In the circuit shown in the adjoining figure, | vedclass

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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 res

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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} \, \Omega^{-1}$$R_p = \frac{12}{10} = 1.2 \, \Omega$Now,this equivalent resistance is in series with the internal resistance $r = 0.3 \, \Omega$.The total resistance of the circuit is $R_{total} = R_p + r = 1.2 \, \Omega + 0.3 \, \Omega = 1.5 \, \Omega$.The electromotive force $(EMF)$ of the battery is $E = 1.5 \, V$.The reading of the ammeter is the total current $I$ flowing through the circuit:$I = \frac{E}{R_{total}} = \frac{1.5 \, V}{1.5 \, \Omega} = 1 \, A$.

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

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