In: Physics
See the diagram :
let i current flows through 5 ohm resistor.
and ii through 4 ohm resistor.
We will apply Kirchoff's junction law at B ,
current (i - i1) flows through 2 ohm resistor.
Now, applying Kirchoff's loop law in the loop ABEFA :
-20 - 4i1 +14 + 2(i - i1) = 0
=> 2i - 6i1 = 6
=> i - 3i1 = 3 ----- (i)
Now applying Kirchoff's loop law in the loop BCDEB :
5i - 36 + 4i1 = 0
=> 5i + 4i1 = 36 ------(ii)
Solving the equations (i) and (ii),
we get,
i = 120/19 A
i1 = 21/19 A
also, i - i1 = 99/19 A
Hence, current through 2 ohm resistor = 99/19 A.
current through 4 ohm resistor = 21/19 A.
current through 5 ohm resistor = 120/19 A.