- #1
jolly_math
- 51
- 5
- Homework Statement
- In the diagram below, the switch S has been open for a long time. It is then suddenly closed. TakeEe = 10.0 V, R1 = 50.0 kΩ, R2 = 100 kΩ, and C = 10.0 µF. Let the switch be closed at t = 5.0 s. Determine the current in the switch as a function of time.
- Relevant Equations
- I(t) = -I/RC * e^(-t/RC)
After the switch is closed, current flows clockwise from the battery to resistor R1 and down through the switch.
I don't understand the reasoning for the following: the current from the capacitor flows counterclockwise and down through the switch to resistor R2. How do I determine the direction of current when a capacitor is charging? Thank you.