00:01
So here i have labeled the resistors and the emfs with their appropriate value corresponding to that variable.
00:13
So for example, r5 is the resistance of this resistor and has a value 5 oms.
00:23
Similarly, this e10, which is the emf, has an emf of 10 volts.
00:31
And for the currents, we have the subscript is just that that's the current corresponding to that particular variable, like that particular resistor or emf.
00:47
So i6 .8 is the current flowing through the 6 .o.
00:52
Resistance resistor.
00:55
And similarly, i12 is the current flowing through the 12 -room resistor.
01:01
And i5 is the current flowing through the 5 volt emf.
01:07
You can call this i5 or i10 whichever one you want to pick.
01:13
So yeah, so now we apply the junction rule to points a and b, and we applied the loop rule to loops 1, 2, 3 that have already labeled here.
01:27
And this will give us 5 equals 5.
01:29
And this will make sense because we have five unknown currents that we need to solve for and therefore we need five equations for that.
01:38
So let's start applying the junction rule at a and b at points a and b.
01:47
So we have i5 equal to i top plus i6 for junction a and then i top plus i 6 .8 to be equal to i 12 because they have the opposite direction so i top is this way and i 6 .8 is like this so at point b these two combine to make i12 and therefore this is the relation so from here we can solve for i top so because i top is common in both of these equations so so doing that, we get i5 minus i6, which is i top equal to i12 minus i6 .8.
03:00
And we can arrange this equation like this, having all the positive signs plus signs, i6.
03:14
So this is our equation from those two junction rules.
03:20
Let's call that equation 1.
03:23
So we got this equation by solving two equations.
03:27
And now, therefore, we have eliminated the variable i -top.
03:32
So we will have four equations now because we have reduced these two into one.
03:38
So now we apply the loop rule.
03:40
So this is for...
03:45
Loop 1 remember if you are going from positive to negative inside a battery, there should be a negative sign before that.
03:56
And if you're going negative to positive in a battery, there should be a positive sign before the emf.
04:04
And for current, if you're going along the direction of current, there should be a negative sign before that.
04:09
And if you're going opposite to the direction of current then there should be a positive sign.
04:19
So those are the simple rules and i have applied them and you can check them if they are correct.
04:31
Also it doesn't really matter how you go, how your loop looks like.
04:36
For example, you can choose this loop 1 to be counterclockwise and you should be able to arrive at this equation.
04:54
So this is for loop 1.
04:57
Let's call that equation 2.
04:58
This is for loop 2 and this is for loop 3.
05:04
So let me just write that so that you don't get confused with the equation number.
05:13
Yeah, and this is the equation that we got from the junction rule.
05:18
Now we will solve these four questions by the method of substitution elimination by the method of elimination and substitution so let's use equation for because i find this equation the simplest out of these four so let's use this to solve for i6 so we will get i6 in terms of i 12 here from this equation and r12 over r6 so since these to basically represent the value that's equal to the subscript.
05:58
So we can plug that.
06:01
You can cross -check that if the resistance have actually these values.
06:05
So we have r -6 equal to twice of, sorry, we have i -6 equal to twice of i -12.
06:15
So now that we have i -6 in terms of i -12, we can eliminate wherever we have i -6 and a that way we will be able to eliminate one more variable so let's combine these two emfs by adding the voltages so we have i5 plus i6 .8 equal to so this is equation one so i have added these two so this will be equal to three i2 so let's call this equation one again because these are practice the same.
07:03
So now we plug that with to the loop one equation to equation two.
07:14
We plug this value of i6 over here.
07:17
So we will get i15 minus i5 r5 minus i 12 r12 equal to zero.
07:29
So i have added these two because e5 has a value 5 volt, e10 is 10 volts, so 5 plus 10 is 12.
07:38
And that's why i have written here to be e15.
07:43
So again, we have e12, which is e4, e4 plus e8.
07:49
And these two add up to 12.
07:51
E12 is equal to minus i12 r12 minus i6 .8...