As per the given circuit,the value of current | vedclass

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(C) $1$. Analyze the diodes: In the circuit,diode $D_1$ is reverse-biased,so it acts as an open circuit. Diode $D_2$ is forward-biased,and diode $D_3$

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$2$

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(C) $1$. Analyze the diodes: In the circuit,diode $D_1$ is reverse-biased,so it acts as an open circuit. Diode $D_2$ is forward-biased,and diode $D_3$ is also forward-biased.$2$. Simplify the circuit: The branch with $D_1$ is inactive. The branches with $D_2$ and $D_3$ are in parallel. The effective resistance of the parallel combination of the two $6 \; \Omega$ resistors is $R_p = (6 	imes 6) / (6 + 6) = 3 \; \Omega$.$3$. Calculate total resistance: This $3 \; \Omega$ equivalent resistance is in series with the $2 \; \Omega$ resistor. Thus,$R_{	ext{total}} = 3 \; \Omega + 2 \; \Omega = 5 \; \Omega$.$4$. Calculate current: Using Ohm's law,$I = V / R_{	ext{total}} = 10 \; V / 5 \; \Omega = 2 \; A$.

As per the given circuit,the value of current through the battery will be $\dots \; A$.

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