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(D) Let the nodes be $P$ (left junction of $A_1, A_2$),$Q$ (middle junction),and $S$ (right junction). Let the potential at $P$ be $V_P$ and at $Q$ be

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$8$

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(D) Let the nodes be $P$ (left junction of $A_1, A_2$),$Q$ (middle junction),and $S$ (right junction). Let the potential at $P$ be $V_P$ and at $Q$ be $V_Q$. Since $A_1$ and $A_2$ have identical resistance $r$,the current through them is determined by the potential difference across them. Let $I_1 = 3 \, A$ and $I_2 = 5 \, A$. $V_P - V_Q = I_1 \cdot r = 3r$ (across $A_1$) and $V_P - V_Q = I_2 \cdot r = 5r$ (across $A_2$). This implies the circuit is oriented such that $A_2$ is at a higher potential. The current through the middle ammeter $A_3$ (resistance $r$) is $I_3 = \frac{V_P - V_Q}{r} = \frac{5r - 3r}{r} = 2 \, A$ flowing from the $A_2$ branch to the $A_1$ branch. Applying Kirchhoff's Current Law at node $Q$: The total current entering the right part of the circuit is $I_1 + I_2 = 3 + 5 = 8 \, A$. Since $A_4$ is in series with the right branch,the current through $A_4$ is the sum of currents from the two branches,which is $8 \, A$.

Four ammeters with identical internal resistance $r$ and a resistor $R$ are connected to a current source as shown. If the readings of $A_1$ and $A_2$ are $3 \, A$ and $5 \, A$ respectively,then the reading of $A_4$ is ............. $A$.

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