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Question

(A) At the knee point,the voltage across the diode is $V_D = 0.7 \, V$.Applying Kirchhoff's Voltage Law $(KVL)$ to the circuit,we have $V_B = I \cdot

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$4.3 \, k\Omega$

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(A) At the knee point,the voltage across the diode is $V_D = 0.7 \, V$.Applying Kirchhoff's Voltage Law $(KVL)$ to the circuit,we have $V_B = I \cdot R + V_D$.Given $V_B = 5 \, V$,$V_D = 0.7 \, V$,and the minimum current $I = 1 \, mA = 1 	imes 10^{-3} \, A$.The voltage across the resistor $R$ is $V_R = V_B - V_D = 5 \, V - 0.7 \, V = 4.3 \, V$.Using Ohm's law,$V_R = I \cdot R$,we get $4.3 \, V = (1 	imes 10^{-3} \, A) \cdot R$.Therefore,$R = \frac{4.3}{1 	imes 10^{-3}} \, \Omega = 4300 \, \Omega = 4.3 \, k\Omega$.Thus,the maximum value of $R$ is $4.3 \, k\Omega$.

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