You have shown a battery as your voltage source, yet you have labelled it '8.5V AC'. This is a misnomer. A battery supplies a DC voltage. If you are refering to DC then your question is valid and the current would reach 63% of its final value (where final value=8.5/1500, ie. 5.7mA), in the period of one time constant.Originally posted by guitar333@Nov 23 2004, 09:18 PM
I am studying for a final but I am stuck. Please help me solve this and tell me what formulas to use so I can do the other questions similar to this one. I am stuck.....
I made it in paint BTW.
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by Jake Hertz