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+// Find current if diode is forwar-biased

+// Basic Electronics

+// By Debashis De

+// First Edition, 2010

+// Dorling Kindersley Pvt. Ltd. India

+// Example 3-1 in page 143

+

+clear; clc; close;

+

+// Given data

+I=29.8*10^-3; // Current in mA

+V=0.208; // Voltage in V

+

+// Calculation

+I=(45-V)/(1.5*10^3);

+printf("I = %0.2e A\n",I);

+printf("For this current,V = 0.2 V\n");

+printf("(a)Therefore I = 29.8 mA\n");

+printf("(b)If battery is inserted with reverse polarity,voltage drop across the 1.5 K resistors is only 15 mV and may be neglected\n");

+printf("(c)In forward direction, I=29.8 mA\n");

+printf("In reverse direction we draw a load line from V=-30 V to I=-30 mA\n");

+y=[-30 -25 -20 -15 -10 -5 0];

+x=[-30 -25 -20 -15 -10 -5 0];

+x=-30-y;

+plot(x,y);

+xlabel('Voltage');

+ylabel('Current');

+title('Current in forward direction');

+I=-30*(20/30);

+printf("Then,I = %0.0f mA\n",I);

+printf("Current=20 mA as there is a 10 V drop");

+

+// Result

+// Graph shows current in reverse direction

+// I' = -20 mA

+// Set axis positions to 'origin' in axis properties to view the graph correctly
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diff --git a/181/CH3/EX3.1/example3_1.txt b/181/CH3/EX3.1/example3_1.txt
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+I = 2.99e-002 A

+For this current,V = 0.2 V

+(a)Therefore I = 29.8 mA

+(b)If battery is inserted with reverse polarity,voltage drop across the 1.5 K resistors is only 15 mV and may be neglected

+(c)In forward direction, I=29.8 mA

+In reverse direction we draw a load line from V=-30 V to I=-30 mA

+Then,I = -20 mA

+Current=20 mA as there is a 10 V drop 
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