How long does it take for a car to discharge through its tires in a pit stop?

In summary: I think I understand now! In summary, at the car's pit stop, it takes 9.5 seconds for the car to discharge through the 4 tyres, so the car is below the critical energy point of 50mJ.
  • #1
menco
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
 
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  • #2
menco said:

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
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 that's 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
menco said:
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
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
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
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
Thanks a lot for your help!
 

1. What is a discharging RC circuit?

A discharging RC circuit is an electronic circuit that consists of a resistor (R) and a capacitor (C) connected in series, with the capacitor initially charged to a certain voltage. When the circuit is closed, the capacitor begins to discharge through the resistor, resulting in a decrease in voltage over time.

2. How does a discharging RC circuit work?

When the circuit is closed, the capacitor starts discharging through the resistor, creating a current flow. The resistor limits this current flow, causing the voltage across the capacitor to decrease exponentially over time. This process continues until the capacitor is fully discharged, and the voltage across it becomes zero.

3. What factors affect the discharging of an RC circuit?

The time it takes for an RC circuit to discharge depends on several factors, including the capacitance of the capacitor, the resistance of the resistor, and the initial voltage across the capacitor. In general, a higher capacitance or lower resistance will result in a longer discharge time.

4. What is the formula for calculating the discharging time of an RC circuit?

The discharging time of an RC circuit can be calculated using the formula t = R x C, where t is the time in seconds, R is the resistance in ohms, and C is the capacitance in farads. This formula assumes that the capacitor is initially fully charged and that the discharging occurs through a single resistor.

5. What are the practical applications of discharging RC circuits?

Discharging RC circuits are commonly used in electronic devices as filters, oscillators, and timing circuits. They are also used in electronic flash units for cameras and as part of power supply circuits. In addition, discharging RC circuits are useful for studying the behavior of capacitors and resistors in an electrical circuit.

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