5 files changed, 93 insertions, 0 deletions
diff --git a/644/CH1/EX1.1/p1.sce b/644/CH1/EX1.1/p1.sce
new file mode 100644
index 000000000..b13997dc7
--- /dev/null
+++ b/644/CH1/EX1.1/p1.sce
@@ -0,0 +1,11 @@
+// Example 1.1 A conductor material has a free- electron density of 10^24 electrons per metre^3.When a voltage is applied, a constant drift velocity of 1.5x10^-2 metre/second is attained by the electrons. If the cross- sectional area of the material is 1 cm^2, calculate the magnitude of the current. Electronic charge is 1.6x10^-19.
+// 1 metre = 100 centimetre
+n = 10^24;// charge density (e/m^3)
+Vd = 1.5*10^-2; //drift velocity attained by electrons(m/s)
+A =  10^-4; // crossectional area of the material (m^2)
+e = 1.6*10^-19; // charge of an electron (coulombs)
+// let i be the magnitude of the current
+// FORMULA : i = nAeVd
+i = prod([n,A,e,Vd]) // calculation
+disp(i,"magnitude of the current(in ampere)= ")
+
diff --git a/644/CH1/EX1.2/p2.sce b/644/CH1/EX1.2/p2.sce
new file mode 100644
index 000000000..6fab4cf33
--- /dev/null
+++ b/644/CH1/EX1.2/p2.sce
@@ -0,0 +1,13 @@
+// Example 1.2 Find the velocity of charge leading to 1 A current which flows in a copper conductor of cross-section 1 cm^2 and length 10 Km. Free  electron density of copper = 8.5X10^28 per m^3. How long will it take the electric charge to travel from one end of the conductor to the other.
+// 1 metre = 100 centimetre
+// 1 kilometre = 1000 metre
+i =1; // value of current (A)
+A = 10^-4;// crossectional area of the conductor (m^2)
+L = 10*10^3;// length of the conductor (m)
+n =8.5*10^28;// charge density (e/m^3)
+// Let V be the velocity of charge (m/S) and t (s) be the time taken by the charge to travel from one end of the conductor to the other
+// FORMULAE: V= i/nAe, where is charge of an electron and t =L/V
+V = i/prod([n,A,e]);// calculation of drift velocity
+t = L/V;//calculation of the time
+disp(V," velocity of the charge (in m/S)=" )
+disp(t,"time taken by the charge to travel conductor of length 10 Km(second)=")
diff --git a/644/CH1/EX1.3/p3.sce b/644/CH1/EX1.3/p3.sce
new file mode 100644
index 000000000..d8309331f
--- /dev/null
+++ b/644/CH1/EX1.3/p3.sce
@@ -0,0 +1,15 @@
+// Example1.3. A coil consists of 2000 turns of copper wire having a cross sectional area  of 0.8 mm^2. The mean length per turn is 80 cm and resistivity of copper is 0.02 micro-ohm- metre. Find the resistance of the coil and power absorbed by the coil when connected across 110 V d.c. supply.
+//1 millimetre = 10^-3 metre
+// 1 micro-ohm = 10^-6 ohms
+N = 2000; // number of turns
+A = 0.8*10^-6;// crossectional area (m^2)
+l = 80*10^-2;// mean length(m)
+p = 0.02*10^-6;// resistivity (ohm-m)
+V = 110; // supply voltage(V)
+// Let R ohms be the resistance of the coil and P watts be the power absorbed
+// FORMULAE: R=p*L/A , where L is the length of the coil ; P= V^2/R
+L= prod([l,N]);// length of the coil(m)
+R =prod([p,L])/A;// calculation of resisrance (ohms)
+P = (V^2)/R;// power absorbed by the coil (Watts)
+disp(R,"resistance of the coil (in ohms)= ")
+disp(P,"power absorbed by the coil(in watts)=") 
diff --git a/644/CH1/EX1.4/p4.sce b/644/CH1/EX1.4/p4.sce
new file mode 100644
index 000000000..d63e962de
--- /dev/null
+++ b/644/CH1/EX1.4/p4.sce
@@ -0,0 +1,23 @@
+// Example1.4 An aluminium wire 7.5 m long is connected in a parallel with a copper wire 6 m long. When a current of 5A is passed through the combination, it is found that the current in the aluminium wire is 3 A. The diameter of the aluminium wire is 1 mm.  Determine the diameter of the copper wire. Resistivity of copper is 0.017 micro-ohm-metre; that of the aluminium is 0.028 micro-ohm- metre.
+// 1 mm =10^-3m
+// 1 micro-ohm = 10^-6 ohm
+// let the numeral 1 represent aluminium and 2 represent copper.
+//Formulae: R=pl/a ; a=%pi*d^2/4
+l1 = 7.5; //length of aluminium wire (m)
+l2 = 6; // length of copper wire (m)
+i = 5; //  total current(amps)
+i1= 3;// current through aluminium wire(amps)
+d1 = 10^-3;// dia. of aluminium wire (m)
+p1 = 0.028*10^-6;// resistivity of aluminium wire(ohm-m)
+p2 = 0.017*10^-6;// resistivity of copper wire(ohm-m)
+// let d2 be the diameter of the copper wire in meters
+i2 = i -i1;// current through copper wire (amps)
+// R1=P1l1/a1 and R2= p2l2/a2, on dividing R2/R1= (p2*l2*a1)/(p1*l1*a2)thus a2= a1*(R1/R2)*(p2/p1)*(l2/l1)----- (equ1)
+// in parallel combination voltage remain same , therefore V= i1R1= i2R2
+//let R2/R1 =m
+m=i2/i1;
+a1= prod([%pi,d1^2])/4;// crossectional area of alluminium wire (m^2)
+a2= prod([a1,m,(p2/p1),(l2/l1)]);// using equ1
+d2=sqrt(prod([4,a2])/%pi); // cal of diameter of copper wire (m)
+ans= prod([d2,10^3])
+disp(ans,"diameter of copper wire(in mm)=")
+\ No newline at end of file
diff --git a/644/CH1/EX1.5/p5.sce b/644/CH1/EX1.5/p5.sce
new file mode 100644
index 000000000..80fd8ea89
--- /dev/null
+++ b/644/CH1/EX1.5/p5.sce
@@ -0,0 +1,31 @@
+//Example 1.5 (a) A rectangular carbon block has dimensions 1.0cmx1.0cmx50cm.(i)What is the resistance measured between the two square ends?  (ii)between two opposing rectangular faces/ resistivity of carbon at 20 degree centigrate is 3.5x10^-5 ohm-m. (b)A current of 5 A exists in a 10 ohm resistance for 4 minutes (i)how many coulombs and (ii)how many  electrons pass through any section of the resistor in this time? charge on electron = 1.6X10^-19 C.
+//(a)
+p = 3.5*10^-5; // resistivity (ohm-m)
+// (i), R=pl/A
+A= 10^-4; //area (m^2)
+l= 50*10^-2;// length (m)
+R= prod([p,(l/A)])
+disp(R,"resistance measured between two square ends (in ohms)=")
+
+//(ii),R=pl/A
+a = 50*10^-4;// area(m^2)
+L = 10^-2;// length(m)
+r = prod([p,(L/a)])
+disp(r,"resistance measured between two rectangular faces(in ohms)=")
+
+//(b)
+clear
+i= 5; // current (amps)
+R = 10;// resistance (ohms)
+t = 4*60;// time (s)
+e = 1.6*10^-19;// charge on electron (C)
+//(i), Q =it
+Q = prod([i,t])
+disp(Q,"charge through 10 ohm resistor in 4 minutes(in coulombs)=")
+
+//(ii), Q = ne
+n = Q/e
+disp(n,"  number of electrons through 10 ohm resistor in 4 minutrs")
+
+  
+ 
+\ No newline at end of file