- #1

Jahnavi

- 848

- 102

## Homework Statement

## Homework Equations

## 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 ?