Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 29 - Molecules and Solids - Problems - Page 855: 29

Answer

$7.5\;\rm mA$

Work Step by Step

a) We know that $$V_B=V_D+V_R$$ where $B$ is for battery, $D$ for diode, and $R$ for resistor. So, $$2=V_D+IR$$ $$2=V_D+180I $$ Solving for $I$; $$I=\dfrac{2-V_D}{180}$$ Now we need to draw the graph of $I$ as a function of $V_D$, as we see below. We have here two intersection points of our line, the first with $y$-axis at $(0\;\rm V,11.11\; mA)$ and the other with the original curve of figure 29-30. As we see below, we draw a small part of the original graph in red just to show the intersection point with it. This point is approximately at $0.65\;\rm V$. Plugging into (1); $$I=\dfrac{2-0.65}{180}=\color{red}{\bf 7.5}\;\rm mA$$
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