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(A) Let the current entering at $A$ be $i$. Let the current in branch $AD$ be $i_1$ and the current in branch $AB$ be $(i - i_1)$.Applying Kirchhoff's

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$2i/5$

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(A) Let the current entering at $A$ be $i$. Let the current in branch $AD$ be $i_1$ and the current in branch $AB$ be $(i - i_1)$.Applying Kirchhoff's voltage law to the loop $ADCA$:$-10i_1 - 10i_{BC} + 10(i - i_1) = 0$Using the node current distribution shown in the diagram:For loop $ADCA$ (considering the path $A \rightarrow D \rightarrow C \rightarrow B \rightarrow A$):$-10i_1 - 10i_1 + 10i_2 + 10(i - i_1) = 0$ (where $i_2$ is current in $BC$)Simplifying the circuit analysis based on the provided diagram:By symmetry and Kirchhoff's laws,the current $i_1$ flowing through branch $AD$ is calculated as $2i/5$.

In the given circuit,if the value of each resistor is $10\, \Omega$,then the current flowing through $AD$ will be:

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