A part of a circuit is shown in the figure. A | vedclass

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(B) Let the potential at the node before $A_1$ be $V_1$ and after the $3 \ \Omega$ resistor be $V_2$. The wire connecting the node before $A_1$ and th

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reading of $A_3$ is $3 \ A$

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(B) Let the potential at the node before $A_1$ be $V_1$ and after the $3 \ \Omega$ resistor be $V_2$. The wire connecting the node before $A_1$ and the node after the $3 \ \Omega$ resistor is a short circuit,so $V_1 = V_2 = 0 \ V$ (taking it as reference).The current through the $6 \ \Omega$ resistor is given as $1.0 \ A$ (reading of $A_1$).The potential drop across the $6 \ \Omega$ resistor is $V = I 	imes R = 1.0 \ A 	imes 6 \ \Omega = 6 \ V$.Since the potential at the start is $0 \ V$,the potential at the node between the $6 \ \Omega$ and $3 \ \Omega$ resistors is $-6 \ V$.Now,consider the $3 \ \Omega$ resistor. The potential difference across it is $V_{diff} = 0 \ V - (-6 \ V) = 6 \ V$.The current through the $3 \ \Omega$ resistor is $I = V / R = 6 \ V / 3 \ \Omega = 2 \ A$.Applying Kirchhoff's Current Law at the node between the $6 \ \Omega$ and $3 \ \Omega$ resistors,the current entering from the $6 \ \Omega$ resistor is $1 \ A$,and the current leaving through the $3 \ \Omega$ resistor is $2 \ A$. Thus,the current entering from the $A_3$ branch must be $I_{A3} = 2 \ A - 1 \ A = 1 \ A$. Wait,looking at the circuit,the current $1 \ A$ flows through $6 \ \Omega$. At the junction,it splits. The current through $3 \ \Omega$ is $2 \ A$. By $KCL$,the current through $A_3$ must be $I_{A3} = I_{3\Omega} + I_{6\Omega} = 2 \ A + 1 \ A = 3 \ A$.Therefore,the reading of $A_3$ is $3 \ A$.

$A$ part of a circuit is shown in the figure. All the ammeters are ideal. If the reading of ammeter $A_1$ is $1.0 \ A$,then:

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