Four ammeters with identical internal resista | vedclass

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(B) Let the current through $A_1$ be $I_1 = 3\ A$ and through $A_2$ be $I_2 = 5\ A$. By Kirchhoff's Current Law $(KCL)$ at the junction between $A_1,

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$3$

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(B) Let the current through $A_1$ be $I_1 = 3\ A$ and through $A_2$ be $I_2 = 5\ A$. By Kirchhoff's Current Law $(KCL)$ at the junction between $A_1, A_2, A_3$, the current through $A_3$ is $I_3 = I_1 + I_2 = 3\ A + 5\ A = 8\ A$ (flowing upwards).At the junction between $A_3, A_4, R$, the current through $A_4$ is $I_4 = I_2 + I_3 = 5\ A + 8\ A = 13\ A$.Let the potential at the left node be $V_0$. The potential after $A_2$ is $V_0 - I_2 r = V_0 - 5r$.The potential after $A_1$ is $V_0 - I_1 r = V_0 - 3r$.The potential difference across $A_3$ is $(V_0 - 3r) - (V_0 - 5r) = 2r$. Since $I_3 = 8\ A$, the resistance of $A_3$ is $r = (2r) / 8 = r/4$, but the problem states all ammeters have resistance $r$. Re-evaluating: The potential at the node after $A_1$ is $V_A = V_0 - 3r$. The potential at the node after $A_2$ is $V_B = V_0 - 5r$. The current $I_3$ flows from $V_A$ to $V_B$ through $A_3$, so $I_3 = (V_A - V_B) / r = (V_0 - 3r - (V_0 - 5r)) / r = 2r / r = 2\ A$.Now, at the node after $A_1$, the current entering is $I_1 = 3\ A$ and $I_3 = 2\ A$ leaves towards $A_3$, so the current through $R$ is $I_R = I_1 - I_3 = 3\ A - 2\ A = 1\ A$.The potential at the node after $A_1$ is $V_A = V_0 - 3r$. The potential at the right end of $R$ is $V_0 - 5r$ (since $A_4$ is in series with the source). Thus, $I_R R = V_A - V_{right} = (V_0 - 3r) - (V_0 - 5r) = 2r$.Since $I_R = 1\ A$, we have $1 	imes R = 2r$, so $R/r = 2$. Wait, checking the provided solution logic: $V_0 - 5r - 7r$ is not standard. Based on the circuit, $R/r = 2$ is the correct physical derivation.

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