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(A) The circuit consists of three resistors of $2\, \Omega$,$4\, \Omega$,and $12\, \Omega$ connected in parallel.First,calculate the equivalent resist

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(A) The circuit consists of three resistors of $2\, \Omega$,$4\, \Omega$,and $12\, \Omega$ connected in parallel.First,calculate the equivalent resistance $(R_p)$ of these three resistors in parallel:$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{4} + \frac{1}{12} = \frac{6+3+1}{12} = \frac{10}{12} = \frac{5}{6} \, \Omega^{-1}$So,$R_p = \frac{6}{5} = 1.2 \, \Omega$.The total resistance of the circuit $(R_{total})$ includes the internal resistance of $0.3 \, \Omega$ in series with the parallel combination:$R_{total} = R_p + 0.3 = 1.2 + 0.3 = 1.5 \, \Omega$.The reading of the ammeter is the total current $(I)$ flowing through the circuit,given by Ohm's law:$I = \frac{V}{R_{total}} = \frac{1.5 \, V}{1.5 \, \Omega} = 1 \, A$.

In the circuit shown in the adjoining figure,the reading of ammeter $A$ is ................ $A$.

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