All the diodes are ideal. The current flowing | vedclass

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(A) $1$. Analyze the circuit: The circuit consists of three parallel branches connected to a $10\,V$ source in series with a $3\,\Omega$ resistor.$2$.

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$1$

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(A) $1$. Analyze the circuit: The circuit consists of three parallel branches connected to a $10\,V$ source in series with a $3\,\Omega$ resistor.$2$. Check diode biasing: - In the middle branch,diode $D_3$ is reverse-biased because its cathode is connected to the positive terminal of the battery. Thus,no current flows through the middle branch.- In the top branch,diodes $D_1$ and $D_2$ are forward-biased. The total resistance of this branch is $1\,\Omega + 2\,\Omega + 1\,\Omega = 4\,\Omega$.$3$. Calculate equivalent resistance: The top branch $(4\,\Omega)$ and the bottom branch $(4\,\Omega)$ are in parallel. Their equivalent resistance is $R_p = \frac{4 	imes 4}{4 + 4} = 2\,\Omega$.$4$. Total circuit resistance: $R_{total} = 3\,\Omega + R_p = 3\,\Omega + 2\,\Omega = 5\,\Omega$.$5$. Total current from the battery: $I = \frac{V}{R_{total}} = \frac{10\,V}{5\,\Omega} = 2\,A$.$6$. Current in the top branch: Since the top and bottom branches have equal resistance ($4\,\Omega$ each),the current splits equally. Current in the top branch $I_{top} = \frac{I}{2} = \frac{2\,A}{2} = 1\,A$. This current flows through the $2\,\Omega$ resistor.

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

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