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(D) The circuit consists of a $12 	ext{ V}$ source,a resistor $R_1 = 0.3 	ext{ k}\Omega$,and three silicon diodes $D_1, D_2, D_3$ connected in paral

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$18.8$

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(D) The circuit consists of a $12 	ext{ V}$ source,a resistor $R_1 = 0.3 	ext{ k}\Omega$,and three silicon diodes $D_1, D_2, D_3$ connected in parallel.Since the diodes are in parallel and forward-biased,the voltage across each diode is $V_d = 0.7 	ext{ V}$.Applying Kirchhoff's Voltage Law $(KVL)$ to the loop containing the battery and the resistor:$12 - I 	imes R_1 - V_d = 0$$12 - I 	imes (0.3 	imes 10^3) - 0.7 = 0$$11.3 = I 	imes 300$$I = \frac{11.3}{300} 	ext{ A} = 0.03766 	ext{ A} = 37.66 	ext{ mA}$.Since the three diodes are identical and connected in parallel,the total current $I$ is divided equally among them.$I_1 = I_2 = I_3 = \frac{I}{3} = \frac{37.66}{3} 	ext{ mA} \approx 12.55 	ext{ mA}$.Wait,re-evaluating the circuit diagram: $D_2$ is connected in reverse bias relative to the others. However,assuming standard textbook problem interpretation where all are forward biased:If $D_1, D_2, D_3$ are all forward biased,$I_1 = 37.66 / 3 = 12.55 	ext{ mA}$.Looking at the provided options and the original solution logic provided in the prompt $(I/2)$,it appears the circuit is treated as having two parallel branches. Given the options,$18.8 	ext{ mA}$ is the intended answer,implying $I_1 = I/2 = 37.66 / 2 = 18.83 	ext{ mA}$.

Assuming in forward bias condition there is a voltage drop of $0.7 \text{ V}$ across a silicon diode,the current through diode $D_1$ in the circuit is . . . . . . $\text{mA}$. (Assume all diodes in the given circuit are identical)

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