Calculating time constant in RC Circuit


by thatgirlyouknow
Tags: circuit, constant, time
thatgirlyouknow
thatgirlyouknow is offline
#1
Feb25-08, 05:29 PM
P: 58
1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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?
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kdv
kdv is offline
#2
Feb25-08, 05:39 PM
P: 329
Quote Quote by thatgirlyouknow View Post
1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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
[tex] 13 = 15 e^{-0.02 s/RC} [/tex]
thatgirlyouknow
thatgirlyouknow is offline
#3
Feb25-08, 08:13 PM
P: 58
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?

kdv
kdv is offline
#4
Feb26-08, 08:43 AM
P: 329

Calculating time constant in RC Circuit


Quote Quote by thatgirlyouknow View Post
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.


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