Discharging this capacitor in an RC bridge circuit

[Karl Karlsson]

Homework Statement

See below

Relevant Equations

Kirchoffs 2 laws

The voltage source in the circuit below has been switched on for a long time when the switch S switches off. How long will it take before the current coming out of the capacitor has become less than 1 mA?

My attempt:

I am far from sure that my solution is correct. This is because i have never done any problem like this one before. I would be really happy if some smart person could review my solution.

Thanks in beforehand!

 

[BvU]

Hi, There are two parts in your exercise:
one is to determine the voltage over the capacitor at the start of unloading
two is to determine how the discharge goes in time

For both you need quantitative information, which appears to be absent. Or does the composer of the exercise want you to provide a huge formula ? If the resistors are big enough, the current never exceeds 1 mA !

relevant equations: see below is meaningless: almost everything is 'below'.

Do you know about voltage dividers and parallel circuits of resistors ?

 

[Karl Karlsson]

Hi, There are two parts in your exercise:
one is to determine the voltage over the capacitor at the start of unloading
two is to determine how the discharge goes in time

For both you need quantitative information, which appears to be absent. Or does the composer of the exercise want you to provide a huge formula ? If the resistors are big enough, the current never exceeds 1 mA !

relevant equations: see below is meaningless: almost everything is 'below'.

Do you know about voltage dividers and parallel circuits of resistors ?

Hi, I edited the thread. We are assuming that current from the capacitor at the start of discharge is at least 1mA

 

[gleem]

You could have solved it more quickly by looking at the steady state after the switch is closed. The currents through R1 , R3 and R2 , R4 become constant defining the voltage difference across the capacitor.

On the discharge The capacitor is shunted by R1+R3 in parallel with R2 + R4 .

 

[Karl Karlsson]

You could have solved it more quickly by looking at the steady state after the switch is closed. The currents through R1 , R3 and R2 , R4 become constant defining the voltage difference across the capacitor.

Interesting, how would you do that? But is my solution correct?

 

[haruspex]

how would you do that?

In steady state the capacitor charge is constant, so no current in the wires connecting it. Easy to calculate the currents through the rest of the circuit and deduce the capacitor voltage.

 

[gleem]

After the switch has been closed for some time the capacitor is fully charged and the currents only pass through the resistors. Since R1,R3 and R2,R4 are connected in parallel you can easily determine the voltages at the positions where the capacitor is connected and determine the potential difference across the capacitor which is U(R3/(R1+R3) - R4/(R2+R4)) which you determined.

After the switch is opened the capacitor will discharge through the equivalent resistance of R1 +R3 in parallel with R2+R4, call it R_eq Using the discharge equation for a capacitor I =V/R_eqe^-t/R_eqC with V = U(R3/(R1+R3) - R4/(R2+R4)) and R_eq =(R1+R3)(R2+R4)/(R1+R2+R3+R4).

Your express looks correct.

 

[Karl Karlsson]

In steady state the capacitor charge is constant, so no current in the wires connecting it. Easy to calculate the currents through the rest of the circuit and deduce the capacitor voltage.

Did not think about that, thanks!

After the switch is opened the capacitor will discharge through the equivalent resistance of R1 +R3 in parallel with R2+R4, call it R_eq Using the discharge equation for a capacitor I =V/R_eqe^-t/R_eqC with V = U(R3/(R1+R3) - R4/(R2+R4)) and R_eq =(R1+R3)(R2+R4)/(R1+R2+R3+R4).

Your express looks correct.

Hi gleem. I have only dealt with the discharge equation for series circuits. Is there any faster way to come up with the expression for that compared to what i did, some general formula for parallell circuits?

 

[gleem]

You have two things connected. They appears to be in parallel but they are also in series too. The current runs through the resistors to discharge the capacitor. The resistors are in fact in series with the capacitor.

 

[hutchphd]

I am surprised that no one has said the magic phrase "Thevenin Equivalent Circuit" here. The "ports" of the circuit are taken at the connection to the capacitor and then the circuit replaced by a R_Thevenin and V_Thevenin in series for final analysis.
I believe this provides a more general and direct method of attack, although the algebra will of course end up looking similar.

 

[BvU]

Hi, I edited the thread. We are assuming that current from the capacitor at the start of discharge is at least 1mA

That adresses one question. Did you ignore the others on purpose ?

Do you see the two voltage dividers sitting there when C is fully loaded ?
Do you see the two parallel paths when C unloads ?