00:01
For this question, we are looking at a capacitor being attached in series, a resistor, and an emf source.
00:13
So it's quite a simple setup.
00:15
We are given all of the values such as the emf, the resistance, as well as the capacitance.
00:27
What you want to find is power dissipated by our resistor, the energy being stored by our capacitor and also the power dissipated by the source right at three different points in time so the first point in time will be just when the circuit is formed right t equals to zero when the current just started to flow number two is when t is equal to infinite so after a very long time and finally the third point in time is when the charge is equals to half of its final value.
01:20
I'm going to start off with the first one, number one when t goes to zero and just started the entire circuit.
01:31
So the capacitor can be ignored, right, can assume that it's just a piece of wire.
01:36
There's no charge stored in it yet or anything so it can just ignore and therefore the power dissipated by our resistor it's just using p equals to v square over r you can take the entire emf of the battery to be the potential drop across the resistor to give us to 460 watts now for how much the capacitor is actually charging, right, the rate of charge.
02:27
We're going to use the equation u equals to q square over 2c.
02:33
And therefore the rate of charging of this electrical energy is du dt.
02:44
D differentiating this, the only one that is dependent on time is q.
02:48
Right, c is a constant so we can keep that.
02:50
Differentiating q, q square, we get 2 q, dqdt.
02:58
You know that the qdt is just the current.
03:03
So this is q times i over c.
03:09
Now because there is no charge, that is stored in the capacitor yet, so q is 0, and therefore this entire thing is 0...