^2}{R_3}\left(1-\frac{R_1}{R_1 + \frac{R_2R_3}{R_2+R_3}}\right)^2}{-\frac{2(R_2+R_3)}{R_2R_3C}}\]\[\boxed{\displaystyle W=\frac{\mathcal E^2 R_2C}{-2(R_2+R_3)}\left(1-\frac{R_1}{R_1 + \dfrac{R_2R_3}{R_2+R_3}}\right)^2}\]Now, just plug given values in and evaluate using a calculator. :-)

Answer 2

One itty bity typo I found: third to last equation: there should be no \(t\) in the denominator of the big fraction, and it should rather be in the exponent, but subsequent calculations aren't affected by that. :)

Answer 3

I get \(1.41\ \mu\text J \) when I plug everything in, though you may wish to double-check that for accuracy.

Answer 4

(Also, I am curious, what level physics class are you in? AP? College-level? This was quite the problem!)

Answer 5

problem solution can be considerably reduced in length as follows: lets start with switched closed:
since we have a dc source, the capacitor of course acts as an open circuit, and we can calculate the voltage v(0) just before (and just after) the circuit is opened from the voltage divider rule:
\[v(0)=\epsilon(R _{\left| \right|}/(R _{\left| \right|}+R1)=10V (2/9)= 20/9 V\]
where I skipped finding R∣∣ as it's easily shown to be 20/7 Ohms.
Next, the switch is open and the capacitor is discharging through R∣∣ according to:
\[v(t)=v(0) e ^{-t/\tau}\], where τ=R∣∣ C
Finally, the energy dissipated through R3 is:
\[E=\int\limits_{0}^{\infty} dt (i _{R _{3}}(t) v(t)) =(v(0)^{2}/R _{3}) \int\limits_{0}^{\infty}dte ^{-2t/\tau}=\left[ v(0)^{2} C R _{\left| \right|}/(2R _{3}\right]\]
plugging in the given values gives the correct answer.

Answer 6

College level,intro physics 2.