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

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(C) $1$. Analyze the circuit: The circuit consists of a $10 \,V$ source,a $2 \,k\Omega$ resistor in series,and two parallel branches. One branch conta

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$5 \,mA, 0$

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(C) $1$. Analyze the circuit: The circuit consists of a $10 \,V$ source,a $2 \,k\Omega$ resistor in series,and two parallel branches. One branch contains a $14 \,k\Omega$ resistor (with current $I_1$),and the other contains a diode in series with a $12 \,k\Omega$ resistor and another diode in parallel.$2$. Check diode biasing: The diode in series with the $12 \,k\Omega$ resistor is forward-biased because the positive terminal of the $10 \,V$ source is connected to its anode. The diode in parallel with the $12 \,k\Omega$ resistor is also forward-biased.$3$. Simplify the circuit: Since the second diode is in parallel with the $12 \,k\Omega$ resistor,it effectively shorts the $12 \,k\Omega$ resistor if we assume ideal diodes (zero resistance in forward bias). Thus,the current $I_2$ through the branch containing the $12 \,k\Omega$ resistor becomes $0$ because the current prefers the path of least resistance (the shorted diode).$4$. Calculate $I_1$: The circuit simplifies to the $10 \,V$ source in series with the $2 \,k\Omega$ resistor and the $14 \,k\Omega$ resistor. The total resistance is $R_{eq} = 2 \,k\Omega + 14 \,k\Omega = 16 \,k\Omega$.$5$. The current $I_1$ is given by $I_1 = V / R_{eq} = 10 \,V / 16 \,k\Omega = 0.625 \,mA$. $6$. Re-evaluating the circuit: If the diodes are not ideal,the parallel diode clamps the voltage. However,based on standard textbook problems of this type,the diode in parallel with the $12 \,k\Omega$ resistor acts as a short circuit,making $I_2 = 0$. Given the options,$I_1$ is calculated as $10 \,V / (2 \,k\Omega + 14 \,k\Omega)$ is not matching. Let's re-examine: If the first diode is forward biased and the second is also forward biased,the second diode shorts the $12 \,k\Omega$ resistor. The current $I_1$ flows through the $14 \,k\Omega$ resistor. The total resistance is $2 \,k\Omega + 14 \,k\Omega = 16 \,k\Omega$. $I_1 = 10/16 = 0.625 \,mA$. Since this is not an option,let's assume the $14 \,k\Omega$ resistor is in parallel with the series combination of the diode and $12 \,k\Omega$ resistor. If the second diode is reverse biased,$I_2 = 0$. If the first diode is forward biased,$I_1 = 10 \,V / (2 \,k\Omega + 14 \,k\Omega) = 0.625 \,mA$. Given the options,the most logical conclusion is that the circuit implies $I_2 = 0$ and $I_1$ is calculated differently or there is a typo in the question values. However,choosing the best fit,$I_2 = 0$ is certain.

List-$I$ (Defects)	List-$II$ (Shown by)
$a$. Frenkel defect	$i$. Non-ionic solids and density of the solid decreases
$b$. Schottky defect	$ii$. Non-ionic solids and density of the solid increases
$c$. Vacancy defect	$iii$. Ionic solids and density of the solid decreases
$d$. Interstitial defect	$iv$. Ionic solids and density of the solid remains constant

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

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