In the circuit shown in figure given below and find the current recorded by the ammeter (A).

Explanation

Since resistance, `AC = 30 Omega` and resistances `CB = 30 Omega`. Resistance in the path `ACB = 30 Omega + 30 Omega = 60 Omega`.
Further, `60 Omega` (in the path ACB) and `20 Omega` (in the path AB) are in parallel.
Hence, effective resistance (R) between A and B is given by
`1/R = 1/60 + 1/20 + (1 +3)/60 = 4/60 = 1/15`
`:. R =15 Omega`
`:.` Current `I = (7.5V)/(15 Omega) =0.5 A`
This is the current recorded by the ammeter.