The Attempt at a Solution
For calculation sake I will consider R = 1Ω . Final answer will be in mA instead of A .
The middle two resistors 3R and 4R are in parallel , so I consider their equivalent resistance 12R/7 in the middle branch .
In the left part of the circuit , suppose i1 current flows in R .
In the right part of the circuit , suppose i2 current flows in 2R .
Current flowing in the middle 12R/7 resistor will be i
Using KCL i = (i1+i2)
Writing KVL in the left loop ,
12 - i1 - 12i/7 = 0
Writing KVL in the right loop ,
24 - 2i2- 12i/7 = 0
Solving for i , i = 6.72
This is the sum of the currents that will flow in the middle 3R and 4R resistors .Since they are in parallel , current will be divided in the ratio 4:3 .This means (1/7)th of the current i =6.72 flows in the middle lower branch ( the one that is asked in the question )
This gives 0.96 which is incorrect .
What is the mistake ?
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