In the following circuit,find I_1 and I_2. | vedclass

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(D) The circuit consists of a $10 \, V$ battery,a $2 \, k\Omega$ resistor,a $14 \, k\Omega$ resistor,a $12 \, k\Omega$ resistor,and two diodes. Let th

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$0, 5 \, mA$

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(D) The circuit consists of a $10 \, V$ battery,a $2 \, k\Omega$ resistor,a $14 \, k\Omega$ resistor,a $12 \, k\Omega$ resistor,and two diodes. Let the node between the $2 \, k\Omega$ resistor and the $14 \, k\Omega$ resistor be $C$. The current $I_1$ flows through the $14 \, k\Omega$ resistor to ground (node $D$). The diode in series with the $12 \, k\Omega$ resistor is forward-biased,while the diode connected across the $12 \, k\Omega$ resistor (between nodes $A$ and $B$) is also forward-biased. Since the diode between $A$ and $B$ is forward-biased,it acts as a short circuit (assuming an ideal diode),effectively making the potential difference across $AB$ equal to $0 \, V$. Consequently,the $12 \, k\Omega$ resistor is shorted out,and no current flows through it. Since the diode between $C$ and $A$ is forward-biased,the potential at $C$ is equal to the potential at $A$,which is $0 \, V$ relative to ground. Thus,the voltage across the $14 \, k\Omega$ resistor is $0 \, V$,so $I_1 = 0 \, A$. The total current from the battery is $I = \frac{10 \, V}{2 \, k\Omega} = 5 \, mA$. Since $I_1 = 0$,all this current must flow through the forward-biased diode branch,so $I_2 = 5 \, mA$.

In the following circuit,find $I_1$ and $I_2$.

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