How Does Opening a Switch Affect a Circuit with a Capacitor and Resistors?

[jrive]

The problem stated that the switch had been previously closed for a while (at rest). So, the cap has been charged to 2V (the same as the source), and there is no more current flowing through the cap.

Then, the switch was opened. The circuit became a "source free" circuit analysis problem, and if you recall, you wrote the equation for i(t) flowing through the circuit at this time using KVL ("the sum of all the voltage drops equals the sum of all the voltage rises in the loop"). Ie.

1/c∫idt +v(0)-i(R+R1)=0.

(notice how all the terms in the equation above a in units of volts, term 1= volts across capacitor, v(0)=initial voltage across capacitor, and term 3=voltage drop across resistors --always good to verify units are consistent in any equation).

You then solved this differential equation which gave you the result in the form of i(t)=K*e^-t/((R1+R)C).

At this point, you have an equation that describes the flow of current in the circuit from the time the switch is open to eternity(while the switch remains open) This is all you need.


From the initial conditions, you can solve for R as I stated in a previous post, since you know the voltage stored in the cap (and the voltage in a capacitor cannot change instantaneously) and you know the value of R1, and the measured current in the loop, 1 amp.

Then, from i(t)=K*e^-t/((R1+R)C)
(you know the value of K (1amp),and now the value of R as computed from the initial conditions), you can solve or C at t=2usecs.

 

[pghaffari]

OK that makes a lot more sense

SO since voltage across capacitor is 2V

We can set this equation for after short is open1/c integral (0 to infinity) of i(t) = 2when i integrate the first one.. I stil have 2 unknow variables, C and R1. how do I get rid of one of them?

i don't know the value of R1

EDIT:
is R1 just 0 since there is NO drop across it?? when C is fullycharged to 2v?

so if i set R1 to 0 in that integration i would get C = 1/2?

 

[rude man]

jrive has summed it up very nicely and I'm handing this over to him.

 

[jrive]

there's nothing to integrate...you have the equation, i(t)=Ke^-t/((R+R1)C).

at t=0, i(t)=K; K=1amp. But, 1 Amp flows because there are 2V stored on the cap and you have a discharge path through the 2 resistors. So,

2V/(R+1ohm)=1, solve for R.

Now,
i(t)=i(t)=1e^-t/((R+R1)C).

the only unknown is C.

You know that the voltage across a resistor is given by V=IR. Since i(t) flows through the resitor R1, mulitply i(t)*R1, equate it to 1/e and solve for C at t=2usec.

 

[pghaffari]

I understand what you did but i still would appreciate it if you could explain afew things to me.

What exactly HAPPENS to this circuit when the switch is open? Also what does it mean to have a discharge path.

So basically I understand this:

Since the circuit has been at rest while switch is closed. Current is constant and the capacitor is fully charged to 2V. (I don't understand why i = 0 , shouldn't it still be a constant value ?)

Then once the switch opens.. what exactly happens to the circuit. Can you explain to me what a circuit behaves like when a switch is open/closed. What causes current to flow after a switch is open (didn't we say current was 0 right before switch was open?)Thank you for everything.

I ended up getting R1 = 1 and C = 1uF

 

[jrive]

When the switch closes, the capacitor charges though the 1ohm resistor. After sometime (as determined bythis equation Vc=Vin(1-e^(-t/(RC))) it is fully charged and since Vc=2V just like the source, current stops flowing through it. Resistor R, though, still conducts the current from the source to ground. However, there is no more current flowing through R1 and C.

When the switch opens, the voltage source in the circuit is now the capacitor, initially at 2V (since that is what it was charged to). So, the capacitor dicharges through the resistors to ground -- that means, current flows from the top terminal of the capacitor, charged to 2V with respect to the bottom terminal, through the resistors to the bottom terminal of the capacitor completing the loop.

As the current flows, the voltage across the capacitor discharges or decays. You can see that relationship when you apply vc=1/c ∫(-i*dt). The negative sign is just convention, since the current is flowing "out" of the capacitor. Remember that you had found i(t) as i(t)=Vinitial/R*e^-t/(RC)), so, this intregral becomes Vc=Vinitial*e^-t/(RC). So, as t gets bigger, the voltage across the capacitor decreases,eventually discharging the cap to 0V (0 V across the capacitor). No more current flows.

If you close the switch again, current begins to flow in two directions. 1 path is through R to ground. The other current path is through the 1ohm resistor and through the capacitor to ground charging the capacitor until the voltage across it matches the source. At that point, the capacitor is fully charged and current stops flowing through the capacitor. just as before, though, current is always flowing through resistor R while the switch is closed, sourced by the 2V voltage source.

 

[pghaffari]

When the switch closes, the capacitor charges though the 1ohm resistor. After sometime (as determined bythis equation Vc=Vin(1-e^(-t/(RC))) it is fully charged and since Vc=2V just like the source, current stops flowing through it. Resistor R, though, still conducts the current from the source to ground. However, there is no more current flowing through R1 and C.

When the switch opens, the voltage source in the circuit is now the capacitor, initially at 2V (since that is what it was charged to). So, the capacitor dicharges through the resistors to ground -- that means, current flows from the top terminal of the capacitor, charged to 2V with respect to the bottom terminal, through the resistors to the bottom terminal of the capacitor completing the loop.

As the current flows, the voltage across the capacitor discharges or decays. You can see that relationship when you apply vc=1/c ∫(-i*dt). The negative sign is just convention, since the current is flowing "out" of the capacitor. Remember that you had found i(t) as i(t)=Vinitial/R*e^-t/(RC)), so, this intregral becomes Vc=Vinitial*e^-t/(RC). So, as t gets bigger, the voltage across the capacitor decreases,eventually discharging the cap to 0V (0 V across the capacitor). No more current flows.

If you close the switch again, current begins to flow in two directions. 1 path is through R to ground. The other current path is through the 1ohm resistor and through the capacitor to ground charging the capacitor until the voltage across it matches the source. At that point, the capacitor is fully charged and current stops flowing through the capacitor. just as before, though, current is always flowing through resistor R while the switch is closed, sourced by the 2V voltage source.

thank you for that and your time! it's clear to me now. (: