In the circuit shown in (fig. 3.51), find the current recorded by ammeter (A).
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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, effective resistance between `A and B` is given
by `(1)/( R)=(1)/(60)+(1)/(20)=(1 + 3)/(60)=(4)/(60)=(1)/(15)` or `R = 15 Omega`
Current, `I = (7.5 V)/(15 Omega) = 0.5 A`.