Resistance from a V-time graph


by max1205
Tags: graph, resistance, vtime
max1205
max1205 is offline
#1
Feb23-08, 10:13 PM
P: 14
1. The problem statement, all variables and given/known data

A 50 uF(microfarad) capacitor that had been charged to 30 V is discharged through a resistor. Figure 31.73 (attachment) shows the capacitor voltage as a function of time. What is the value of the resistance?


2. Relevant equations

Q = Qo e^(-t/RC), o = initial

Qo = CV, o = initial

I = Io e^(-t/RC), o = initial

3. The attempt at a solution

Original V = 30, the final V at 3ms is approximately 7V. Thus the final charge is 0.233 (7/30) of the initial charge.

Q = Qo e^(-t/RC)
ln (7/30) = -3/(R)(50 uF)

R = 0.041 ohms

The right answer is 36.4 ohms.

Any suggestions?
Attached Thumbnails
figure 31.73.jpg  
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naele
naele is offline
#2
Feb23-08, 10:36 PM
P: 202
[tex]R = \frac {-t}{\ln\frac{V}{Vo}C}[/tex]. However that gives me 41.22 ohm so maybe there's a typo in the problem?

edit: added latex
edit2: actually you didn't solve for 1/R like I first thought, you're just off by 3 orders of magnitude because you probably forgot to convert the milliseconds to seconds.
max1205
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#3
Feb24-08, 08:13 PM
P: 14
does anyone have an idea?

naele
naele is offline
#4
Feb24-08, 08:59 PM
P: 202

Resistance from a V-time graph


Well you estimated the 7V at 3ms since apparently all you had to go by was that graph. If you plug in 36.4 ohms as the resistance into the equation you get something like 5.8V which makes sense. Kind of a lame question if they don't give exact values.
max1205
max1205 is offline
#5
Feb24-08, 11:03 PM
P: 14
thanks!
JosieNutter
JosieNutter is offline
#6
May24-08, 02:24 PM
P: 10
Don't estimate - you can see exactly what V is at 2ms. R is constant.

V = V_0 * e^(-t/RC)

Solve for R an input data from the graph where t=2ms.


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