Resistance from a V-time graph

In summary, the value of the resistance in this scenario can be calculated using the equation R = -t/ln(V/Vo)C, where V is the voltage at a given time (2ms in this case) and Vo is the initial voltage (30V). By plugging in the values from the graph, the calculated resistance should be approximately 36.4 ohms, which matches the correct answer.
  • #1
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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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  • #2
[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
does anyone have an idea?
 
  • #4
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
thanks!
 
  • #6
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.
 

1. What is a V-time graph?

A V-time graph, also known as a voltage-time graph, is a visual representation of how the voltage changes over time in an electrical circuit. It is typically plotted with voltage on the y-axis and time on the x-axis.

2. How do I interpret a V-time graph?

To interpret a V-time graph, look at the slope of the line. The steeper the slope, the higher the voltage change over a given time interval. A flat line indicates a constant voltage, while a curved line indicates a changing voltage.

3. What does resistance look like on a V-time graph?

Resistance on a V-time graph appears as the slope of the line. The greater the resistance, the shallower the slope will be. This is because resistance restricts the flow of current, resulting in a smaller change in voltage over time.

4. How can I calculate resistance from a V-time graph?

To calculate resistance from a V-time graph, you can use Ohm's Law: R = V/I. Measure the change in voltage (V) and current (I) over a given time interval and plug them into the formula to calculate resistance.

5. Can a V-time graph show different types of resistance?

Yes, a V-time graph can show different types of resistance. A linear graph with a constant slope indicates a fixed resistance, while a curved graph with varying slopes indicates a changing resistance. Additionally, the shape of the graph can also indicate whether the resistance is ohmic (linear) or non-ohmic (curved).

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