Current at Resistor 2: How to Calculate using Capacitor Voltage and Ohm's Law

[Tiago3434]

Homework Statement

In Fig. 27-66, R1= 10.0 kΩ, R2 = 15.0 kΩ, C = 0.400 µF, and the ideal battery has emf E = 20.0 V. First, the switch is closed a long time so that the steady state is reached. Then the switch is opened at time t = 0. What is the current in resistor 2 at t = 4.00 ms?

This is question 8 of the following pdf if you need the picture (I don´t have a clue about how to include a picture in the question)
http://www.phys.ufl.edu/~majewski/2049/solns/hw5/hw5_solutions.pdf
It´s also question 65 of chapter 27 of Halliday

{Schematic added by mentor}

Homework Equations

The Attempt at a Solution

I understand the solution the way it´s done through the voltage drop across the capacitor being the same as through resistor 2, but I don´t get why couldn´t i just find the current in the circuit at time t=4ms, given that all the charge in one of the plates passes through resistor 2? I know this method right here is wrong, because it yields a different answer, but could someone tell me why?

 

[Isaac0427]

given that all the charge in one of the plates passes through resistor 2?

How do you know that?

 

[Tiago3434]

Isaac0427 Well, if the we disconnect the key, won´t all the charge built up in one of the plates be forced to pass through resistor 2?

 

[Isaac0427]

Isaac0427 Well, if the we disconnect the key, won´t all the charge built up in one of the plates be forced to pass through resistor 2?

Yes, eventually. But not in 4.00 ms! You do not know how much charge passes through the plate at 4.00 ms (until you get the voltage). But even then it would only give you average current, but you want instantaneous current.

 

[Abhishek kumar]

Homework Statement

In Fig. 27-66, R1= 10.0 kΩ, R2 = 15.0 kΩ, C = 0.400 µF, and the ideal battery has emf E = 20.0 V. First, the switch is closed a long time so that the steady state is reached. Then the switch is opened at time t = 0. What is the current in resistor 2 at t = 4.00 ms?

This is question 8 of the following pdf if you need the picture (I don´t have a clue about how to include a picture in the question)
http://www.phys.ufl.edu/~majewski/2049/solns/hw5/hw5_solutions.pdf
It´s also question 65 of chapter 27 of Halliday
View attachment 215421{Schematic added by mentor}

Homework Equations

The Attempt at a Solution

I understand the solution the way it´s done through the voltage drop across the capacitor being the same as through resistor 2, but I don´t get why couldn´t i just find the current in the circuit at time t=4ms, given that all the charge in one of the plates passes through resistor 2? I know this method right here is wrong, because it yields a different answer, but could someone tell me why?

What exactly u can't understand in this solution?

 

[gneill]

I understand the solution the way it´s done through the voltage drop across the capacitor being the same as through resistor 2, but I don´t get why couldn´t i just find the current in the circuit at time t=4ms, given that all the charge in one of the plates passes through resistor 2? I know this method right here is wrong, because it yields a different answer, but could someone tell me why?

Perhaps you could elaborate your attempted solution with the math associated with it so that we can see exactly what it is you've tried?

 

[Tiago3434]

Wouldn´t the expression i=dq/dt=-(q/RC)e^-t/RC give me the instantaneous current? With q being the equilibrium charge

 

[Isaac0427]

Wouldn´t the expression i=dq/dt=-(q/RC)e^-t/RC give me the instantaneous current? With q being the equilibrium charge

That’s exactly what the resource you showed did.

 

[Tiago3434]

The source calculated the voltage drop across the capacitor as a function of time and then used ohm´s law. I proposed going directly to the current using the above expression.

 

[Isaac0427]

The source calculated the voltage drop across the capacitor as a function of time and then used ohm´s law. I proposed going directly to the current using the above expression.

Combine the two steps that the source took. Remember that V=q/C.