Current of RC Circuit After Switch is Opened

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Homework Help Overview

The discussion revolves around an RC circuit problem involving a capacitor and a switch that has been closed for an extended period. Participants are exploring the behavior of the circuit when the switch is opened at time t=0, specifically focusing on the charge on the capacitor and the initial current flow.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to determine the initial current flow after the switch is opened, questioning whether this current is solely due to the battery or influenced by the capacitor's charge. There is also a discussion about the implications of the switch being closed for a long time and its effect on current flow.

Discussion Status

Some participants have provided insights into the interpretation of the problem, emphasizing the role of the capacitor's residual charge after the switch is opened. There is an ongoing exploration of the initial conditions and the flow of current in the circuit.

Contextual Notes

Participants note that the switch has been closed for a long time, which may imply certain assumptions about the circuit's behavior prior to t=0. There is also mention of the circuit diagram being referenced, which is not visible in the discussion.

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Homework Statement


See attached image for circuit diagram and givens.
The switch has been closed for a very long time.
a) What is the charge on the capacitor?
b) The switch is opened at t=0 s. What current initially flows?

Homework Equations


V = IR
Q=CV
I=I0e-t/(RC)

The Attempt at a Solution


Since the switch has been open for a long time, no current flows through the 10 Ohm resistor. Therefore,
I = 100V/(60 Ohm + 40 Ohm) = 1 Amp, V = 1Amp*40 Ohm, and Q = 2uF * 40V = 80uC.

So my question is: Is the current that initially flows after the switch opens at t=0 s just due to the battery? So it would just be 1Amp? I'm having trouble understanding how to find the current at part b.
 

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Hello MUU, :welcome:

First step is to establish where the current flows. Through an open switch ? Or elsewhere ?
 
Hello @Math-U-Up. Welcome to PF!

Math-U-Up said:

Homework Statement


See attached image for circuit diagram and givens.
The switch has been closed for a very long time.
a) What is the charge on the capacitor?
b) The switch is opened at t=0 s. What current initially flows?

Homework Equations


V = IR
Q=CV
I=I0e-t/(RC)

The Attempt at a Solution


Since the switch has been open for a long time, no current flows through the 10 Ohm resistor. Therefore,
I = 100V/(60 Ohm + 40 Ohm) = 1 Amp, V = 1Amp*40 Ohm, and Q = 2uF * 40V = 80uC.

So my question is: Is the current that initially flows after the switch opens at t=0 s just due to the battery? So it would just be 1Amp? I'm having trouble understanding how to find the current at part b.
I think you might have misread the problem statement. According to the wording in the problem statement, initially, the switch has been closed for a very long time. That means that current has been flowing through the battery for a very long time.

At time t = 0, the switch is opened, meaning that after t = 0 no current flows through the battery, and any current flowing in the circuit is the result of whatever residual charge is left in the capacitor.

At least that's the way that I interpret the problem.

(A "closed" switch means that the switch can conduct current. An "open" switch does not conduct current.)
 
The idea in PF is that the thread poster discovers that after a small nudge from a helper ...
 

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