00:01
Now in this question, let first calculate the equivalent resistance.
00:06
So here, r1 is given as 2 oom, this is 4 om, this is 6 oom, this is 5 om and this is 8 ome.
00:19
So these two are parallel, r1 and r2.
00:22
So one can write 1 by r -daz equals to 1 by 2 plus 1 by 4.
00:29
So, r -dice will be equal to 8x or 4 by 3.
00:35
So one can delete this and assume the equivalent resistance of 1 .33.
00:46
Similarly, one can also calculate the equivalent resistance between r4 and r3 and let's say it is represented by r -daband -dash and in that case it will be 1 by 6 plus 1 by 5 whole inverse and it will be equal to 30 divided by 11.
01:15
So after doing this we will get an equivalent resistance.
01:20
Let's suppose an equivalent resistance of r1, r2, r3, r5 represented by this one will be equal to 12 om.
01:36
So one can simplify that this is the battery of 9 volt and equivalent resistance will be equals to approximately 12 om.
01:46
So from here, one can calculate the current drawn that will be equals to v by r.
01:52
9 by 12 that is equal to 0 .75 amper.
01:58
So this is the amount of current drawn into the circuit.
02:04
So one can here first write 12 om and current is 0 .75.
02:14
Now coming to this and r2, their equivalent resistance was 4 byte 3.
02:25
So we can write i, i1 here will be equals to v by r.
02:40
Now i1 is here .75, so one can write .75 into r.
02:49
Now r equals to 4 by 3, so one can write 3 by 4 into 4 .3.
02:56
3 by 4 is 0 .75 and r is 4 by 3.
03:00
This will cancel out and v equals to 1 om.
03:04
So from here one can calculate that v1, if the potential difference between this v1, that is, let's say, this is the v1 across the potential difference across the resistance r1, so, v1 will be equals to 1 volt.
03:29
Similarly, this r1 and r2 are parallel connected.
03:34
So in parallel connection, the voltage remains same.
03:37
So, v2 will be also 1 volt.
03:39
That means the voltage difference width across this r2 resistance will be also equals to 1 volt and current drawn is equal to.
03:51
Now in this case, let's say that current drawn is i1 from r1 and i from r2.
03:59
So we can write since voltage is same i1 into r1 equals to i2 into r2 so i1 into r1 is 2 oom and r2 is i2 so this gives i1 equals to 2 i2 now we know that some of the current i1 plus i2 must be equals to 0 .75 because current once coming in this direction then again reunited with each other...