00:01
Hello everyone.
00:02
In the present problem it is given that source voltage b is 12 volt.
00:07
R1 is 2 .29 om.
00:09
R2 is 4 .58 o.
00:12
R3 is 11 .4 .6 .87 o.
00:18
We have to find the current that flows through each of the four resistors.
00:24
Coming to the solution, we have since r2 r3 and r4 are in parallel the equivalent resistance.
01:01
Ra is given by 1 divided by ra is equal to 1 divided by r2 plus 1 divided by r3 plus 1 divided by r3 plus 1 divided by r4.
01:24
Substituting the values, we have 1 divided by r .a is equal to 1 divided by 4 .58 plus 1 divided by 11 .45 plus 1 divided by 6 .87.
01:44
Solving, we have 1 divided by r .a is equal to 0 .455.
01:52
110 therefore ra is equal to 2 .21680.
02:02
Further, ra and r1 are in series and current remain same in series configuration.
02:42
Therefore, current through r1 is given as i1 is equal to b divided by ra plus r1.
03:04
This is equal to 12 divided by 2 .2168 plus 2 .29.
03:16
Hence i1 is equal to 2 .66 .26 amper.
03:26
Next, voltage across ra is given as ba is equal to i1 multiplied by ra...