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Discharging RC circuit

  • Thread starter menco
  • Start date
  • #1
43
0

Homework Statement


When the cehicl stop's in the team's pit stop, consider the time as being t=0. If the potential difference between the car and ground is 30kV, the car-ground capacitance is 500pF and the resistance of each tyre is 100Gohm, determine the time it takes for the car to discharge through the tires so that it is below the critical energy point of 50mJ


Homework Equations


Q=CV
T=RC
U=1/2 CV^2
Q=CV_in exp^-t/RC

The Attempt at a Solution


Q=CV
=(500x10^-12)(300x10^3)
=15x10^-6 C

T=RC
=(100x10^9)(500x10^-12)
=50s

U=1/2 CV^2
(50x10^-3)=1/2(500x10^-12)v^2
v=14142.1 v

Q=CV_in exp^-t/RC
t=-37.6s
 

Answers and Replies

  • #2
ehild
Homework Helper
15,477
1,854

Homework Statement


When the cehicl stop's in the team's pit stop, consider the time as being t=0. If the potential difference between the car and ground is 30kV, the car-ground capacitance is 500pF and the resistance of each tyre is 100Gohm, determine the time it takes for the car to discharge through the tires so that it is below the critical energy point of 50mJ


Homework Equations


Q=CV
T=RC
U=1/2 CV^2
Q=CV_in exp^-t/RC

The Attempt at a Solution


Q=CV
=(500x10^-12)(300x10^3)
=15x10^-6 C

T=RC
=(100x10^9)(500x10^-12)
=50s

U=1/2 CV^2
(50x10^-3)=1/2(500x10^-12)v^2
v=14142.1 v

Q=CV_in exp^-t/RC
t=-37.6s
The resistance of each tire is 100 Gohm. Hpw many tyres the car has? How are their resistance connected between the car and ground?

You are serious saying that the time needed for the discharge is negative???


ehild
 
  • #3
43
0
Yeah I wasnt sure if I should factor in the 4 tyres as I wasn't sure if maybe it just takes the path of least resistance through one tyre. Every time i calculate I get a negative value for time, I know it's not suppose to be so thats why I am lost

100/4 = 25Gohm

So T=RC
=(25x10^9)(500x10^-12)
=12.5s

Therefore Q=CV_in exp^-t/RC
t = (-12.5) ln ((15x10^-6)/(500x10^-12)(14142.5))
=-9.4s
 
  • #4
ehild
Homework Helper
15,477
1,854
Therefore Q=CV_in exp^-t/RC
t = (-12.5) ln ((15x10^-6)/(500x10^-12)(14142.5))
=-9.4s
Q is the charge at t, which corresponds to the given minimum energy, when the voltage is 14142 V. You substituted the initial charge.

ehild
 
  • #5
43
0
I think I see now so Q=Q_max exp^-t/RC

Q_max = (500x10^-12)(30x10^3) = 15x10^-6

Q = (500x10^-12)(14142.5) = 7x10^-6

t=(-12.5) ln ((7x10-6)/(15x10^-6))
= 9.5s
 
  • #6
43
0
I think I see now so Q=Q_max exp^-t/RC

Q_max = (500x10^-12)(30x10^3) = 15x10^-6

Q = (500x10^-12)(14142.5) = 7x10^-6

t=(-12.5) ln ((7x10-6)/(15x10^-6))
= 9.5s
 
  • #7
ehild
Homework Helper
15,477
1,854
You can also see that as Q=CV, Q(t)=Qine-t/(RC)
V(t)C=VinCe-t/(RC).
The voltage also decreases exponentiallywith time:
V(t)=Vine-t/(RC), so can get t form the initial and final voltage :

14142 = 30000e-t/12.5,

If you calculate t from this equation, the result does not contain the rounding errors you made when calculating the charges. Your t is off by 0.1.

ehild
 
  • #8
43
0
Thanks a lot for your help!
 

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