00:01
Hello students in this question we are given a circuit and we have to find current in each register so let us see the circuit so this is a circuit and here suppose we have i1 from this battery so this i1 will move here and this i from i1 here it goes i2 so in this branch it will go i1 minus i2 and then here it will be i2 then here it will be again i1 so in the first loop if we see and we apply kirchhoff's law so applying applying kirchhoff's law we get so according to kirchof's law we have 58 is equal to 120 i1 120 i1 plus 82 i1 plus 64 multiplied by i 1 minus i 2 so this will come out to be here 133 i1 minus 32 i2 here we have minus 32 i2 is equal to 29 this is our first equation now if we apply kitchoff's law in second in second part so we have applying kirchhoff's law.
01:47
Kirchof's law.
01:52
So we get, here we get, 3 is equal to 25 i2 minus 64 multiply by i1 minus i2 plus 110 i2.
02:09
So from here we get the equation minus 64 i1 plus 199 i2.
02:19
Is equal to 3 this is our second equation now on solving on solving and to simultaneously we get we get so here we get the value of i1 as 0 .24 ampiers and the value of i2 as 0 .092 amperes now if we write our answer, so we were required to find the current in each register.
03:02
If we find that current in, current in 120 oom register, so current in 120 oom register, so value of i1 is 0 .24 ampers.
03:20
Now current in, current in, so here we have current in 82 oom.
03:30
Will be again i1 so if we see current in 82 om register will be 0 .24 ampers now here current in 64 oom register current in 64 oom register it is i1 minus i2 here it will be i1 minus i2 so we write here so it is 0 .148 amperors.
04:12
Now we have, if we talk about current in 110 ome resistor, so it is i2...