# RC Circuit Energy Problem

1. Mar 16, 2014

### majormaaz

1. The problem statement, all variables and given/known data
The switch in the figure below has been in position a for a very long time. It is suddenly flipped to position b for 1.65 ms, then back to a. How much energy is dissipated by the 50 Ω resistor? (Assume = 65 V.)

2. Relevant equations
ΔVloop = 0 Volts
Q/V = C
V = IR

3. The attempt at a solution
I understand that if the 125 Ω resistor wasn't there, the capacitor would have the full 65 Volts across it, but there is a voltage drop across the 125 Ω resistor, so the capacitor has less than 65 Volts on it.

I set up an equation starting from the bottom left corner of the circuit, 0 = 65 Volts - (125 Ω)*I - Q/(20μF), and I tried to solve it by bringing the Q/(20μF) to the other side and replacing I with V/R, but the V in V/R is the voltage across the resistor, not the battery's voltage.

In short, I'm stuck.

2. Mar 16, 2014

### Staff: Mentor

With the switch in position a, and "after a very long time", what do you expect the value of the current to be?

3. Mar 16, 2014

### majormaaz

The current would be 0 Amps then...? So the capacitor does reach a fully charged state i.e. Vc = EMF? If I may ask, what's the point of the 125 Ω resistor? It's not going to factor into the time constant, is it?

4. Mar 16, 2014

### Staff: Mentor

The 125 Ω resistor limits the charging current, otherwise (assuming ideal components and wiring) the voltage source would need to supply infinite current... not a good thing for a practical circuit. As a side effect, it also results in a time constant for the charging of the capacitor so that you could, if you wanted to, make a good estimate of what might be considered "a very long time" Otherwise, its particular value (125 Ω) is not relevant to the problem at hand.

5. Mar 16, 2014

### majormaaz

Ok, let me just log my progress here.

Since there's no current after a long time, my loop equation drops down to 0 = V - Q/C, where with a V of 65 Volts and C of 20 μF, I end up with a Q of 1300 μC.

Since the capacitor is fully charged and discharged for 1.65 ms, I can use the equation Q = Q0(e-t/τ), where τ = RC. With t = 1.65 e -3 secs, Q0 of 1300 μC and a τ of (50Ω)(20μF) = 0.001 secs, I ended up with 249.66 μC of charge on the capacitor.

Q/C = V, and 249.66 μC / 20 μF = 12.48 Volts left on the capacitor. That means (65 - 12.48) V = 52.52 Volts have passed through the 50 Ω resistor.

I'm pretty sure P = V2/R, and using 52.522/50 gives me 55.16 Watts. A Watt is a J/s, so multiplying by 1.65 e -3 seconds got me 0.091 J or 91.0 mJ.

However, that approach was ultimately wrong. Is there anything I'm missing?

6. Mar 17, 2014

### willem2

This last sentence doesn't make any sense, Potential, voltage or volts do not pass through a resistor. You start out with 65 V across the resistor and end up with 12.48 V. The voltage across the resistor isn't constant, nor is the power dissipated in it.

To get the power dissipated across the resistor during the 1.65 ms you need an integral:

$$\int_0^{0.00165} { \frac { (V(t))^2}{R} dt }$$

V(t) is easy to find from Q(t) wich you already have.

This integral is easy, but it's even easier to find the power dissipated by the resistor from the loss of energy of the capacitor.

7. Mar 17, 2014

### majormaaz

To be honest, this is the first time I'm seeing an integral like this being used for a RC circuit, but it does make sense, and I did get the right answer once I sorted out the integral.

With respect to the 'loss of energy of the capacitor', I'm sure it's 1/2CV02, but how would the time factor in? I understand the integral, but I'd like to know how this way works for future reference.

8. Mar 17, 2014

### willem2

The capacitor loses energy between t=0 and t=0.00165. There's only one place where this energy can go.

9. Mar 17, 2014

### majormaaz

Wow, that was simple. I kinda dumbed out there, but thanks for the help!