For the given circuit,the reading of the amme | vedclass

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(C) To find the current through the ammeter,we can simplify the circuit using Kirchhoff's laws or by simplifying the network.Let the potential at the

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(C) To find the current through the ammeter,we can simplify the circuit using Kirchhoff's laws or by simplifying the network.Let the potential at the node between the $2 \ \Omega$ resistor and the ammeter be $V_x$,and the potential at the other side of the ammeter be $V_y$. However,a simpler approach is to use nodal analysis.Let the potential at the junction to the left of the ammeter be $V_1$ and to the right be $V_2$.Applying Kirchhoff's Current Law $(KCL)$ at the junction to the left of the ammeter:$\frac{V_1 - 2}{2} + \frac{V_1 - V_2}{2} = 0 \implies 2V_1 - V_2 = 2$Applying $KCL$ at the junction to the right of the ammeter:$\frac{V_2 - V_1}{2} + \frac{V_2 - 2}{2} = 0 \implies 2V_2 - V_1 = 2$Solving these two equations:$V_1 = V_2 = 2 \ V$.Since the potential difference across the ammeter is $V_1 - V_2 = 0$,the current $I$ through the ammeter is $0 \ A$. Wait,re-evaluating the circuit diagram: The two $2 \ V$ cells are connected in parallel with their respective $2 \ \Omega$ internal resistances. The equivalent $EMF$ $E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{2/2 + 2/2}{1/2 + 1/2} = 2 \ V$. The equivalent internal resistance $r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 	imes 2}{2 + 2} = 1 \ \Omega$.The total resistance in the loop is $R_{total} = r_{eq} + 2 \ \Omega = 1 + 2 = 3 \ \Omega$.The current $I = \frac{E_{eq}}{R_{total}} = \frac{2}{3} \ A$.

For the given circuit,the reading of the ammeter in amperes is........

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