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(A) Let $i$ be the current in the diode and $V$ be the voltage drop across it. According to Kirchhoff's voltage law for the given circuit:$i 	imes 10

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$0.01$

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(A) Let $i$ be the current in the diode and $V$ be the voltage drop across it. According to Kirchhoff's voltage law for the given circuit:$i 	imes 100 + V = 8$Rearranging the equation to express $i$ in terms of $V$:$i = -\frac{1}{100}V + \frac{8}{100}$$i = -(0.01)V + 0.08$Comparing this with the equation of a straight line $y = mx + c$,where $y = i$ and $x = V$,the slope $m$ is given by $-\frac{1}{R}$.The magnitude of the slope of the load line is $|m| = \frac{1}{R} = \frac{1}{100} = 0.01 	ext{ } \Omega^{-1}$.

$A$ source voltage of $8V$ drives the diode in the figure through a current-limiting resistor of $100 \Omega$. Then the magnitude of the slope of the load line on the $V-I$ characteristics of the diode is

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