In the circuit shown in the adjoining figure, | vedclass

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

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(A) The circuit consists of three resistors $2\, \Omega$,$4\, \Omega$,and $12\, \Omega$ connected in parallel.First,calculate the equivalent resistance $(R_p)$ of these three resistors:$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{4} + \frac{1}{12} = \frac{6 + 3 + 1}{12} = \frac{10}{12}$$R_p = \frac{12}{10} = 1.2\, \Omega$Now,the total resistance $(R_{total})$ of the circuit 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 total voltage $(V)$ is $1.5\, V$.Using Ohm's law,the current $(I)$ measured by the ammeter is:$I = \frac{V}{R_{total}} = \frac{1.5}{1.5} = 1\, A$

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

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