Solve RC Circuit Problem: Find Time of Current > 50mA

In summary, if you remove a 120 microfarad capacitor from a camera flash unit, the voltage will be 210 V and the current across your chest will exceed the 50 mA danger level for .184 seconds.
  • #1
kcombs
5
0

Homework Statement


Suppose a 120 micro-farad capacitor from a camera flash unit retains voltage of 210 V when an unwary student removes it from the camera. If the student accidentally touches the two terminals with his hands, and if the resistance of his body between his hands is 1.8 kilo-ohms, for how long will the current across his chest exceed the danger level of 50 mA?

Homework Equations


q(t)=Qe^(-t/(RC))
i(t)=(-Q/(RC))e^(-t/(RC))
V=q/C
i=(dq)/(dt)

The Attempt at a Solution


I'm not sure where the voltage comes into this question. I assume I don't just ignore that piece of given information. I wrote out all the equation, just like above, and I'm not sure how to put them together to find the correct time.

Thanks in advance!
 
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  • #2
What's the initial current?
 
  • #3
gneill said:
What's the initial current?

No value is given.
 
  • #4
kcombs said:
No value is given.
So you will have to calculate it from what is given...
 
  • #5
gneill said:
So you will have to calculate it from what is given...
Okay. I can calculate that, but how does voltage come into the picture?
 
  • #6
kcombs said:
Okay. I can calculate that, but how does voltage come into the picture?
What is the initial voltage across the load resistance?
 
  • #7
gneill said:
What is the initial voltage across the load resistance?
Actually, I'm not sure it's correct but this is what I did:
q=VC
i(t)=(V/R)e^(-t/(RC)) which, due to t=0, simplifies to i(t)=V/R
When I plugged in numbers I got .117 A
 
  • #8
kcombs said:
Actually, I'm not sure it's correct but this is what I did:
q=VC
i(t)=(V/R)e^(-t/(RC)) which, due to t=0, simplifies to i(t)=V/R
When I plugged in numbers I got .117 A
Yes, the initial current is indeed given by V/R , and your equation for i(t), namely,

i(t)=(V/R)e^(-t/(RC))

looks good.

So now you have an equation for i(t); can you solve for the time when i(t) = 50 mA?
 
  • #9
gneill said:
Yes, the initial current is indeed given by V/R , and your equation for i(t), namely,

i(t)=(V/R)e^(-t/(RC))

looks good.

So now you have an equation for i(t); can you solve for the time when i(t) = 50 mA?

I just got t=.184 s, which is the correct answer. Thanks for you help! I was definitely overthinking that!
 
  • #10
Well done. Glad I could help!
 

What is an RC circuit?

An RC circuit is a circuit that contains both a resistor (R) and a capacitor (C). The capacitor stores energy in the form of an electric field, while the resistor controls the flow of current in the circuit.

How do you solve an RC circuit problem?

To solve an RC circuit problem, you need to use Kirchhoff's laws and the equations for resistors and capacitors in series and parallel. You also need to know the initial conditions of the circuit, such as the voltage across the capacitor and the current flowing through the circuit.

What is the time constant in an RC circuit?

The time constant in an RC circuit is a measure of how quickly the capacitor charges or discharges. It is calculated by multiplying the resistance (R) and the capacitance (C) in the circuit.

What is the significance of finding the time when the current is greater than 50mA in an RC circuit?

In an RC circuit, the time when the current reaches a certain value (in this case, greater than 50mA) is important because it helps us understand the behavior of the circuit and how quickly it charges or discharges. It can also be used to determine the time it takes for a circuit to reach a certain state.

Can you provide an example of solving an RC circuit problem to find the time when the current is greater than 50mA?

For example, if we have an RC circuit with a resistance of 100 ohms and a capacitance of 10 microfarads, we can use the formula for the time constant (RC) to calculate that it takes 1 millisecond for the capacitor to charge or discharge. From there, we can use Kirchhoff's laws and the equations for resistors and capacitors in series and parallel to find the time when the current reaches 50mA.

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