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/CH32/EX32.02/32_02.sce | 30 ++++++++++++++++++++++++++++++
 1 file changed, 30 insertions(+)
 create mode 100755 608/CH32/EX32.02/32_02.sce

(limited to '608/CH32/EX32.02/32_02.sce')

diff --git a/608/CH32/EX32.02/32_02.sce b/608/CH32/EX32.02/32_02.sce
new file mode 100755
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+++ b/608/CH32/EX32.02/32_02.sce
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+//Problem 32.02:Use the superposition theorem to determine the current in the 4 ohm resistor of the network shown in Figure 32.11.
+
+//initializing the variables:
+V1 = 12; // in volts
+V2 = 20; // in volts
+R1 = 5; // in ohm
+R2 = 4; // in ohm
+R3 = 2.5; // in ohm
+R4 = 6; // in ohm
+R5 = 2; // in ohm
+
+//calculation:
+//Removing the 20 V source gives the network shown in Figure 32.12.
+//Currents I1 and I2 are shown labelled in Figure 32.12
+Re1 = (R4*R5/(R4 + R5)) + R3
+Re2 = Re1*R2/(Re1  + R2) + R1
+I1 = V1/Re2
+I2 = (R2/(Re1 + R2))*I1
+//Removing the 12 V source from the original network gives the network shown in Figure 32.14.
+//Currents I3, I4 and I5 are shown labelled in Figure 32.14.
+Re3 = (R1*R2/(R1 + R2)) + R3
+Re4 = Re3*R4/(Re3 + R4) + R5
+I3 = V2/Re4
+I4 = (R4/(Re3 + R4))*I3
+I5 = (R1/(R1 + R2))*I4
+//Superimposing Figure 32.14 on Figure 32.12 shows that the current flowing in the 4 ohm resistor is given by
+Ir4 = I5 - I2
+
+printf("\n\n Result \n\n")
+printf("\ncurrent in the 4 ohm resistor of the network is %.3f A",Ir4)
\ No newline at end of file
-- 
cgit