All the diodes are ideal. The current flowing | vedclass

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(A) In the given circuit,the diodes $D_1$ and $D_2$ are in forward bias because their $p$-terminals are connected to the positive terminal of the batt

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$1$

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(A) In the given circuit,the diodes $D_1$ and $D_2$ are in forward bias because their $p$-terminals are connected to the positive terminal of the battery through the $1 \, \Omega$ resistor and their $n$-terminals are connected to the negative terminal.The diodes $D_3$ and $D_4$ are in reverse bias because their $p$-terminals are connected to the negative terminal of the battery,so they act as open circuits.The equivalent resistance of the upper branch (containing $D_1$ and $D_2$) is $R_{up} = 1 \, \Omega + 2 \, \Omega + 1 \, \Omega = 4 \, \Omega$.The middle branch has a resistance of $4 \, \Omega$.These two branches are in parallel,so their equivalent resistance is $R_p = \frac{4 \, \Omega 	imes 4 \, \Omega}{4 \, \Omega + 4 \, \Omega} = 2 \, \Omega$.The total resistance of the circuit including the internal $3 \, \Omega$ resistor is $R_{total} = 2 \, \Omega + 3 \, \Omega = 5 \, \Omega$.The total current from the battery is $I = \frac{V}{R_{total}} = \frac{10 \, V}{5 \, \Omega} = 2 \, A$.Since the resistance of the upper branch and the middle branch are equal ($4 \, \Omega$ each),the current $I$ divides equally between them.Therefore,the current flowing through the $2 \, \Omega$ resistor is $I_{up} = \frac{I}{2} = \frac{2 \, A}{2} = 1 \, A$.

All the diodes are ideal. The current flowing in the $2 \, \Omega$ resistor connected between the diodes $D_1$ and $D_2$ is .......... $A$

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