00:01
In this problem we are going to calculate the current passing through the register r3, which we call i 3.
00:08
Let we have r1 equals to 3 .0, r2 equals to 4 .0, r3 equals to 12m, r4 equals to 6 .0, and r5 equals to 2 .0.
00:26
As we know that 2r4s are in parallel combination, so we can write, 1 by re1 equals to 1 by r4 plus 1 by r4.
00:36
Now writing the value for the r4 we get 1 by re1 equals to 1 divided by 6 .0 plus 1 divided by 6 .0.
00:46
So this will give us re1 equals to 3 .0.
00:50
Let's call it equation 1.
00:51
Now for series combination we can write re2 equals to re1 plus r2.
00:58
Let's write the values for this re1 and r2.
01:02
Re2 equals to 3 .0 plus 3 .0.
01:05
This is equals to 6 .0.
01:09
Let's call it equation 2.
01:12
As r2 and r3 are in parallel combination so we can write 1 by re3 equals to 1 divided by r2 plus 1 by r3.
01:20
Let's write the values for this r2 and r3.
01:23
1 by r3 equals to 1 divided by 4 .0 plus 1 divided by 12 om.
01:29
So this will give us re3 equals to 3 .0.
01:32
Let's call it equation 3.
01:35
Now for series combination we can write re4 equals to re3 plus r5.
01:41
Now setting the values of this re3 and r5 into this equation we get re4 equals to 3 .0 plus 2 .0...