Calculating time constant in RC Circuit

AI Thread Summary
The discussion centers on calculating the time constant of an RC circuit using oscilloscope data. The voltage at peak (15 V) is noted, but confusion arises regarding the correct time reference for measurements. Participants clarify that the time should start at the peak voltage, not at 20 ms, as V cannot be the same at both points. The exponential decay formula is emphasized, and it is suggested to measure voltage at a different time to find the time constant accurately. Properly resetting the time reference is crucial for solving the equation correctly.
thatgirlyouknow
Messages
57
Reaction score
0

Homework Statement



You see this trace on your oscilloscope for an RC circuit. The vertical scale is 5volt/square and the horizontal scale is 50ms/square. What is the time constant of the RC circuit?

Homework Equations



Vc(t) = Vo(e^(-t/RC)
Vr(t) = -Vo * e^(-t/RC)
t halftime = Tln(2) = .69T

The Attempt at a Solution



So when the voltage is at its peak (15 V), that's Vo. and if I take that as V(t), it's V at 20 ms (approximately). Substituting this:

15 V = 15V(e^(.02/RC))
1 = e^-.02/RC
ln(1) = -.02/RC
0 = -.02/RC

Cross multiplying to solve here doesn't help me. Is there another formula I can use?
 

Attachments

  • graph.jpg
    graph.jpg
    18 KB · Views: 1,098
Physics news on Phys.org
thatgirlyouknow said:

Homework Statement



You see this trace on your oscilloscope for an RC circuit. The vertical scale is 5volt/square and the horizontal scale is 50ms/square. What is the time constant of the RC circuit?

Homework Equations



Vc(t) = Vo(e^(-t/RC)
Vr(t) = -Vo * e^(-t/RC)
t halftime = Tln(2) = .69T

The Attempt at a Solution



So when the voltage is at its peak (15 V), that's Vo. and if I take that as V(t), it's V at 20 ms (approximately). Substituting this:

15 V = 15V(e^(.02/RC))
1 = e^-.02/RC
ln(1) = -.02/RC
0 = -.02/RC

Cross multiplying to solve here doesn't help me. Is there another formula I can use?

The left side of the equation is V(t) so you need to read off the value of V at 20 ms. For example, if it's 13 volts, then you will solve
13 = 15 e^{-0.02 s/RC}
 
Yeah, that's what I did..I said that the V at 20 ms was 15 (the peak). Should I just pick another arbitrary time?
 
thatgirlyouknow said:
Yeah, that's what I did..I said that the V at 20 ms was 15 (the peak). Should I just pick another arbitrary time?

But V cannot be equal to 15 volts at 0 second AND at 20 ms!

V_0 is the voltage at t=0

I think your mistake is that you did not reset t=0 at the point where the decreasing exponential starts. If you measure V_0 at the peak, you must set t=0 at that point.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
4
Views
1K
Replies
2
Views
2K
Replies
2
Views
2K
Replies
4
Views
2K
Replies
4
Views
2K
Replies
5
Views
7K
Back
Top