6 files changed, 102 insertions, 0 deletions
diff --git a/Working_Examples/154/CH3/EX3.3/ch3_3.sce b/Working_Examples/154/CH3/EX3.3/ch3_3.sce
new file mode 100755
index 0000000..e72c536
--- /dev/null
+++ b/Working_Examples/154/CH3/EX3.3/ch3_3.sce
@@ -0,0 +1,13 @@
+clc
+disp("Example 3.3")
+printf("\n")
+printf("Given")
+disp("Equivalent resistance of three resistors is 750 ohm")
+disp("values of two resistors are 40 ohm and 410 ohm")
+Req=750;R1=40;R2=410;
+
+//For series resistance 
+disp("Req=R1+R2+R3")
+//On solving for R3
+R3=Req-R1-R2
+printf("Value of third ohmic resistor is %dohm\n",R3)
diff --git a/Working_Examples/154/CH3/EX3.4/ch3_4.sce b/Working_Examples/154/CH3/EX3.4/ch3_4.sce
new file mode 100755
index 0000000..26e7932
--- /dev/null
+++ b/Working_Examples/154/CH3/EX3.4/ch3_4.sce
@@ -0,0 +1,17 @@
+clc
+disp("Example 3.4")
+printf("\n")
+printf("Given")
+disp("values of two capacitors are 2uF and 10uF")
+C1=2*10^-6;C2=10*10^-6;
+//For two capacitors in series
+disp("Ceq=(C1*C2)/(C1+C2)")
+//On solving for Ceq
+Ceq=((C1*C2)/(C1+C2))*10^6
+printf("Value of equivalent capacitance is %3.2fuF\n",Ceq)
+
+disp("If C2=10pF")
+C2=10*10^-12;
+
+Ceq=((C1*C2)/(C1+C2))*10^12
+printf("Value of equivalent capacitance is %3.2fpF\n",Ceq)
+\ No newline at end of file
diff --git a/Working_Examples/154/CH3/EX3.5/ch3_5.sce b/Working_Examples/154/CH3/EX3.5/ch3_5.sce
new file mode 100755
index 0000000..9ca5084
--- /dev/null
+++ b/Working_Examples/154/CH3/EX3.5/ch3_5.sce
@@ -0,0 +1,18 @@
+clc
+disp("Example 3.5")
+printf("\n")
+printf("Given")
+disp("a)")
+disp("values of two resistors are 60 ohm and 60 ohm")
+R1=60;R2=60;
+disp("If resistors are parallel")
+Req=(R1*R2)/(R1+R2)
+printf("Value of equivalent resistance is %dohm\n",Req)
+
+disp("b)")
+disp("values of three equal resistors are 60 ohm")
+R1=60;R2=60;R3=60;
+disp("If resistors are parallel")
+x=1/R1+1/R2+1/R3
+Req=1/x;
+printf("Value of equivalent resistance is %dohm\n",Req)
diff --git a/Working_Examples/154/CH3/EX3.6/ch3_6.sce b/Working_Examples/154/CH3/EX3.6/ch3_6.sce
new file mode 100755
index 0000000..5156eb9
--- /dev/null
+++ b/Working_Examples/154/CH3/EX3.6/ch3_6.sce
@@ -0,0 +1,11 @@
+clc
+disp("Example 3.6")
+printf("\n")
+printf("Given")
+
+disp("values of two inductors are 3mH and 6 mH")
+L1=3*10^-3;L2=6*10^-3;
+disp("If inductors are parallel")
+Leq=((L1*L2)/(L1+L2))*10^3
+printf("Value of equivalent inductance is %3.1fmH\n",Leq)
+
diff --git a/Working_Examples/154/CH3/EX3.7/ch3_7.sce b/Working_Examples/154/CH3/EX3.7/ch3_7.sce
new file mode 100755
index 0000000..72aa564
--- /dev/null
+++ b/Working_Examples/154/CH3/EX3.7/ch3_7.sce
@@ -0,0 +1,19 @@
+clc
+disp("Example 3.7")
+printf("\n")
+printf("Given")
+disp("Total resistance of three resistors is 50 ohm")
+R=50;
+disp("Output voltage is 10 percent of the input voltage")
+//Let v be input voltage and v1 be output voltage
+//Let v1/v=V
+V=0.1;
+//As V=R1/(Total resistance)
+//Solving for R1
+R1=V*R;
+//As R=R1+R2
+//Solving for R2
+R2=R-R1;
+printf("R1=%dohm\n R2=%dohm\n",R1,R2)
+
+
diff --git a/Working_Examples/154/CH3/EX3.8/ch3_8.sce b/Working_Examples/154/CH3/EX3.8/ch3_8.sce
new file mode 100755
index 0000000..f744f0f
--- /dev/null
+++ b/Working_Examples/154/CH3/EX3.8/ch3_8.sce
@@ -0,0 +1,24 @@
+clc
+disp("Example 3.8")
+printf("\n")
+printf("Given")
+disp("Total current is 30mA")
+disp("Branch currents are 20mA and 10mA")
+disp("Equivalent resistance is equal to or greater than 10 ohm")
+
+//From Fig 3.6
+//Current flowing through R1 be i1 and let it be equal to 10mA
+//Current flowing through R2 be i2 and let it be equal to 20mA
+i1=10*10^-3;i2=20*10^-3;
+i=30*10^-3;
+
+//Let R1/(R1+R2)=X1        (1)
+//Let R2/(R1+R2)=X2        (2)
+X1=i1/i;
+X2=i2/i;
+//Let R1*R2(R1+R2)=Y        (3)
+//Given that 
+printf("\n Given")
+disp("R1*R2(R1+R2)>=10")
+//Solving (1),(2) and (3) we get
+printf("R1>=%dohm\nR2>=%dohm\n",15,30)