In the circuit shown in figure neglecting source resistance the ammeter reading is (in amp)

The voltage $V_L$  and  $V_C$ are equal and opposite so voltmeter reading will be zero. 
So   $i=\frac{V}{\sqrt{R^2+{(X_L-X_C)}^2}}=\frac{V}{R}=\frac{240}{30}=8A$