In the circuit shown here, `E_(1) = E_(2) = E_(3) = 2 V` and `R_(1) = R_(2) = 4 ohms`. The current flowing between point `A` and `B` through battery `E_(2)` is

Options

A. zero
B. 2amp from A to B
C. 2 amp from B to A
D. none of the above

Answer

2

Explanation

Circuit can be redrawn as
Total emf `=2+2=4V`
so `1=(4)/(2)=2A`