For the circuit given below, find the current (in A) in the `1 Omega` resistor ? Assume the batteries are ideal.

Answer

2.48

Explanation

The circuit given in the problem can be redrawn as shown below.

The three branches are connected between the same points P and Q and hence they are in parallel. For such a circuit the potential difference between the points P and Q is:
`V_(QP) = (E_(1)//E_(2) + E_(2)//R_(2) + E_(3)//R_(3))/(1//R_(1) + 1//R_(2) + 1//R_(3))`
Where `E_(1), E_(2)` and `E_3` are the emf of the batteries present in the first, second and the third branch. Similarly, `R_(1), R_(2)` and `R_3` are the values of resistance in each of the respective branches.
`V_(QP) = (6//3 + 0//1 + 9//5)/(1//3 + 1//1 + 1//5) = 57/23` V
The value of current through the `1 Omega` resistance is:
`i = (V_(QP))/1 = 57/23 A = 2.48 A`