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/CH31/EX31.01/31_01.sce | 36 ++++++++++++++++++++++++++++++++++++
 608/CH31/EX31.02/31_02.sce | 40 ++++++++++++++++++++++++++++++++++++++++
 608/CH31/EX31.03/31_03.sce | 38 ++++++++++++++++++++++++++++++++++++++
 608/CH31/EX31.04/31_04.sce | 21 +++++++++++++++++++++
 608/CH31/EX31.05/31_05.sce | 22 ++++++++++++++++++++++
 608/CH31/EX31.06/31_06.sce | 24 ++++++++++++++++++++++++
 608/CH31/EX31.07/31_07.sce | 44 ++++++++++++++++++++++++++++++++++++++++++++
 608/CH31/EX31.08/31_08.sce | 36 ++++++++++++++++++++++++++++++++++++
 608/CH31/EX31.09/31_09.sce | 41 +++++++++++++++++++++++++++++++++++++++++
 9 files changed, 302 insertions(+)
 create mode 100755 608/CH31/EX31.01/31_01.sce
 create mode 100755 608/CH31/EX31.02/31_02.sce
 create mode 100755 608/CH31/EX31.03/31_03.sce
 create mode 100755 608/CH31/EX31.04/31_04.sce
 create mode 100755 608/CH31/EX31.05/31_05.sce
 create mode 100755 608/CH31/EX31.06/31_06.sce
 create mode 100755 608/CH31/EX31.07/31_07.sce
 create mode 100755 608/CH31/EX31.08/31_08.sce
 create mode 100755 608/CH31/EX31.09/31_09.sce

(limited to '608/CH31')

diff --git a/608/CH31/EX31.01/31_01.sce b/608/CH31/EX31.01/31_01.sce
new file mode 100755
index 000000000..86228e4ec
--- /dev/null
+++ b/608/CH31/EX31.01/31_01.sce
@@ -0,0 +1,36 @@
+//Problem 31.01: Use mesh-current analysis to determine the current flowing in (a) the 5 ohm resistance, and (b) the 1ohm  resistance of the d.c. circuit shown in Figure 31.2.
+
+//initializing the variables:
+V1 = 4; // in volts
+V2 = 5; // in volts
+R1 = 3; // in ohm
+R2 = 5; // in ohm
+R3 = 4; // in ohm
+R4 = 1; // in ohm
+R5 = 6; // in ohm
+R6 = 8; // in ohm
+
+//calculation:
+//The mesh currents I1, I2 and I3 are shown in Figure 31.2. Using Kirchhoff’s voltage law in 3 loops
+//three eqns obtained
+//(R1 + R2)*I1 - R2*I2 = V1
+//-1*R2*I1 + (R2 + R3 + R4 + R5)*I2 - R4*I3 = 0
+// -1*R4*I2 + (R4 + R6)*I3 = -1*V2
+//using determinants
+d1 = [V1 -1*R2 0; 0 (R2 + R3 + R4 + R5) -1*R4; -1*V2 -1*R4 (R4 + R6)]
+D1 = det(d1)
+d2 = [(R1 + R2) V1 0; -1*R2 0 -1*R4; 0 -1*V2 (R4 + R6)]
+D2 = det(d2)
+d3 = [(R1 + R2) -1*R2 V1; -1*R2 (R2 + R3 + R4 + R5) 0; 0 -1*R4 -1*V2]
+D3 = det(d3)
+d = [(R1 + R2) -1*R2 0; -1*R2 (R2 + R3 + R4 + R5) -1*R4; 0 -1*R4 (R4 + R6)]
+D = det(d)
+I1 = D1/D
+I2 = D2/D
+I3 = D3/D 
+IR2 = I1 - I2
+IR4 = I2 - I3
+
+printf("\n\n Result \n\n")
+printf("\n (a)current in the 5 ohm resistance is %.2f A",IR2)
+printf("\n (b)current in the 1 ohm resistance is %.2f A",IR4)
\ No newline at end of file
diff --git a/608/CH31/EX31.02/31_02.sce b/608/CH31/EX31.02/31_02.sce
new file mode 100755
index 000000000..16f18b47f
--- /dev/null
+++ b/608/CH31/EX31.02/31_02.sce
@@ -0,0 +1,40 @@
+//Problem 31.02: For the a.c. network shown in Figure 31.3 determine, using mesh-current analysis, (a) the mesh currents I1 and I2 (b) the current flowing in the capacitor, and (c) the active power delivered by the 100/_0° V voltage source.
+
+//initializing the variables:
+rv = 100; // in volts
+thetav = 0; // in degrees
+R1 = 5; // in ohm
+R2 = -1*4*%i; // in ohm
+R3 = 4; // in ohm
+R4 = %i*3; // in ohm
+
+//calculation:
+//voltages
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//Currents I1, I2 with their directions are shown in Figure 31.03.
+//Two loops are chosen. The choice of loop directions is arbitrary.
+//using kirchoff rule in 2 loops
+//two eqns obtained
+//(R1 + R2)*I1 - R2*I2 = V
+//-1*R2*I1 + (R3 + R2 + R4)*I2 = 0
+//using determinants
+d1 = [V -1*R2; 0 (R3 + R2 + R4)]
+D1 = det(d1)
+d2 = [(R1 + R2) V; -1*R2 0]
+D2 = det(d2)
+d = [(R1 + R2) -1*R2; -1*R2 (R3 + R2 + R4)]
+D = det(d)
+I1 = D1/D
+I2 = D2/D
+I1mag = (real(I1)^2 + imag(I1)^2)^0.5
+//Current flowing in capacitor
+Ic = I1 - I2
+//Source power P
+phi = atan(imag(I1)/real(I1))
+P = V*I1mag*cos(phi)
+Icmag = (real(Ic)^2 + imag(Ic)^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n (a)current, I1 is %.2f + (%.2f)i A, current, I2 is  %.2f + (%.2f)i A",real(I1), imag(I1),real(I2), imag(I2))
+printf("\n (b)current in the capacitor is %.2f A",Icmag)
+printf("\n (c)Source power P is %.2f W",P)
\ No newline at end of file
diff --git a/608/CH31/EX31.03/31_03.sce b/608/CH31/EX31.03/31_03.sce
new file mode 100755
index 000000000..19cc488b6
--- /dev/null
+++ b/608/CH31/EX31.03/31_03.sce
@@ -0,0 +1,38 @@
+//Problem 31.03: A balanced star-connected 3-phase load is shown in Figure 31.4. Determine the value of the line currents IR, IY and IB using mesh-current analysis.
+
+//initializing the variables:
+rv1 = 415; // in volts
+rv2 = 415; // in volts
+thetav1 = 120; // in degrees
+thetav2 = 0; // in degrees
+R = 3 + %i*4; // in ohm
+
+//calculation:
+//voltages
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+//Two mesh currents I1 and I2 are chosen as shown in Figure 31.4.
+//Two loops are chosen. The choice of loop directions is arbitrary.
+//using kirchoff rule in 2 loops
+//two eqns obtained
+//2*R*I1 - R*I2 = V1
+//-1*R*I1 + 2*R*I2 = V2
+//using determinants
+d1 = [V1 -1*R; V2 2*R]
+D1 = det(d1)
+d2 = [2*R V1; -1*R V2]
+D2 = det(d2)
+d = [2*R -1*R; -1*R 2*R]
+D = det(d)
+I1 = D1/D
+I2 = D2/D
+I1mag = (real(I1)^2 + imag(I1)^2)^0.5
+//line current IR
+IR = I1
+//line current IB
+IB = -1*I2
+//line current IY
+IY = I2 - I1
+
+printf("\n\n Result \n\n")
+printf("\n current, IR is %.2f + (%.2f)i A, current, IB is  %.2f + (%.2f)i A and current, IY is %.2f + (%.2f)i A",real(IR), imag(IR),real(IB), imag(IB),real(IY), imag(IY))
\ No newline at end of file
diff --git a/608/CH31/EX31.04/31_04.sce b/608/CH31/EX31.04/31_04.sce
new file mode 100755
index 000000000..a5ad7b13c
--- /dev/null
+++ b/608/CH31/EX31.04/31_04.sce
@@ -0,0 +1,21 @@
+//Problem 31.04: For the network shown in Figure 31.8, determine the voltage VAB, by using nodal analysis.
+
+//initializing the variables:
+ri = 20; // in amperes
+thetai = 0; // in degrees
+R1 = 10; // in ohm
+R2 = %i*3; // in ohm
+R3 = 4; // in ohm
+R4 = 16; // in ohm
+
+//calculation:
+//current
+I = ri*cos(thetai*%pi/180) + %i*ri*sin(thetai*%pi/180)
+//Figure 31.8 contains two principal nodes (at 1 and B) and thus only one nodal equation is required. B is taken as the reference node and the equation for node 1 is obtained as follows. Applying Kirchhoff’s current law to node 1 gives:
+//IX + IY = I
+V1 = I/((1/R4) +(1/(R2  +R3)))
+IY = V1/(R2 + R3)
+VAB = IY*R3
+
+printf("\n\n Result \n\n")
+printf("\n voltage VAB is %.2f + (%.2f)i V",real(VAB), imag(VAB))
\ No newline at end of file
diff --git a/608/CH31/EX31.05/31_05.sce b/608/CH31/EX31.05/31_05.sce
new file mode 100755
index 000000000..7a612c1d8
--- /dev/null
+++ b/608/CH31/EX31.05/31_05.sce
@@ -0,0 +1,22 @@
+//Problem 31.05: Determine the value of voltage VXY shown in the circuit of Figure 31.9.
+
+//initializing the variables:
+rv1 = 8; // in volts
+rv2 = 8; // in volts
+thetav1 = 0; // in degrees
+thetav2 = 90; // in degrees
+R1 = 5; // in ohm
+R2 = %i*6; // in ohm
+R3 = 4; // in ohm
+R4 = 3; // in ohm
+
+//calculation:
+//voltages
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+//The circuit contains no principal nodes. However, if point Y is chosen as the reference node then an equation may be written for node X assuming that current leaves point X by both branches
+VX = [(V1/(R1 + R3) + V2/(R2 + R4))/(1/(R1 + R3) + 1/(R2 + R4))]
+VXY = VX
+
+printf("\n\n Result \n\n")
+printf("\n voltage VXY is %.2f + (%.2f)i V",real(VXY), imag(VXY))
\ No newline at end of file
diff --git a/608/CH31/EX31.06/31_06.sce b/608/CH31/EX31.06/31_06.sce
new file mode 100755
index 000000000..9c6f82966
--- /dev/null
+++ b/608/CH31/EX31.06/31_06.sce
@@ -0,0 +1,24 @@
+//Problem 31.06: Use nodal analysis to determine the current flowing in each branch of the network shown in Figure 31.10.
+
+//initializing the variables:
+rv1 = 100; // in volts
+rv2 = 50; // in volts
+thetav1 = 0; // in degrees
+thetav2 = 90; // in degrees
+R1 = 25; // in ohm
+R2 = 20; // in ohm
+R3 = 10; // in ohm
+
+//calculation:
+//voltages
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+//There are only two principal nodes in Figure 31.10 so only one nodal equation is required. Node 2 is taken as the reference node.
+//The equation at node 1 is I1 + I2 + I3 = 0
+Vn1 = [(V1/R1 + V2/R3)/(1/R1 + 1/R2 + 1/R3)]
+I1 = (Vn1 - V1)/R1
+I2 = Vn1/R2
+I3 = (Vn1 - V2)/R3
+
+printf("\n\n Result \n\n")
+printf("\n current, I1 is %.2f + (%.2f)i A, current, I2 is  %.2f + (%.2f)i A and current, I3 is %.2f + (%.2f)i A",real(I1), imag(I1),real(I2), imag(I2),real(I3), imag(I3))
\ No newline at end of file
diff --git a/608/CH31/EX31.07/31_07.sce b/608/CH31/EX31.07/31_07.sce
new file mode 100755
index 000000000..cd396bd9e
--- /dev/null
+++ b/608/CH31/EX31.07/31_07.sce
@@ -0,0 +1,44 @@
+//Problem 31.07: In the network of Figure 31.11 use nodal analysis to determine (a) the voltage at nodes 1 and 2, (b) the current in the j4 ohm inductance, (c) the current in the 5 ohm resistance, and (d) the magnitude of the active power dissipated in the 2.5 ohm resistance.
+
+//initializing the variables:
+rv1 = 25; // in volts
+rv2 = 25; // in volts
+thetav1 = 0; // in degrees
+thetav2 = 90; // in degrees
+R1 = 2; // in ohm
+R2 = -1*%i*4; // in ohm
+R3 = 5; // in ohm
+R4 = %i*4; // in ohm
+R5 = 2.5; // in ohm
+
+//calculation:
+//voltages
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+//The equation at node 1
+//Vn1*(1/R1 + 1/R2 + 1/R3) - Vn2/R3 = V1/R1
+//The equation at node 2
+//Vn1*(-1/R3) + Vn2*(1/R4 + 1/R5 + 1/R3) = V2/R5
+//using determinants
+d1 = [V1/R1 -1/R3; V2/R5 (1/R4 + 1/R5 + 1/R3)]
+D1 = det(d1)
+d2 = [(1/R1 + 1/R2 + 1/R3) V1/R1; -1/R3 V2/R5]
+D2 = det(d2)
+d = [(1/R1 + 1/R2 + 1/R3) -1/R3; -1/R3 (1/R4 + 1/R5 + 1/R3)]
+D = det(d)
+Vn1 = D1/D
+Vn2 = D2/D
+//current in the j4 ohm inductance is given by:
+I4 = Vn2/R4
+//current in the 5 ohm resistance is given by:
+I3 = (Vn1 - Vn2)/R3
+//active power dissipated in the 2.5 ohm resistor is given by
+P5 = R5*((Vn2 - V2)/R5)^2
+//magnitude of the active power dissipated
+P5mag = (real(P5)^2 + imag(P5)^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n (a) the voltage at nodes 1 and 2 is %.2f + (%.2f)i V and %.2f + (%.2f)i V",real(Vn1), imag(Vn1),real(Vn2), imag(Vn2))
+printf("\n (b)the current in the j4 ohm inductance is %.2f + (%.2f)i A",real(I4), imag(I4))
+printf("\n (c)the current in the 5 ohm resistance is %.2f + (%.2f)i A",real(I3), imag(I3))
+printf("\n (d) magnitude of the active power dissipated in the 2.5 ohm resistance is %.2f W",P5mag)
\ No newline at end of file
diff --git a/608/CH31/EX31.08/31_08.sce b/608/CH31/EX31.08/31_08.sce
new file mode 100755
index 000000000..a71af8f54
--- /dev/null
+++ b/608/CH31/EX31.08/31_08.sce
@@ -0,0 +1,36 @@
+//Problem 31.08: In the network shown in Figure 31.12 determine the voltage VXY using nodal analysis
+
+//initializing the variables:
+ri = 25; // in amperes
+thetai = 0; // in degrees
+R1 = 4; // in ohm
+R2 = %i*3; // in ohm
+R3 = 5; // in ohm
+R4 = %i*10; // in ohm
+R5 = %i*20; // in ohm
+
+//calculation:
+//current
+I = ri*cos(thetai*%pi/180) + %i*ri*sin(thetai*%pi/180)
+//Node 3 is taken as the reference node.
+//At node 1,
+//V1*(1/(R1 + R2) + 1/R3) - V2/R3 = I
+//The equation at node 2
+//V1*(-1/R3) + V2*(1/R4 + 1/R5 + 1/R3) = 0
+//using determinants
+d1 = [I -1/R3; 0 (1/R4 + 1/R5 + 1/R3)]
+D1 = det(d1)
+d2 = [(1/(R1 + R2) + 1/R3) I; -1/R3 0]
+D2 = det(d2)
+d = [(1/(R1 + R2) + 1/R3) -1/R3; -1/R3 (1/R4 + 1/R5 + 1/R3)]
+D = det(d)
+V1 = D1/D
+V2 = D2/D
+//the voltage between point X and node 3 is
+VX = V1*R2/(R1 + R2)
+//Thus the voltage
+VY = V2
+VXY = VX - VY
+
+printf("\n\n Result \n\n")
+printf("\n voltage VXY is %.2f + (%.2f)i V",real(VXY), imag(VXY))
\ No newline at end of file
diff --git a/608/CH31/EX31.09/31_09.sce b/608/CH31/EX31.09/31_09.sce
new file mode 100755
index 000000000..3d0f14dcd
--- /dev/null
+++ b/608/CH31/EX31.09/31_09.sce
@@ -0,0 +1,41 @@
+//Problem 31.09: Use nodal analysis to determine the voltages at nodes 2 and 3 in Figure 31.13 and hence determine the current flowing in the 2 ohm resistor and the power dissipated in the 3 ohm resistor.
+
+//initializing the variables:
+V = 8; // in volts
+R1 = 1; // in ohm
+R2 = 2; // in ohm
+R3 = 3; // in ohm
+R4 = 4; // in ohm
+R5 = 5; // in ohm
+R6 = 6; // in ohm
+
+//calculation:
+//In Figure 31.13, the reference node is shown at point A.
+//At node 1,
+//V1*(1/R1 + 1/R6 + 1/R5) - V2/R1 - V3/R5 = V/R5
+//The equation at node 2
+//V1*(-1/R1) + V2*(1/R2 + 1/R1 + 1/R3) - V3/R3 = 0
+//At node 3
+// - V1/R5 - V2/R3 + V3*(1/R4 + 1/R3 + 1/R5) = -1*V/R5
+//using determinants
+d1 = [V/R5 -1/R1 -1/R5; 0 (1/R2 + 1/R1 + 1/R3) -1/R3; -1*V/R5 -1/R3 (1/R4 + 1/R3 + 1/R5)]
+D1 = det(d1)
+d2 = [(1/R1 + 1/R6 + 1/R5) V/R5 -1/R5; -1/R1 0 -1/R3; -1/R5 -1*V/R5 (1/R4 + 1/R3 + 1/R5)]
+D2 = det(d2)
+d3 = [(1/R1 + 1/R6 + 1/R5) -1/R1 V/R5; -1/R1 (1/R2 + 1/R1 + 1/R3) 0; -1/R5 -1/R3 -1*V/R5]
+D3 = det(d3)
+d = [(1/R1 + 1/R6 + 1/R5) -1/R1 -1/R5; -1/R1 (1/R2 + 1/R1 + 1/R3) -1/R3; -1/R5 -1/R3 (1/R4 + 1/R3 + 1/R5)]
+D = det(d)
+Vn1 = D1/D
+Vn2 = D2/D
+Vn3 = D3/D 
+//the current in the 2 ohm resistor
+I2 = Vn2/R2
+//power dissipated in the 3 ohm resistance
+P3 = R3*((Vn2 - Vn3)/R3)^2
+
+printf("\n\n Result \n\n")
+printf("\n voltage at node 2 is %.2f V",Vn2)
+printf("\n voltage at node 3 is %.2f V",Vn3)
+printf("\n (a)current through 2 ohm resistor is %.2f A",I2)
+printf("\n (b)power dissipated in the 3 ohm resistor is %.2f W",P3)
\ No newline at end of file
-- 
cgit