00:01
In this question, this is the rough circuit that was given in the textbook before the question.
00:08
Now, what we want to find over here is the emf, given in the battery over here, such that the current passing through the 5 -volt battery is 2m.
00:21
We want this current to be 2ms.
00:26
And the way we're going to do that is we're going to first simplify the very disgusting parallel portion.
00:32
Which is over here, the parallel part of the circuit.
00:39
We can simplify that by considering these two 60 -oom circuits first, being parallel with each other.
00:48
So being the same resistance, two of this in parallel will just divide the overall resistance by 2.
00:55
So we can simplify this into just one resistor of resistance 60 over 2, which is 30 -ooms.
01:04
So this will now be 30.
01:12
And the 10 om and 20 oms are basically in series with each other.
01:17
So we can simplify this into a single resistor of 30.
01:29
Now we have 3 of the 30 oms in parallel with each other.
01:35
So basically we just take 30 divide by 3.
01:38
And our overall circuits.
01:42
Now becomes much cleaner here with just one 10 oms right and see we see that the 5 oms 10 omen and the 5 ome over here right this one two three all in series with each other so we can actually combine them all right and we'll combine them further to give us a single 20 oom resistor.
02:25
So this is a much simpler circuit that we can work with for our kerchief's loop rule and our junction rule which we are needed to find what is the emf of our battery over here.
02:39
So we're going to use two different loops.
02:44
But first of all we need to label the current passing through.
02:49
We're going to assume that the current through here will be in this direction we call this i1 and the other current passing through here called this i2 we're going to have two loops one loop we're going to consider the loop inside right bypassing the 15 oom resistor and then coming out bypassing the 5 volt and to the 20 oom so this is our first loop second loop we're going to go below this loop over here and come out so this will be our second loop this will bypass our 20 -oom resistor this 20 -oom through our current i1 from the junction rule looking at this particular junction the incoming current will be 2 amps and the outgoing current is just i1 and i2 so this our junction rule equation now looking at the loop rows.
04:07
For the first loop, the inner loop, we start off from the 5 volt battery, so from minus to plus, so we add the emf, it's 5.
04:19
We bypass the 20 -o resistor in the same direction, this is our current.
04:25
So we'll subtract the potential drop, which is 20 times 2.
04:29
We bypass the 10 volts battery from plus to minus, so we will have to subtract that emf.
04:38
We bypass the inner, the unknown emf from plus to minus again, so we subtract the emf.
04:47
Bypass the 15 -oom resistor in the same direction as i2, so we subtract 15 i2.
04:55
Create everything to 0.
04:59
This is our first loop.
05:01
Now going past the second loop outside, the first few are the same.
05:08
Because the external circuit is the same.
05:11
Now passing through the 20 om, we bypass the same.
05:16
Direction is the current, so we subtract the potential drop...