Resistance from a V-time graph

  • Thread starter max1205
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  • #1
max1205
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Homework Statement



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?


Homework Equations



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

Qo = CV, o = initial

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

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?
 

Attachments

  • figure 31.73.jpg
    figure 31.73.jpg
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Answers and Replies

  • #2
naele
202
1
[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.
 
Last edited:
  • #3
max1205
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does anyone have an idea?
 
  • #4
naele
202
1
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.
 
  • #5
max1205
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thanks!
 
  • #6
JosieNutter
10
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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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