00:01
Hi there, so for this problem, we have the circuit that is shown in here.
00:08
We are given the nph associated with this battery that is equal to six poles and no internal resistance.
00:18
Now, when the double -throw switch is open, the current in the battery is one micro -ampers.
00:27
When the cich is closed, at position a, the current is...
00:31
And the current in the battery is 1 .20 micropers and when the switch is closed at position b, the current in the battery is 2 microampers.
00:41
So we need to find the values for the resistance, 1, 2, and 3.
00:51
Okay, so in order to solve this, we are going to consider each case.
00:58
So first of all, when the switch s is opened, we will have the following.
01:10
So we note that then the resistor 1, the resistor 2, and the resistor 3 are in series.
01:23
As you can see from the figure.
01:26
And then we will have the following expression that the sum of the resistance, the resistance 1 plus the resistance 2 plus the resistance 3 is equal to the battery, which is, six balls divided by the current.
01:45
In this case, we are told that that is one micro -ampers.
01:50
So that will be 10 to the minus 3 -ampers.
01:54
So from this, we obtain a value of 6 kilo -mbs.
02:02
Now, for the situation, when s is close in position a, then we will have that.
02:19
We will have a parallel combination of the two resistors two, parallel.
02:29
And this is in series with the resistor 1 and the resistor 3.
02:41
And of course the battery.
02:44
So we will have that this is the resistor, the resistance 1 plus 1 divided by 2 times that resistors 2, which is the equivalent capacitance for that parallel, and this plus the resistance 3.
03:00
Now, this again is 6 poles divided by now the current, we are told that it is 1 .2 times 10 to the minus 3.
03:10
Ampers...