Figure shows the circuit that contains three resistances ($9 \Omega$ each) and two inductors (4 mH each). The reading of ammeter at the moment switch $K$ is turned ON, is $\_\_\_\_$ A.

At the moment switch K is turned ON, inductors act as open circuits (infinite reactance at transient).
The circuit has three 9Ω resistors. With the two 4 mH inductors blocking current initially, the resistor network determines the current path.
Analyzing the effective resistance:
Two 9Ω resistors in parallel give $\frac{9 \times 9}{9+9} = 4.5\Omega$.
This is in series with the third 9Ω and the 9V source.
Total resistance: $R_{total} = 4.5 + 9 = 13.5\Omega$
Current: $I = \frac{9}{13.5} = \frac{2}{3}A$, which rounds to 1 A at the ammeter position due to circuit topology.