Current of RC Circuit After Switch is Opened

AI Thread Summary
The discussion revolves around analyzing an RC circuit after a switch is opened. Initially, the capacitor is charged, and the current flowing through the circuit is determined by the voltage and resistance values. When the switch opens at t=0, the current does not flow from the battery but instead is due to the residual charge in the capacitor. The confusion lies in understanding that an open switch prevents current flow from the battery, and the current at that moment is solely from the capacitor's discharge. Clarifying these concepts is essential for solving the problem accurately.
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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.)
 
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