Find the transient current in this circuit

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The discussion focuses on analyzing a circuit with a capacitor and resistors to determine the transient currents I1, I2, and I3 at various states: immediately after the switch is closed, after sufficient time has passed, and immediately after the switch is opened. Initially, the capacitor acts as a short circuit, resulting in I1 being 1 mA and I2 being 0 mA. After sufficient time, the capacitor behaves like an open circuit, leading to I3 being 0 mA and I2 equaling I1. When the switch is opened, the charged capacitor acts like a battery, with the voltage across it being equal to the voltage across the parallel resistor, affecting the currents accordingly. The final analysis confirms the relationships between the currents and the behavior of the capacitor throughout the circuit's operation.
  • #31
Helly123 said:
Maybe you mean the bottow is positive, so current go from bottom (+) to upper (-) , counterclockwise
No, the top plate of the capacitor is connected to the positive terminal of the battery according to your diagram. So the top plate is at higher potential than the bottom plate. Current flows in the direction of I2 in your drawing and opposite to I3.
 
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  • #32
kuruman said:
No, the top plate of the capacitor is connected to the positive terminal of the battery according to your diagram. So the top plate is at higher potential than the bottom plate. Current flows in the direction of I2 in your drawing and opposite to I3.
Ok. Thanks sir. You save me
 

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