Find the ammeter reading in the circuit

Answer

``4A`

Explanation

The resistance of `(3 + 7) Omega , 5 Omega and 10 Omega` are in parallel between point `C and D`. Their equivalent resistance `R'` is given by
`(1)/(R ) = (1)/(10) + (1)/(5) + (1)/(10) = (4)/(10) = (2)/(5) or R' = (5)/(2) Omega `
`R'` is in series with `1 Omega` resistance
Total resistance `R = (5)/(2) + 1 = (7)/(2) Omega`
Current `I = (epsilon)/(R ) = (14)/(7//2) = 4 A`