From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 608/CH13/EX13.22/13_22.sce | 24 ++++++++++++++++++++++++
 1 file changed, 24 insertions(+)
 create mode 100755 608/CH13/EX13.22/13_22.sce

(limited to '608/CH13/EX13.22/13_22.sce')

diff --git a/608/CH13/EX13.22/13_22.sce b/608/CH13/EX13.22/13_22.sce
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+++ b/608/CH13/EX13.22/13_22.sce
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+//Problem 13.22: Find the value of the load resistor RL shown in Figure 13.51(a) that gives maximum power dissipation and determine the value of this power.
+
+//initializing the variables:
+V = 15; // in volts
+R1 = 3; // in ohms
+R2 = 12; // in ohms
+
+//calculation:
+//Resistance RL is removed from the circuit as shown in Figure 13.51(b).
+//The p.d. across AB is the same as the p.d. across the 12 ohm resistor.
+E = (R2/(R1 + R2))*V
+//Removing the source of e.m.f. gives the circuit of Figure 13.51(c),
+//from which resistance, r
+r = R1*R2/(R1 + R2)
+//The equivalent Th´evenin’s circuit supplying terminals AB is shown in Figure 13.51(d), from which, current I = E/(r + RL)
+//For maximum power, RL = r
+RL = r
+I = E/(r + RL)
+//Power, P, dissipated in load RL, P
+P = RL*I^2
+
+printf("\n\n Result \n\n") 
+printf("\n (a) the value of the load resistor RL is %.1f ohm",RL)
+printf("\n (b) maximum power dissipation = %.0f W",P)
\ No newline at end of file
-- 
cgit