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

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(D) In the given circuit,the first diode is forward-biased,while the second diode (in parallel with the $12\ k\Omega$ resistor) is reverse-biased beca

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$0, 5\ mA$

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(D) In the given circuit,the first diode is forward-biased,while the second diode (in parallel with the $12\ k\Omega$ resistor) is reverse-biased because its p-terminal is at a lower potential than its n-terminal.Since the second diode is reverse-biased,it acts as an open circuit. Thus,the current $I_2$ flows through the $12\ k\Omega$ resistor.However,looking at the circuit diagram,the $14\ k\Omega$ resistor is connected in parallel with the series combination of the diode and the rest of the circuit. The diode is forward-biased. The potential at the node after the $2\ k\Omega$ resistor is determined by the voltage divider rule.Actually,the diode is forward-biased,so it acts as a short circuit (assuming an ideal diode). The $14\ k\Omega$ resistor is connected in parallel with the branch containing the diode. Since the diode is ideal and forward-biased,it effectively shorts the $14\ k\Omega$ resistor,making the potential across it $0\ V$. Therefore,$I_1 = 0$.The total resistance of the circuit is $2\ k\Omega$. The current drawn from the $10\ V$ battery is $I = \frac{10\ V}{2\ k\Omega} = 5\ mA$. Since the $14\ k\Omega$ resistor is shorted,all this current flows through the diode as $I_2$. Thus,$I_1 = 0$ and $I_2 = 5\ mA$.

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

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