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. `2 amp` from `A` to `B`
C. `2 amp` from `B` to `A`
D. None of the above

Answer

B

Explanation

(b) The given circuit can be redrawn
`E_(eq) = (E_(1) R_(2) + E_(2) R_(1))/(R_(1) + R_(2)) = (2 xx 4 + 2 xx 4)/(4 + 4) = 2 V` and
`R_(eq) = (4)/(2) 2 Omega`
current `i= (2 + 2)/(2) = 2A` from `A` to `B` through `E_(2)`.