Please include all steps/reasoning.
How does the current from the battery compare when the switch is open (and has been open for at least five minutes) and when the switch is closed (and has been closed for five minutes) in each of the circuits below? Explain.
Please see attached file for full problem description.
1. a) When the switch is open, we have 6 kOhms in parallel with 4 kOhms, which is (remember "product over sum" for parallel resistors or series capacitors) 24/10 = 2.4 kOhms. So the battery current is 9V/2.4kOhms = 3.75 mA.
When the switch is closed, we have 4 kOhms in parallel with 1 kOhms, which is 4/5 = 0.8 kOhms, and we have 2 kOhms in parallel with 3 kOhms, which is 6/5 = 1.2 kOhms. These two parallel resistances are in series with each other, so the total resistance is 2.0 kOhms. So the battery current is 9V/2kOhms = 4.5 mA.
The battery current is less when the switch is open. There are no capacitors in the circuit so we do not need to worry about how long the switch has been open or closed - in theory everything would change infinitely fast when the switch is ...
Solution computes the battery current for switch open and switch closed conditions for each of the given circuits, and explains how the conclusions are drawn.