How Do You Solve Complex RC Circuit Problems?

[horsedeg]

Homework Statement

In the figure below, suppose the switch has been closed for a time interval sufficiently long for the capacitor to become fully charged. (Assume R1 = 11.0 kΩ, R2 = 22.0 kΩ, R3 = 4.00 kΩ, and C = 11.0 μF.)

(a) Find the steady-state current in each resistor.
(b) Find the charge Qmax on the capacitor.
(c) The switch is now opened at t = 0. Write an equation for the current in R2 as a function of time. (Use the following as necessary: t. Do not enter units in your answers. Assume the current is in microamperes, and t is in seconds.)
(d) Find the time interval required for the charge on the capacitor to fall to one-fifth its initial value.

Homework Equations

V=IR
Q=CV

The Attempt at a Solution

Found (a) and (b) pretty easily. I just have trouble with understanding part (c). I'm sure part (d) will be even easier if I can find this equation.

So when the switch opens, there is no voltage source anymore, and that entire left branch pretty much disappears, no? The capacitor starts discharging through R2 and then R3 slowly. The current will stop later and then charge will just be distributed throughout that area, I think.

I don't really know how to find the formula, though. For one I'm not sure where to start. Also, there seems to be a generic formula that let's you find current over time using I₀ and the time constant RC. However, do you have to derive that formula every time for each specific situation? Because that's what it feels like, even though every answer seems to just use the same exact one.

Also, I apologize for making two threads in a row, these are just both things I have a hard time understanding.

 

[ehild]

Homework Statement

In the figure below, suppose the switch has been closed for a time interval sufficiently long for the capacitor to become fully charged. (Assume R1 = 11.0 kΩ, R2 = 22.0 kΩ, R3 = 4.00 kΩ, and C = 11.0 μF.)

(a) Find the steady-state current in each resistor.
(b) Find the charge Qmax on the capacitor.
(c) The switch is now opened at t = 0. Write an equation for the current in R2 as a function of time. (Use the following as necessary: t. Do not enter units in your answers. Assume the current is in microamperes, and t is in seconds.)
(d) Find the time interval required for the charge on the capacitor to fall to one-fifth its initial value.

Homework Equations

V=IR
Q=CV

The Attempt at a Solution

Found (a) and (b) pretty easily. I just have trouble with understanding part (c). I'm sure part (d) will be even easier if I can find this equation.

So when the switch opens, there is no voltage source anymore, and that entire left branch pretty much disappears, no? The capacitor starts discharging through R2 and then R3 slowly. The current will stop later and then charge will just be distributed throughout that area, I think.

I don't really know how to find the formula, though. For one I'm not sure where to start. Also, there seems to be a generic formula that let's you find current over time using I₀ and the time constant RC. However, do you have to derive that formula every time for each specific situation? Because that's what it feels like, even though every answer seems to just use the same exact one.

Also, I apologize for making two threads in a row, these are just both things I have a hard time understanding.

You are right, when the switch opens, you have only a charged capacitor connected to two resistors in series. The resistors can be replaced with a single one, and then you have the problem "capacitor discharging through resistor", you know the solution of. You can refer to that solution, or derive it again, as you like.

 

[horsedeg]

You are right, when the switch opens, you have only a charged capacitor connected to two resistors in series. the resistors can be replaced with a simple one, and then you have the problem "capacitor discharging through resistor", you know the solution of. You can refer to that solution, or derive it again, as you like.

Is there ever a time where it's not safe to use the same solution? Highly dependent on the situation I guess?

 

[ehild]

Is there ever a time where it's not safe to use the same solution? Highly dependent on the situation I guess?

If you can transform the circuit to a single capacitor and single resistor connected, the solution is the same