RC Circuits: Measuring Capacitance

Click For Summary
The discussion revolves around a misunderstanding of how to measure capacitance in RC circuits. The original poster incorrectly assumed that the slope of the T versus R graph directly indicates capacitance. However, the correct approach involves measuring the time constant τ, which is related to capacitance through the formula τ = RC. Instead, the poster measured T_{1/2}, which requires a different derivation to relate it to R and C. Clarifying this distinction is essential for accurate capacitance measurement in RC circuits.
amwil
Messages
3
Reaction score
0
Homework Statement
To measure the capacitance C of a capacitor, you attach the capacitor to a battery and wait until it is fully charged. You then disconnect the capacitor from the battery and let it discharge through a resistor of resistance R. You measure the time T1/2 that it takes the voltage across the resistor to decrease to half its initial value at the instant that the connection to the capacitor is first completed. You repeat this for several different resistors. You plot the data as T1/2 versus R and find that they lie close to a straight line that has slope 4.00 μF.
What is the capacitance C of the capacitor?
Relevant Equations
T=RC
I thought that if the slope of the T versus R graph is 4.00μF, then the capacitance would be 4.00μF but it says that's wrong. I know I'm doing something wrong but I'm not sure how to approach this problem. Can someone help?
 
Physics news on Phys.org
If you have measured the time constant ##\tau = RC## and plotted ##\tau## vs. R, you would indeed find the slope to correspond to ##C##, but you didn't measure ##\tau##. You measured ##T_{1/2}##. Try deriving an expression for ##T_{1/2}## in terms of ##R## and ##C##.
 
  • Like
Likes malawi_glenn and DaveE
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
19
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K