00:01
Hi there, so for this problem we have a capacitor of capacitance that is given and that capacitance is equal to six of five point five microfarats and is initially on charge.
00:29
It is connected in series with a switch of negligible resistance a resistors of resistance are that it is also given.
00:40
And that value is 10 .5 kilooms.
00:53
Now, with that set, we are also given the information that there is a battery which provides a potential difference.
01:03
That battery is bb, this one right here, and that it has a value of 70, so question a of this problem ask about to calculate the time constant for the circuit in seconds.
01:22
Now we know that the time constant of an rc circuit as in this case is given by just the product between the resistance and the capacitance.
01:35
So we just need to simply substitute those values.
01:38
So for the resistance, we have 10 .5 kilo, which means 10 to the 3 oms.
01:48
And this times the capacitance, which is 5 .5 micro, which means 10 to the minus 6, ferrets.
02:05
And now we just need to simply do this product right here, so that will be 10 .5 times 5 .5.
02:17
So from this we obtain a value of 0 .058 seconds.
02:29
So that's the solution for part a of this problem.
02:35
Now for part b, we are asked about after a very long time after the switch has been closed.
02:45
What is the voltage drop bc across the capacitor in terms of the potential b? with that said, we know that that is going to just be given by the potential that we are given, which is 75 volts.
03:10
That's because we know that initially the capacitor is uncharged.
03:17
When the switch is closed, at the first instance, the voltage drop across the resistor is equal to the voltage of the battery, while the voltage drop across the capacitor is zero.
03:29
Then the current starts to flow in the circuit, and so the charge of the capacitor increases, and as a consequence, the voltage drop across the capacitor increases as well.
03:41
Now, after a fairly long time, as in this case that we are asked about at this condition, at that time, the capacitor has been completely charged so that the voltage drop across the capacitor has become equal to the voltage of the battery.
04:00
So that's why it is 75 balls.
04:03
Now, for part c of this problem, we are asked about to calculate the charge queue on the capacitor a very long time after the switch has been closed.
04:17
Now, in that, and we know that as we said at point b, a very long time after the switch has been closed, the capacitor is fully charged.
04:30
And so its charge is just given by the product between the capacitance and the potential difference.
04:39
So that will be 5 .5 times 10 to the minus 6 pharats.
04:44
And this times the potential difference that we are given for this, which is 75 degrees, 75 balls, sorry.
04:53
And then the value that we obtained from this product is equal to 4 .125 times 10 to the minus 4 columns.
05:09
And that's, well, we can also write this in micro, and that will be 412 .5 microlums.
05:24
So that's the solution for par c of this problem.
05:31
Yes, now for part d of this problem, we are asked about to calculate the current a very long time after the switch has been closed.
05:47
We know that the answer for this is that the current is going to be zero.
05:53
Because as we said at point b, a very long time after the switch has been closed, no more current is flowing through it through the circuit because the capacitor is now fully charged.
06:06
So the potential drop across it is equal to the voltage of the battery and can store more charge.
06:14
On the contrary, the voltage drop across the resistor is now zero.
06:19
So no current can flow through the circuit.
06:23
Now, for part e of this problem, we are asked about to calculate the time after which the current through the resistor is one -third of its maximum value.
06:39
So we need to find that value.
06:42
So we start with the equation for the current in this type of circuit.
06:47
So the current depends on the time, and that will be the initial current times the exponential of minus the time divided by the time constant that we determined from before.
07:04
Now, the initial current is the maximum value that we can have for the current, so that is the potential given divided by the resistance.
07:14
So we know that that is going to be 75 balls divided by the resistance, which is 10.
07:22
5 kilo times 10 to the 3 bowls...