D_1 and D_2 are two ideal diodes. Find the va | vedclass

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(B) In the given circuit,the $12\,V$ battery is connected such that its positive terminal is connected to the cathode side of diode $D_2$ and the anod

Vedclass · Accepted Answer

$2.4\,A, 0\,A$

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(B) In the given circuit,the $12\,V$ battery is connected such that its positive terminal is connected to the cathode side of diode $D_2$ and the anode side of diode $D_1$.For diode $D_1$,the anode is at a higher potential than the cathode,so $D_1$ is forward-biased and acts as a short circuit (zero resistance).For diode $D_2$,the anode is at a lower potential than the cathode,so $D_2$ is reverse-biased and acts as an open circuit (infinite resistance).Since $D_2$ is reverse-biased,no current flows through the branch containing $D_2$,so $i_2 = 0\,A$.The circuit simplifies to a series combination of the $12\,V$ battery,the $1\,\Omega$ resistor,and the $4\,\Omega$ resistor (since $D_1$ is ideal).The total resistance in the circuit is $R_{eq} = 1\,\Omega + 4\,\Omega = 5\,\Omega$.The current $i_1$ is given by Ohm's law: $i_1 = \frac{V}{R_{eq}} = \frac{12\,V}{5\,\Omega} = 2.4\,A$.Thus,$i_1 = 2.4\,A$ and $i_2 = 0\,A$.

$D_1$ and $D_2$ are two ideal diodes. Find the values of $i_1$ and $i_2$.

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