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(A) In the given circuit,diodes $D_1$ and $D_3$ are connected in forward bias,while diode $D_2$ is connected in reverse bias.Therefore,the branch cont

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$1.5$

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(A) In the given circuit,diodes $D_1$ and $D_3$ are connected in forward bias,while diode $D_2$ is connected in reverse bias.Therefore,the branch containing $D_2$ acts as an open circuit.The circuit simplifies to two parallel branches connected to the $10\,V$ battery.The first branch consists of a $10\,\Omega$ resistor in series with $D_1$ and another $10\,\Omega$ resistor,giving a total resistance of $R_1 = 10\,\Omega + 10\,\Omega = 20\,\Omega$.The second branch consists of a $10\,\Omega$ resistor in series with $D_3$,giving a resistance of $R_2 = 10\,\Omega$.The equivalent resistance $(R_{eq})$ of these two parallel branches is given by:$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{20} + \frac{1}{10} = \frac{1+2}{20} = \frac{3}{20}$$R_{eq} = \frac{20}{3}\,\Omega$Using Ohm's law,the current $(I)$ through the battery is:$I = \frac{V}{R_{eq}} = \frac{10}{20/3} = \frac{30}{20} = 1.5\,A$.

In the given circuit,the current $(I)$ through the battery will be $..........\,A$.

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