1. Feb 22, 2014

### hvthvt

1. The problem statement, all variables and given/known data

The figure shows the circuit of a flashing lamp.There is a current through the lamp (see the photo) only when the potential difference across it reaches the breakdown voltage VL; the capacitance DIScharges completely through the lamp.

(a)A lamp has breakdown voltage VL=72V and is wired to a 95V ideal battery and a capacitor of 0.15 μF. The lamp flashes two times per second. Sketch how the voltage across the capacitor changes over 3 seconds. Neglect the period of time that it takes to discharge.
(b) What resistance is needed for two flashes per second?
(c) How much energy is released through the lamp per flash?

2. Relevant equations

E=0.5CV^2
V=ε(1-e^-(t:RC))

3. The attempt at a solution
I have solved (b), i got 3.25*10^6 ohm. However, I am having difficulty with the sketch. How does it look like, should I use the second formula I mention in 2. Relevant equations? What should I do with the information ' discharge' ?
Furthermore, as I already figured out the resistance needed for two flashes, is this relevant for how mudh energy is released? Or is it just a basic solution like: E=0.5*0.15*10^-6*72^2 ?

2. Feb 22, 2014

### CWatters

You didn't post the circuit but I can guess what it might look like.

Have a look at the curve for the voltage on a capacitor during the charge...

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html

When the voltage reaches 72V the lamp will conduct and this will discharge the capacitor. The problem statement says "Neglect the period of time that it takes to discharge" so you can assume the voltage falls from 72V to 0V in zero time and then starts rising again.

3. Feb 22, 2014

### CWatters

Hint: Where does the energy stored in the capacitor go?

4. Feb 23, 2014

### hvthvt

O, I am very sorry. I forgot to include the circuit itself. Here it is!

So, I think that the graph starts from
V=72V for t=0 This is the moment when the lamp starts flashing and the capacitor starts discharging, right? and it reaches zero in a couple of seconds? So from left to right it decreases. Is this right?
However, can anybody help me with my problem of not knowing what to do with 3 seconds? How do I know how much of the capacitor has discharged in 3 sec?

For (c), the energy stored in the capacitor leaves the capacitor and goes to light the flashing lamp right? Is this simply E=Pt?

I would get for the energy E=U*I*t=72V*22μA*0.5=80mJ or E=0.5*C*V^2=39mJ (by using V=72V)

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• ###### Schermafbeelding 2014-02-22 om 22.18.15.png
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5. Feb 23, 2014

### CWatters

No they say to neglect the discharge time so you should assume the duration of the flash is very short and that it discharges the capacitor instantaneously. The problem statement doesn't provide enough data to work out the real discharge time anyway.

I would start the graph at t=0 and V=0 (eg pretend the voltage source is connected at t=0 and before that point the capacitor is discharged).

So the first part would show a charge cycle from 0V towards 95V. It never gets to 95V because at 72V the lamp flashes and the capacitor discharges instantaneously to 0V and then starts charging again. Repeat.

The problem says the lamp flashes twice a second so the time taken to get to 72V is half a second. There will be exactly 6 charge cycles each taking 0.5 seconds in the 3 seconds they ask you to draw.

6. Feb 23, 2014

### CWatters

The current isn't a constant 22uA so that method can't be used (at least not easily).

Find an equation for the energy stored in a capacitor given just the capacitance (0.15μF) and the voltage (72V).

7. Feb 23, 2014

### hvthvt

Well, then I would say.. Wcharging=∫$\frac{q}{C}$dq=$\frac{1}{2}$ Q^2:C =0.5QV2=0.5CV2 ??? I don't see how I can integrate it with the current varying..

8. Feb 23, 2014

### CWatters

9. Feb 23, 2014

### CWatters

Sorry I just spotted that you had the right equation in your post here (just check your working I made it 0.38mJ)....