Hi everyone,
Please see the attached file for reference to a question I've recently been given as part of a test.
The question involves a capacitor charging from a supply at the changeover of a switch and I would normally find this kind of question very easy - you just use the formula VC = VS . 1- e-t/rc and transpose if you need any of the other variables (taking natural logs of both sides if necessary)
However, this particular question has confused me because you've got a capacitor charding via a fixed potential divider circuit. With the values given, this means that the capacitor can only charge to a maximum of 9V, because when fully charged the capacitor will be an open circuit and at that time the output of the potential divider will be 9V... right?
The first part of the question asks what the initial current will be. Well, if we're assuming that the capacitor starts off completely discharged, then initially it will be a short circuit so the current would be the supply voltage (18V) divided by the resistance of R1. Negligable current would flow through R2 due to the short circuit of the capacitor.
And this is where I start becoming confused. As the capacitor charges up, it will become more and more resistive. But then it's in parallel with R2, and I was unsure how this would effect the formula I originally quoted. The rest of the question goes on to ask how long it'd take to charge to a particular voltage, what voltage would be across the cap after a certain period of time etc but I'm guessing you can't use that formula anymore? If you can, what do you put for R in the equation?!?!?!
If only he had left out R2, I'd have flown that question.
Brian