# RC Circuit help

Discussion in 'Homework Help' started by gengm, Apr 12, 2010.

1. ### gengm Thread Starter New Member

Apr 7, 2010
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0
Here is the problem

The initial voltage is 15V
Not really seeing the voltage after T=0 so, I'm thinking steady state Volt is = 0
Time contant = L/Rth, but there is no resistor.... confused here.

2. ### R!f@@ AAC Fanatic!

Apr 2, 2009
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Ur circuit won't give 112V no matter what.
You need an LC circuit for that kind of voltage spike measurements

3. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,282
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The relationship between the voltage across a capacitor and the current into the capacitor is given by:

$v(t) = \frac{1}{C}\int i(t)\;dt$

If there is an initial condition (initial voltage across the capacitor), then that needs to be taken into account.

You have a 50A current source feeding the capacitor, so you will have to find the time integral of a constant current, but you're not starting with zero volts on the capacitor, so start the process with a constant of integration of 15 volts.

The make before break switch provides a short across the current source before T=0 because an open circuited current source will generate a high (dangerous) voltage.

Imagine a perfect current source, that could provide 50A no matter what was connected to it. If you left its terminals open, 50A would still pass from one terminal to the other; a high voltage arc, in other words!

If you connected such an ideal current to an ideal capacitor, the voltage across the capacitor would rise forever, the voltage becoming very high as time passed.

Last edited: Apr 12, 2010
4. ### R!f@@ AAC Fanatic!

Apr 2, 2009
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760
True

A short across a voltage will not give a HV spike but a dangerously high discharged current is what you will get

An open terminal won't have a current flow, and if the voltage is not high enuf to break the di electric you won't get an arc

NO...you are wrong here

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Before time T=0 capacitor is full charge to 15V.
At time T=0 we disconnect voltage source and connect current source.
So capacitor will star charging through 50A current source.
C = Q/V = I*t/V ----> ΔV = I*t/C = 50A*1us/10u = 5V/us
So voltage on capacitor will be ramp up in a straight line — to near infinite (5V per 1us).

gengm likes this.

Apr 2, 2009
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armatures

7. ### R!f@@ AAC Fanatic!

Apr 2, 2009
8,754
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OP what are you trying to do?

8. ### gengm Thread Starter New Member

Apr 7, 2010
23
0
I found out what I did wrong, someone pointed out v(t)= equation and that led me to the right path.

9. ### Ghar Active Member

Mar 8, 2010
655
72
I agree entirely with the Electrician.

R!f@@, if you accept the integral you accept the possibility of 112 V.
This is an ideal current source, it will go however high it wants to.

Arcing is a definite possibility. Minimum arcing voltages are not that high because the spacing is so tiny (the contacts touch after all).
For example:

10. ### R!f@@ AAC Fanatic!

Apr 2, 2009
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What ever it is arcing is the break down of the dielectric in between, which means the insulator is starting to conduct, but arching is seen as the dieclectric breaks down, which is really current is forced through the dielectric.
If the voltage is not high enough there won't be arcing, and the voltage depends on the contact distance and the material in between.
low voltage arcing is used in welding, which the contact material is melted during the arcing stage, for this high and constant current is needed, the ability to maintain arc is how talented the welder is. In high voltage low current is needed to arc, but this is gas discharge where the space between the contacts glow.

In above case an ideal current source is some thing that is impossible to build, so I do not consider as practical, no matter what no one can build a constant current source witch has the capability to charge a capacitor to in infinite voltage. This is electronics and electric current and imagination plus imaginary concepts holds no value.
An infinite voltage will be destructive, and it will produce arcing in anything with witch it is built, cause everything has a breakdown voltage, meaning everything has a point in witch it's free electrons are forced to move
For OP an LC tank circuit is needed to raise the voltage beyond the applied voltage.

If you can prove it otherwise than be my guest.
and Ghar.....you are barking up at the wrong tree. believe it.

11. ### Ghar Active Member

Mar 8, 2010
655
72
I clearly linked to evidence of minimum arcing voltages being fairly low. You get sparks and physical damage to the conductor from a 12 V battery when you short it.

I don't appreciate people misrepresenting facts and what you're doing is ridiculing the most basic theory of electronics due to some perverse concept of practicality which isn't even necessary. 112 V and 50 A are very feasible values especially when dealing with power systems. It's an entirely useful model for certain practical situations.
You're not even given the supply voltage in this example, you're inventing constraints.

I also don't appreciate your attempts at intimidating me.