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(A) The circuit is a Wheatstone bridge. The condition for no current to flow through the $10\,\Omega$ resistor (the galvanometer arm) is that the brid

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$10$

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(A) The circuit is a Wheatstone bridge. The condition for no current to flow through the $10\,\Omega$ resistor (the galvanometer arm) is that the bridge must be balanced.For a balanced Wheatstone bridge,the ratio of resistances in opposite arms must be equal:$\frac{R}{3} = \frac{4}{6}$$\Rightarrow R = 3 	imes \frac{4}{6} = 2\,\Omega$When the bridge is balanced,the $10\,\Omega$ resistor can be removed from the circuit.Now,the circuit consists of two parallel branches:Branch $1$: $(R + 4)\,\Omega = (2 + 4)\,\Omega = 6\,\Omega$Branch $2$: $(3 + 6)\,\Omega = 9\,\Omega$The equivalent resistance $R_{	ext{eq}}$ of these two parallel branches is:$R_{	ext{eq}} = \frac{6 	imes 9}{6 + 9} = \frac{54}{15} = 3.6\,\Omega$The total current $I$ measured by the ammeter is:$I = \frac{V}{R_{	ext{eq}}} = \frac{36}{3.6} = 10\,A$

Current measured by the ammeter $(A)$ in the given circuit when no current flows through the $10\,\Omega$ resistance is . . . . . . $A$.

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