Consider the circuit in the attachment.
If the switch has been open for a long time, then the potential at point A is 0V and the current is 0A. When the switch is closed, the potential at point A is Vs and current starts to flow.
So instead of the Vs voltage source and the switch, I can use a voltage source that is 0V before t=0, and then at t=0 it becomes Vs. So the input voltage is :
v = Vs*u(t), where u(t) is the step input voltage
And thus I can analize this circuit with the Laplace method or other methods of solving differential equations.
Now, my problem.
If the switch has been closed for a long time, then the potential at point A is Vs and the current is maximum. When the switch is opened, the potential at point A is not ground. So instead of the Vs voltage source and the switch, I cannot use a voltage source that is Vs before t=0, and then at t=0 it becomes 0V:
v=-Vs*u(t)+Vs
So, how can I analyze this circuit?
If the switch has been open for a long time, then the potential at point A is 0V and the current is 0A. When the switch is closed, the potential at point A is Vs and current starts to flow.
So instead of the Vs voltage source and the switch, I can use a voltage source that is 0V before t=0, and then at t=0 it becomes Vs. So the input voltage is :
v = Vs*u(t), where u(t) is the step input voltage
And thus I can analize this circuit with the Laplace method or other methods of solving differential equations.
Now, my problem.
If the switch has been closed for a long time, then the potential at point A is Vs and the current is maximum. When the switch is opened, the potential at point A is not ground. So instead of the Vs voltage source and the switch, I cannot use a voltage source that is Vs before t=0, and then at t=0 it becomes 0V:
v=-Vs*u(t)+Vs
So, how can I analyze this circuit?
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