00:01
Hi there! so for this problem, we are told that an 8 micro -farrats capacitor, so we are given its capacitance that is equal to 8 micro -pharets, that is the same as 8 times 10 to the minus 6 pharats, and it is initially uncharged and connected in series with a 5 -a -mns resistor.
00:33
So its resistance is given and that is equal to five oms.
00:42
And an f source that is also given that is equal to 70 volts.
00:54
N negligible internal resistance.
00:57
So the question is, at what instant when the resistor is dissipating electrical energy at a rate that is given? so we are given that the power is equal to 300 watts.
01:13
How much energy has been stored in the capacitor? now, first in this problem, we know that the total nph is the sum of the nth due to the capacitor and the nph due to the resistor, the nph of the capacitor plus the nph of the resistor.
01:41
Now the emph generated by the capacitor, we know that that is just simply the charge stored in that capacitor divided by the capacitance.
01:54
And the emph stored in the resists in the resistor, we can express it in terms of the power as the square root of the part between the power and the resistance.
02:14
So with that said, the total nph for this circuit is just simply the charge divided by the capacitance plus the square root of the priority between the power and the resistance.
02:29
So from this, what we can do is to solve for the charge, so we will obtain that the charge is equal to the capacitance times the total nph minus the square root of the power times the resistance.
02:51
So we know that generally the energy stored in the capacitor that we are going to call e .c is equal to the charge square divided by two times the capacitance.
03:10
So what we are going to do now is to substitute the expression that we found for the charge.
03:15
So that will be the capacitance square times the total mf minus the square root of the power times the resistance and all of that to the square...