# Heat in a resistor (circuit with inductor and capacitor)

1. Sep 1, 2014

### Rugile

1. The problem statement, all variables and given/known data
A source of ε = 10V, capacitor of C = 5 μf, inductor of L = 15 mH and r = 10 Ω and a resistor of R = 100 Ω are connected as in the schematic (attached). How much heat will dissipate in the resistor after the switch is opened?

2. Relevant equations

dQ = I2Rdt
KVL

3. The attempt at a solution

First of all I was quite confused by the question, so I just assumed that the switch is opened after a long time, when the capacitor and the inductor are fully charged. Then the voltage across the capacitor is U0 = ε and the current in the inductor is I0 = ε / (R+r). We can write the equation (KVL):
$L\frac{dI}{dt} + I(R+r) + U_C = 0$, where UC is the voltage across the capacitor. I get stuck here - I suppose I could write second order differential equation involving the function of q: $L \frac{d^2 q}{dt^2} + (R+r) \frac{dq}{dt} + \frac{C}{q} = 0$, but I'm not sure what to do with that either. Any hints?

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2. Sep 1, 2014

### td21

Step 1: After switch is opened, LC oscillation occurs as described by your equation.
Step 2: Heat loss through R and r. Your task is to find out the heat loss thorugh R. At the end, all energy is dissipated through heat loss. So I suggest you find the ratio between heat loss through R: heat loss through r. I also suggest you find the initial energy.
Step 3: The ratio can be found from the ratio of R:r.

3. Sep 1, 2014

### Rugile

Okay, so it is enough to find the initial energy of the system and then calculate the heat loss in R using the ratio? Can I say that the initial energy of the system (after opening the switch) was $E = \frac{CU^2}{2} + \frac{LI^2}{2}$, where $I = \frac{\epsilon}{R+r}$ and U = epsilon = EMF ?

4. Sep 2, 2014

What switch?

5. Sep 3, 2014

### Staff: Mentor

The one marked with letter 'J'.

6. Sep 3, 2014

### Rugile

Yes, exactly.

Does anyone have any idea if the equation of initial energy written above is correct? I have doubts about the current in the inductor, since when we open the switch it disappears (gradually), doesn't it?

7. Sep 3, 2014

### BvU

O2: Very subtle...
Rugile's post #3 seems pretty good to me: keep going !

And post #6: yes, it dampens out quickly. But the exercise only wants the total, so no need to solve the D.E.

8. Sep 3, 2014

### Staff: Mentor

That looks right. That's the stored energy before the jumper to the battery is removed.

After the jumper is removed, the inductor current becomes the capacitor current, the two now being in series.