24 files changed, 423 insertions, 0 deletions
diff --git a/1430/CH4/EX4.1/exa4_1.sce b/1430/CH4/EX4.1/exa4_1.sce
new file mode 100644
index 000000000..4501bd5da
--- /dev/null
+++ b/1430/CH4/EX4.1/exa4_1.sce
@@ -0,0 +1,15 @@
+// Example 4.1
+// Node Analysis with One Unknown
+// From figure 4.3
+G_11=1/6+1/(5+7)+1/4; // Sum of all conductance connected at node 1
+i_s1=18/6+(-60/4); // Net equivalent source current into node 1
+ // From node equation G_11*v_1=i_s1
+ v_1=i_s1/G_11; // Node voltage v_1
+ // Using Ohm's Law
+ i_a=(18-v_1)/6; //Current through 6-Ohm resistor
+ i_b=v_1/(5+7);// Current through 5-Ohm resistor
+ i_c=(v_1-(-60))/4; // Current through 4-Ohm resistor
+ disp(v_1,"Node Voltage(in Volts)=")
+ disp(i_a,"Branch Current through 6-Ohms(in Amps)=")
+ disp(i_b,"BRanch Current through 5-Ohms(in Amps)=")
+ disp(i_c,"Branch Current through 4-Ohms(in Amps)=")
diff --git a/1430/CH4/EX4.1/exa4_1.txt b/1430/CH4/EX4.1/exa4_1.txt
new file mode 100644
index 000000000..c1131cfba
--- /dev/null
+++ b/1430/CH4/EX4.1/exa4_1.txt
@@ -0,0 +1,19 @@
+ 
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 4\exa4.1.sce', -1)
+ 
+ Node Voltage(in Volts)=   
+ 
+  - 24.  
+ 
+ Branch Current through 6-Ohms(in Amps)=   
+ 
+    7.  
+ 
+ BRanch Current through 5-Ohms(in Amps)=   
+ 
+  - 2.  
+ 
+ Branch Current through 4-Ohms(in Amps)=   
+ 
+    9.  
+ 
diff --git a/1430/CH4/EX4.10/exa4_10.sce b/1430/CH4/EX4.10/exa4_10.sce
new file mode 100644
index 000000000..1cdb677ea
--- /dev/null
+++ b/1430/CH4/EX4.10/exa4_10.sce
@@ -0,0 +1,16 @@
+// Example 4.10
+// Mesh Analysis with a CCCS
+// By inspection of figure 4.34,we find the resistance matrix to be
+R=[10+4,-4;-4,4+7+3];
+// i_a= i_1-i_2
+//v_s=[6;(-24*i_1+24*i_2)]
+//v_s=[6;0]+[0,0;-24,24]*[i_1;i_2]   -------equation 1
+// Comparing Equation 1 with , [v_s]=[v_s_tilda]+[R_tilda]*[i]
+v_s_tilda=[6;0];
+R_tilda=[0,0;-24,24];
+// using Equation [R-R_tilda]*[i]=[v_s_tilda]
+i=(R-R_tilda)\v_s_tilda;
+i_1=i(1,1);
+i_2=i(2,1);
+disp(i_1,"Current in Mesh 1(in Amps)=")
+disp(i_2,"Current in Mesh 2(in Amps)=")
diff --git a/1430/CH4/EX4.10/exa4_10.txt b/1430/CH4/EX4.10/exa4_10.txt
new file mode 100644
index 000000000..31e054b52
--- /dev/null
+++ b/1430/CH4/EX4.10/exa4_10.txt
@@ -0,0 +1,11 @@
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 4\exa4.10.sce', -1)
+ 
+ Current in Mesh 1(in Amps)=   
+ 
+    1.  
+ 
+ Current in Mesh 2(in Amps)=   
+ 
+    2.  
+
diff --git a/1430/CH4/EX4.11/exa4_11.sce b/1430/CH4/EX4.11/exa4_11.sce
new file mode 100644
index 000000000..6a0b5f521
--- /dev/null
+++ b/1430/CH4/EX4.11/exa4_11.sce
@@ -0,0 +1,15 @@
+// Example 4.11
+// Mesh Analysis of a Current Amplifier
+R=[37 -1 0;-1 41 -4;0 -4 19]; // Resistance matrix
+//v_a=6*i_1 , v_b=4*(i_3-i_2)
+// [v_s]=[30*i_s;0;0]+[0 0 0;-864 0 0 ; 0 96 -96]*[i_1;i_2;i_3]
+i_s=10^-3; // Assumption
+R_tilda=[0 0 0;-864 0 0;0 96 -96];
+v_s_tilda=[30*i_s;0;0];
+// Using Equation, [R-R_tilda][i]=[v_s_tilda]
+i=(R-R_tilda)\v_s_tilda
+i_1=i(1,1);
+i_2=i(2,1);
+i_3=i(3,1);
+A_i=i_3/i_s; // Gain Of Current Amplifier
+disp(A_i,"Gain of Current Amplifier is=")
diff --git a/1430/CH4/EX4.11/exa4_11.txt b/1430/CH4/EX4.11/exa4_11.txt
new file mode 100644
index 000000000..460cf4438
--- /dev/null
+++ b/1430/CH4/EX4.11/exa4_11.txt
@@ -0,0 +1,7 @@
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 4\exa4.11.sce', -1)
+ 
+ Gain of Current Amplifier is=   
+ 
+  - 10.  
+
diff --git a/1430/CH4/EX4.16/exa4_16.sce b/1430/CH4/EX4.16/exa4_16.sce
new file mode 100644
index 000000000..aaecd40e5
--- /dev/null
+++ b/1430/CH4/EX4.16/exa4_16.sce
@@ -0,0 +1,16 @@
+// Example 4.16
+// Transformation of a Resistive Bridge
+// From figure 4.49(b)
+R_a=3;// Resistances of Wye Network
+R_b=12;
+R_c=4;
+R_2=(R_a*R_b)+(R_c*R_b)+(R_a*R_c);
+// Equivalent Delta Resistances are,
+R_ab=R_2/R_c; 
+R_bc=R_2/R_a;
+R_ca=R_2/R_b;
+// From Figure 4.49(c)
+R_par_1=(R_bc*8)/(R_bc+8);// Combining R_bc & 8
+R_par_2=(R_ca*2)/(R_ca+2);// Combining R_ca & 2
+R_eq=(24*(R_par_1+R_par_2))/(24+R_par_2+R_par_1);// Equivalent Resistance
+disp(R_eq,"Equivalent Resistance of the network(in Ohms)=")
diff --git a/1430/CH4/EX4.16/exa4_16.txt b/1430/CH4/EX4.16/exa4_16.txt
new file mode 100644
index 000000000..da513ee90
--- /dev/null
+++ b/1430/CH4/EX4.16/exa4_16.txt
@@ -0,0 +1,6 @@
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 4\exa4.16.sce', -1)
+ 
+ Equivalent Resistance of the network(in Ohms)=   
+ 
+    6.  
diff --git a/1430/CH4/EX4.2/exa4_2.sce b/1430/CH4/EX4.2/exa4_2.sce
new file mode 100644
index 000000000..75238a12b
--- /dev/null
+++ b/1430/CH4/EX4.2/exa4_2.sce
@@ -0,0 +1,25 @@
+// Example 4.2
+// Matrix Node Analysis with two Unknowns
+// From figure 4.6
+G_11= 1/6+1/15+1/12+1/60; // Sum of conductance connected to node 1
+G_22=1/5+1/12+1/60;// Sum of conductance connected to node 2
+G_12=1/12+1/60;
+G_21=G_12;  //Equivalent conductance connecting node 1 & 2
+// Conductance matrix is given by
+G=[G_11,-G_12;-G_21,G_22];
+i_s11=30/6+1;// Net Equivalent source current into node 1
+i_s21=50/5-1;// Net Equivalent source current into node 2
+i_s=[i_s11;i_s21]; // Current Vector
+// Using Matrix node Equation G*v=i
+v=G\i_s;
+v_1=v(1,1);
+v_2=v(2,1);
+v_12=v_1-v_2; // Voltage across current source
+i_a=(30-v_1)/6;// Current through 30 source
+i_b=(50-v_2)/5; // Current through 50V source
+p_50= 50*i_b;// Power supplied by 50V source
+p_30=30*i_a;// Power supplied by 30V source
+p_1=v_12*1;// Power supplied by 1A current source
+disp(p_50,"Power supplied by 50V source(in Watt)=")
+disp(p_30,"Power supplied by 30V source(in Watt)=")
+disp(p_1,"Power supplied by 1A source(in Watt)=")
diff --git a/1430/CH4/EX4.2/exa4_2.txt b/1430/CH4/EX4.2/exa4_2.txt
new file mode 100644
index 000000000..d51b6d15b
--- /dev/null
+++ b/1430/CH4/EX4.2/exa4_2.txt
@@ -0,0 +1,15 @@
+ 
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 4\exa4.2.sce', -1)
+ 
+ Power supplied by 50V source(in Watt)=   
+ 
+    100.  
+ 
+ Power supplied by 30V source(in Watt)=   
+ 
+    0.  
+ 
+ Power supplied by 1A source(in Watt)=   
+ 
+  - 10.  
+ 
diff --git a/1430/CH4/EX4.3/exa4_3.sce b/1430/CH4/EX4.3/exa4_3.sce
new file mode 100644
index 000000000..ea0cad621
--- /dev/null
+++ b/1430/CH4/EX4.3/exa4_3.sce
@@ -0,0 +1,24 @@
+// Example 4.3
+// Matrix Node Analysis with Three Unknown
+// From Figure 4.7
+G_11= 1/4+1/2+1/10; // Sum of Conductance at node 1
+G_12=1/10// Equivalent Conductance connecting node 1 & 2
+G_13= 0; // Equivalant Conductance connecting node 1 & 3
+G_21=G_12; // Symmetry Property of Conductance Matrix
+G_22= 1/10+1/5; // Sum of conductance at node 2
+G_23=1/5; // Equivalent Conductance connecting node 2 & 3
+G_31=G_13; // Symmetry Property of Conductance Matrix
+G_32=G_23; // Symmetry Property of Conductance Matrix
+G_33=1/5+1/20; // Sum of Conductance at node 3
+G=[G_11,-G_12,-G_13;-G_21,G_22,-G_23;-G_31,-G_32,G_33]; // Conductance Matrix
+i_s11= 30/2+3; // Net Equivalent source current into node 1
+i_s21= -1; // Net Equivalent source current into node 2
+i_s31=-3; // Net Equivalent source current into node 3
+i_s=[i_s11;i_s21;i_s31]; // Current Vector
+v=G\i_s;
+v_1=v(1,1);
+v_2=v(2,1);
+v_3=v(3,1);
+disp(v_1," Voltage at node 1(in Volts)=")
+disp(v_2,"Voltage at node 2(in Volts)=")
+disp(v_3," Voltage at node 3(in Volts)=")
diff --git a/1430/CH4/EX4.3/exa4_3.txt b/1430/CH4/EX4.3/exa4_3.txt
new file mode 100644
index 000000000..f55ccf33d
--- /dev/null
+++ b/1430/CH4/EX4.3/exa4_3.txt
@@ -0,0 +1,15 @@
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 4\exa4.3.sce', -1)
+ 
+  Voltage at node 1(in Volts)=   
+ 
+    20.  
+ 
+ Voltage at node 2(in Volts)=   
+ 
+  - 10.  
+ 
+  Voltage at node 3(in Volts)=   
+ 
+  - 20.  
+
diff --git a/1430/CH4/EX4.4/exa4_4.sce b/1430/CH4/EX4.4/exa4_4.sce
new file mode 100644
index 000000000..19e46e7f1
--- /dev/null
+++ b/1430/CH4/EX4.4/exa4_4.sce
@@ -0,0 +1,22 @@
+// Example 4.4
+// Matrix Node Analysis with source Conversion
+// From figure 4.11(a), The floating 48V source with series resistance has been converted into a current source with parallel resistance
+// From figure 4.11(b),
+G_11=1/4+1/2+1; // Sum of conductance at node 1
+G_12=1; // Equivalent conductance between node 1 & 2
+G_21=G_12; // Symmetry Property of Conductance matrix
+G_22=1/3+1+1/6; // Sum of conductance at node 2
+G=[G_11,-G_12;-G_21,G_22]; // Conductance Matrix
+i_s11=-24/4; // Equivalent source current at node 1
+i_s21=15/3+(-24)/6+48/6;// Equivalent source current at node 2
+i_s=[i_s11;i_s21]; // Current Matrix
+// Using Matrix Node Equation
+// G*v=i
+v=G\i_s;
+v_1=v(1,1);
+v_2=v(2,1);
+i_a=(15-v_2)/(3*10^3);// Current through 3-kOhm resistor
+i_b=-v_1/(2*10^3); // Current through 2-kOhm resistor
+disp(v_1,v_2)
+disp(i_a,"Current through 3-kOhm resistor(in Amps)=")
+disp(i_b,"Current through 2-kOhm resistor(in Amps)=")
diff --git a/1430/CH4/EX4.4/exa4_4.txt b/1430/CH4/EX4.4/exa4_4.txt
new file mode 100644
index 000000000..a1e32e7a1
--- /dev/null
+++ b/1430/CH4/EX4.4/exa4_4.txt
@@ -0,0 +1,44 @@
+ ans  =
+ 
+    1.  
+ 
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 4\exa4.4.sce', -1)
+ 
+ Current through 3-kOhm resistor(in Amps)=   
+ 
+    0.0005385  
+ 
+ Current through 2-kOhm resistor(in Amps)=   
+ 
+    0.0055385  
+ 
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 4\exa4.4.sce', -1)
+ 
+    13.384615  
+ 
+  - 11.076923  
+ 
+ Current through 3-kOhm resistor(in Amps)=   
+ 
+    0.0005385  
+ 
+ Current through 2-kOhm resistor(in Amps)=   
+ 
+    0.0055385  
+ 
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 4\exa4.4.sce', -1)
+ 
+    6.  
+ 
+  - 5.075D-16  
+ 
+ Current through 3-kOhm resistor(in Amps)=   
+ 
+    0.003  
+ 
+ Current through 2-kOhm resistor(in Amps)=   
+ 
+    2.538D-19  
+ 
+-->exit
+-->exec('SCI/etc/scilab.quit','errcatch',-1);quit;
diff --git a/1430/CH4/EX4.5/exa4_5.sce b/1430/CH4/EX4.5/exa4_5.sce
new file mode 100644
index 000000000..f9f6da522
--- /dev/null
+++ b/1430/CH4/EX4.5/exa4_5.sce
@@ -0,0 +1,16 @@
+// Example 4.5
+// Node analysis with a Supernode
+// From figure 4.15, Applying KCL at Supernode
+disp("((v_1-30)-v_2)/2-1+(v_1-v_2)/10+(v_1-50)/5=0 -------- Equation 1")
+//Applying KCL at node 2
+disp("(v_2-v_1)/10+(v_2-(v_1-30))/2+v_2-7=0 --------- Equation 2")
+disp("Rearrangement then yields a pair of equations in standard form,")
+disp("0.8v_1-0.6v_2=26")
+disp("-0.6v_1+1.6v_2=-8")
+G=[0.8,-0.6;-0.6,1.6]; // Conductance Matrix
+i=[26;-8]; // Current Matrix
+v=G\i;
+v_1=v(1,1);
+v_2=v(2,1);
+disp(v_1,"Voltage at Node 1(in Volts)=")
+disp(v_2,"Voltage at Node 2(in Volts)=")
diff --git a/1430/CH4/EX4.5/exa4_5.txt b/1430/CH4/EX4.5/exa4_5.txt
new file mode 100644
index 000000000..03f5dbb40
--- /dev/null
+++ b/1430/CH4/EX4.5/exa4_5.txt
@@ -0,0 +1,21 @@
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 4\exa4.5.sce', -1)
+ 
+ ((v_1-30)-v_2)/2-1+(v_1-v_2)/10+(v_1-50)/5=0 -------- Equation 1   
+ 
+ (v_2-v_1)/10+(v_2-(v_1-30))/2+v_2-7=0 --------- Equation 2   
+ 
+ Rearrangement then yields a pair of equations in standard form,   
+ 
+ 0.8v_1-0.6v_2=26   
+ 
+ -0.6v_1+1.6v_2=-8   
+ 
+ Voltage at Node 1(in Volts)=   
+ 
+    40.  
+ 
+ Voltage at Node 2(in Volts)=   
+ 
+    10.  
+ 
diff --git a/1430/CH4/EX4.6/exa4_6.sce b/1430/CH4/EX4.6/exa4_6.sce
new file mode 100644
index 000000000..3dd315c1a
--- /dev/null
+++ b/1430/CH4/EX4.6/exa4_6.sce
@@ -0,0 +1,24 @@
+// Example 4.6
+// Matrix Mesh Analysis with Two Unknowns
+// From Figure 4.22(a)
+R_11=6+15; // Sum of Resistance around the mesh 1
+R_12=15;// Equivalent Resistance shared by mesh 1 & 2
+R_21=R_12; // Symmetry Property of Resistance Matrix
+R_22=15+5+(60*12)/(60+12); // Sum of Resistance around the mesh 2
+R=[R_11,-R_12;-R_21,R_22]; // Resistance Matrix
+v_s_11=30; // Net Equivalent source Voltage that drives i_1 in mesh 1.
+v_s_21=-50-(60*12)/(60+12); // Net Equivalent source Voltage that drives i_2 in mesh 2
+v_s=[v_s_11;v_s_21]; // Voltage Vector
+// Form Matrix Mesh Equation R*i=v
+i=R\v_s; // Current Vector
+i_1=i(1,1);
+i_2=i(2,1);
+p_50=50*(-i_2);// Power supplied by 50V source
+p_30=30*i_1;// Power supplied by 30V source
+v_a=(60*12*(1+i_2))/(60+12); // Voltage across 1A current source
+disp(i_1,"Current in Mesh 1(in Amps)=")
+disp(i_2,"Current in mesh 2(in Amps)=")
+disp(p_50,"Power supplied by 50V source(in Watt)=")
+disp(p_30,"Power supplied by 30V source(in Watt)=")
+disp(v_a,"Voltage across Current source(in Volts=")
+
diff --git a/1430/CH4/EX4.6/exa4_6.txt b/1430/CH4/EX4.6/exa4_6.txt
new file mode 100644
index 000000000..17b6d6637
--- /dev/null
+++ b/1430/CH4/EX4.6/exa4_6.txt
@@ -0,0 +1,23 @@
+ 
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 4\exa4.6.sce', -1)
+ 
+ Current in Mesh 1(in Amps)=   
+ 
+    0.  
+ 
+ Current in mesh 2(in Amps)=   
+ 
+  - 2.  
+ 
+ Power supplied by 50V source(in Watt)=   
+ 
+    100.  
+ 
+ Power supplied by 30V source(in Watt)=   
+ 
+    0.  
+ 
+ Voltage across Current source(in Volts=   
+ 
+  - 10.  
+
diff --git a/1430/CH4/EX4.7/exa4_7.sce b/1430/CH4/EX4.7/exa4_7.sce
new file mode 100644
index 000000000..d05ca4903
--- /dev/null
+++ b/1430/CH4/EX4.7/exa4_7.sce
@@ -0,0 +1,25 @@
+// Example 4.7
+// Matrix Mesh Analysis with Three Unknown
+// From Figure 4.23
+R_11= 10+8+3; // Sum of Resistance around mesh 1
+R_12=3; // Equivalent Resistance shared by mesh 1 & 2
+R_13=0; // Equivalent resistance shared by mesh 1 & 3
+R_21=R_12; // Symmetry Property of Resistance Matrix
+R_22=3+6;// Sum of resistance around mesh 2
+R_23=6; // Equivalent Resistance shared by mesh 2 & 3
+R_31=R_13;// Symmetry Property of Resistance Matrix
+R_32=R_23;// symmetry Property of Resistance Matrix
+R_33=6+1;// Sum of resistance around mesh 3
+R=[R_11,-R_12,-R_13;-R_21,R_22,-R_23;-R_31,-R_32,R_33]; // Resistance Matrix
+v_s_11= 20+10*4;// Net Equivalent Source Voltage Driving current i_1 in Mesh 1
+v_s_21=12;// Net Equivalent Source Voltage Driving current i_2 in Mesh 2
+v_s_31=-20;// Net Equivalent Source Voltage Driving current i_3 in Mesh 3
+v_s=[v_s_11;v_s_21;v_s_31];// Voltage Vector
+// Using Matrix Node Equation, R*i=v
+i=R\v_s; // Current Vector
+i_1=i(1,1);
+i_2=i(2,1);
+i_3=i(3,1);
+disp(i_1*10^-3,"Current in Mesh 1(in Amps)=")
+disp(i_2*10^-3,"Current in Mesh 2(in Amps)=")
+disp(i_3*10^-3,"Current in Mesh 3(in Amps)=")
diff --git a/1430/CH4/EX4.7/exa4_7.txt b/1430/CH4/EX4.7/exa4_7.txt
new file mode 100644
index 000000000..4b5e62db5
--- /dev/null
+++ b/1430/CH4/EX4.7/exa4_7.txt
@@ -0,0 +1,15 @@
+ 
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 4\exa4.7.sce', -1)
+ 
+ Current in Mesh 1(in Amps)=   
+ 
+    0.003  
+ 
+ Current in Mesh 2(in Amps)=   
+ 
+    0.001  
+ 
+ Current in Mesh 3(in Amps)=   
+ 
+  - 0.002  
+
diff --git a/1430/CH4/EX4.8/exa4_8.sce b/1430/CH4/EX4.8/exa4_8.sce
new file mode 100644
index 000000000..fe4f7238c
--- /dev/null
+++ b/1430/CH4/EX4.8/exa4_8.sce
@@ -0,0 +1,20 @@
+// Rxample 4.8
+// Matrix Mesh Analysis with Source Conversion
+// From figure 4.27(c)
+R_11=4+2+8; // Sum of resistance around mesh 1
+R_12=2; // Equivalent Meshes shared by mesh 1 & 2
+R_21=R_12; // Symmetry Property of Reistance matrix
+R_22=6+2+10; // Sum of reistance around mesh 2
+R=[R_11,-R_12;-R_21,R_22]; // Resistance Matrix
+v_s_11=20+8*7; // Net Equivalent Source Voltage that drives current i_1 in Mesh 1
+v_s_21=18+6*7; // Net Equivalent Source Voltage that drives current i_2 in Mesh 2
+v=[v_s_11;v_s_21]; // Voltage Vector
+// Using Matrix Mesh Equation, R*i=v
+i=R\v; // Current Vector
+i_1=i(1,1);
+i_2=i(2,1);
+v_a=8*(7-i_1);// Ohm's Law
+i_x=3-i_2; //from KCL
+i_b=7+i_x; // from KCL
+disp(v_a,"Voltage across 8-Ohm Resistor(in Volts)=")
+disp(i_b,"Current through 6-Ohm Resistor(in Amps)=")
diff --git a/1430/CH4/EX4.8/exa4_8.txt b/1430/CH4/EX4.8/exa4_8.txt
new file mode 100644
index 000000000..ff24ffa74
--- /dev/null
+++ b/1430/CH4/EX4.8/exa4_8.txt
@@ -0,0 +1,11 @@
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 4\exa4.8.sce', -1)
+ 
+ Voltage across 8-Ohm Resistor(in Volts)=   
+ 
+    8.  
+ 
+ Current through 6-Ohm Resistor(in Amps)=   
+ 
+    6.  
+
diff --git a/1430/CH4/EX4.9/exa4_9.sce b/1430/CH4/EX4.9/exa4_9.sce
new file mode 100644
index 000000000..00debd3c2
--- /dev/null
+++ b/1430/CH4/EX4.9/exa4_9.sce
@@ -0,0 +1,8 @@
+// Example 4.9
+// Mesh Analysis with a Supernode
+// From figure 4.31, Applying KVL in Supermesh we get,
+disp("6*(i_1-5)+10*i_1+3*(i_1+4)-20=0")
+// Rearrangements gives
+disp("(6+10+3)*i_1=6*5-(3*4)+20")
+i_1=linsolve((6+10+3),-((6*5)-(3*4)+20))// Linear equation solver
+disp(i_1,"Current through the Upper Portion of the Supermesh(in Amps)=")
diff --git a/1430/CH4/EX4.9/exa4_9.txt b/1430/CH4/EX4.9/exa4_9.txt
new file mode 100644
index 000000000..d830d694a
--- /dev/null
+++ b/1430/CH4/EX4.9/exa4_9.txt
@@ -0,0 +1,10 @@
+
+-->exec('C:\Users\sangeet\Documents\Scilab\Circuits\Chapter 4\exa4.9.sce', -1)
+ 
+ 6*(i_1-5)+10*i_1+3*(i_1+4)-20=0   
+ 
+ (6+10+3)*i_1=6*5-(3*4)+20   
+ 
+ Current through the Upper Portion of the Supermesh(in Amps)=   
+ 
+    2.