diff options
Diffstat (limited to '2345')
76 files changed, 1031 insertions, 0 deletions
diff --git a/2345/CH15/EX15.1/Ex15_1.sce b/2345/CH15/EX15.1/Ex15_1.sce new file mode 100755 index 000000000..169315e96 --- /dev/null +++ b/2345/CH15/EX15.1/Ex15_1.sce @@ -0,0 +1,18 @@ +//consumer voltage and energy loss
+//Example 15.1(pg 392)
+clc
+clear
+R=0.2//total resistance of cable in ohms
+I=200//current in A
+t=100//time in hours
+V=240//voltage in volts
+c=0.8//cost of electrical energy in Rs per unit
+V1=I*R//voltage drop in the cable
+//(i)consumer voltage
+Vc=V-V1
+//(ii)Power loss in the cable
+P=I*I*R//in watts
+E=P*t/1000//energy loss in kWh
+C=E*c//cost of energy loss in Rs.
+printf('(i)Consumer voltage is %3.1f Volts \n',Vc)
+printf('(ii)cost of energy loss is Rs %3.2f ',C)
diff --git a/2345/CH15/EX15.10/Ex15_10.sce b/2345/CH15/EX15.10/Ex15_10.sce new file mode 100755 index 000000000..82187aa3c --- /dev/null +++ b/2345/CH15/EX15.10/Ex15_10.sce @@ -0,0 +1,13 @@ +//Finding resistance
+//Example 15.10(pg 398)
+clc
+clear
+rho=1.7*(10^-6)//resistivity of copper in ohm-cm
+l=5//length in metres
+t=0.005//thickness in m
+D=0.08//external diameter in m
+d=D-(2*t)//internal diameter in m
+a=%pi*(D^2-d^2)/4//cross section area in cm^2
+R=rho*l/a//resistance of copper tube in ohm
+R1=R/(10^-4)//resistance in micro-ohm
+printf('Thus the resistance of copper tube is %3.2f micro-ohm',R1)
diff --git a/2345/CH15/EX15.11/Ex15_11.sce b/2345/CH15/EX15.11/Ex15_11.sce new file mode 100755 index 000000000..81b247e91 --- /dev/null +++ b/2345/CH15/EX15.11/Ex15_11.sce @@ -0,0 +1,11 @@ +//Conductivity and conductance
+//Example 15.11(pg 399)
+clc
+clear
+rho=1.7*(10^-8)//resistivity in ohm-m
+K=1/rho//conductivity in mho/m
+a=0.125*(10^-4)//cross sectional area of cable in m^2
+l=2000//length of cable in meters
+G=K*a/l//conductance
+printf('Thus conductivity of cable is %e mho/metres \n',K)
+printf('and conductance of cable is %3f mho',G)
diff --git a/2345/CH15/EX15.12/Ex15_12.sce b/2345/CH15/EX15.12/Ex15_12.sce new file mode 100755 index 000000000..4d69a26ba --- /dev/null +++ b/2345/CH15/EX15.12/Ex15_12.sce @@ -0,0 +1,9 @@ +//Finding resistivity
+//Example 15.12(pg 399)
+clc
+clear
+V=0.05//volume in m^3
+l=300//length in m
+R=0.0306//resistance of conductor in ohm
+rho=R*V/(l^2)//resistivity of conducting material
+printf('Thus resistivity of conducting material is %e ohm-m',rho)
diff --git a/2345/CH15/EX15.13/Ex15_13.sce b/2345/CH15/EX15.13/Ex15_13.sce new file mode 100755 index 000000000..c5f5513e7 --- /dev/null +++ b/2345/CH15/EX15.13/Ex15_13.sce @@ -0,0 +1,11 @@ +//Finding resistivity
+//Example 15.12(pg 399)
+clc
+clear
+rho=0.67*(10^-6)//resistivity in ohm-inch
+m=39.4//1meter = 39.4inch
+m2=1525//1 meter2=1525 square inch
+rhoc=rho*m/m2//resistivity of copper in ohm/m^3
+rho1=rhoc/(10^-6)
+printf('Thus resistivity of copper is %e ohm/m^3',rhoc)
+printf('/n which is equal to %2.4f micro-ohm/m^3',rho1)
diff --git a/2345/CH15/EX15.14/Ex15_14.sce b/2345/CH15/EX15.14/Ex15_14.sce new file mode 100755 index 000000000..69b4a7fbe --- /dev/null +++ b/2345/CH15/EX15.14/Ex15_14.sce @@ -0,0 +1,13 @@ +//Finding resistance
+//Example 15.14(pg. 400)
+clc
+clear
+R1=0.12//old conductor resistance in ohm
+d1=15//diameter of old conductor in cm
+d2=0.4*d1//diameter of new conductor in cm
+a1=%pi*(d1^2)/4//area of cross section of old conductor
+a2=%pi*(d2^2)/4//area of cross section of new conductor
+//R=rho*l/a=rho*V/a^2
+//Henec R is proportional to 1/a^2
+R2=R1*((a1/a2)^2)//resistance of new conductor
+printf('Thus resistance of new conductor is %2.4f ohm',R2)
diff --git a/2345/CH15/EX15.15/Ex15_15.sce b/2345/CH15/EX15.15/Ex15_15.sce new file mode 100755 index 000000000..ae0c135a8 --- /dev/null +++ b/2345/CH15/EX15.15/Ex15_15.sce @@ -0,0 +1,11 @@ +//Finding resistance
+//Example 15.15(pg. 401)
+clc
+clear
+lab=10//la=10*lb ratio of length of A to length of B.
+Aab=1/2//Aa=1/2*Ab ratio of area of A to area of B
+RHOab=1/2//RHOa=2*RHOb ratio of resistivity of A to resistivity of B
+Ra=2//resistance of A in ohm
+Rb=(Ra*Aab)/(lab*RHOab)//resistance of B in ohm
+//Since Ra=RHOa*la/Aa and Rb=RHOb*lb/Ab so from ratio of two we get Rb
+printf('Thus resistance of resistor B is %2.2f ohm',Rb)
diff --git a/2345/CH15/EX15.16/Ex15_16.sce b/2345/CH15/EX15.16/Ex15_16.sce new file mode 100755 index 000000000..a4f415448 --- /dev/null +++ b/2345/CH15/EX15.16/Ex15_16.sce @@ -0,0 +1,14 @@ +//Finding resistance
+//Example 15.16(pg. 402)
+clc
+clear
+RHOo=10.3*(10^-6)//resistivity of platinum wire at 0 degree in ohm-cm
+d=0.0074//diameter of platinum wire
+a=%pi*(d^2)/4//area of cross section of platinum wire in sq cm
+Ro=4//resistance of wire in ohm
+l=Ro*a/RHOo//length of wire in cm
+alphao=0.0038
+t=100//temp in degree C
+R100=Ro*(1+(alphao*t))
+printf('Thus length of wire required is %3.2f cms\n',l)
+printf('and Resistance of wire at 100 degreeC is %2.2f ohms',R100)
diff --git a/2345/CH15/EX15.17/Ex15_17.sce b/2345/CH15/EX15.17/Ex15_17.sce new file mode 100755 index 000000000..95d4f69d8 --- /dev/null +++ b/2345/CH15/EX15.17/Ex15_17.sce @@ -0,0 +1,11 @@ +//Finding resistance
+//Example 15.17(pg. 403)
+clc
+clear
+Ra=1//resistance of A in ohm
+lab=20//ratio of length of A to length of B
+Aab=1/3//ratio of area of A to area of B
+//resistivity is same for both wires
+Rb=Ra*(Aab/lab)//resistance of wire B in ohm
+//since Ra=rho*la/Aa and Rb=rho*lb/Ab so from ratio of both we get Rb
+printf('Thus resistance of wire B is %2.4f omhs',Rb)
diff --git a/2345/CH15/EX15.19/Ex15_19.sce b/2345/CH15/EX15.19/Ex15_19.sce new file mode 100755 index 000000000..d05c3c386 --- /dev/null +++ b/2345/CH15/EX15.19/Ex15_19.sce @@ -0,0 +1,11 @@ +//Finding potential difference
+//Example 15.19(pg. 405)
+clc
+clear
+I1=2/(2+3)//current across 2V battery in circuit EBD in A
+Vbe=3*I1//voltage dropp across BE in V
+I2=4/(5+3)//current across 4V battery in circuit AFC in A
+Vaf=3*I2//voltage dropp across AF in V
+V=Vbe+4-Vaf//sum of potential drops starting from E and ending at F
+//V is the P.D. between E and F
+printf('Thus the P.D. between E and F is %2.1f Volts',V)
diff --git a/2345/CH15/EX15.2/Ex15_2.sce b/2345/CH15/EX15.2/Ex15_2.sce new file mode 100755 index 000000000..03a645516 --- /dev/null +++ b/2345/CH15/EX15.2/Ex15_2.sce @@ -0,0 +1,17 @@ +//Resistance and BOT units
+//Example 15.2(pg 393)
+clc
+clear
+Vi=220//voltage in volts supplied by dynamo
+Vo=200//voltage in volts required for lighting
+I=40//current in Amperes
+Pi=Vi*I//power output of dynamo
+Po=Vo*I//power consumed for lighting
+L=Pi-Po//line losses
+R=L/(I^2)//resistance of lines since line losses=I^2*R
+t=10//time in hrs
+N=(Po*t)/1000//no of units of consumed in B.O.T units
+Nw=(L*t)/1000//No of units wasted in B.O.T units
+printf('(i)Resistance of lines is %3.1f Ohms \n',R)
+printf('(ii)No. of B.O.T units consumed in 10hrs is %3.2f B.O.T units\n',N)
+printf('(iii)No. of B.O.T units wasted in 10hrs is %3.2f B.O.T units\n',Nw)
diff --git a/2345/CH15/EX15.20/Ex15_20.sce b/2345/CH15/EX15.20/Ex15_20.sce new file mode 100755 index 000000000..10e1de2af --- /dev/null +++ b/2345/CH15/EX15.20/Ex15_20.sce @@ -0,0 +1,9 @@ +//Finding current
+//Example 15.20(pg. 405)
+clc
+clear
+//Let current in XA=I, in XY=I1, in AY=I-40, in YB=I-40+I1-60, in BX=I+I1-150.
+//By Kirchhoff's second law, in circuit XAYA I-I1=20
+// and in circuit XAYBX 25I+15I1=1950
+I1=(1950-500)/(15+25)//in Amperes
+printf('Thus the current in branch XY is I1=%2.2f Amps',I1)
diff --git a/2345/CH15/EX15.21/Ex15_21.sce b/2345/CH15/EX15.21/Ex15_21.sce new file mode 100755 index 000000000..6342c129a --- /dev/null +++ b/2345/CH15/EX15.21/Ex15_21.sce @@ -0,0 +1,12 @@ +//Finding loss
+//Example 15.21(pg. 407)
+clc
+clear
+A=30//area of hysteresis material in cm^2
+s1=0.4//scale is 1cm=0.4Wb/m^2
+s2=400// and 1cm=400AT/m
+V=1.2*(10^-3)
+f=50//frequency in Hz
+H=A*s1*s2//hysteresis loss/m^3/cycle in joules
+Hp=H*V*f//hysteresis power loss in Watts
+printf('Thus hysteresis power loss is %3.2f Watts',Hp)
diff --git a/2345/CH15/EX15.22/Ex15_22.sce b/2345/CH15/EX15.22/Ex15_22.sce new file mode 100755 index 000000000..c175f9c27 --- /dev/null +++ b/2345/CH15/EX15.22/Ex15_22.sce @@ -0,0 +1,12 @@ +//Finding energy loss
+//Example 15.22(pg. 407)
+clc
+clear
+d=7500//density of iron in kg/m^3
+w=12//weight of iron in kgm
+V=w/d//volume of iron in m^3
+f=25//frequency in Hz
+N=3600*f//number of cycle per hour
+A=300//area in joules/m^3
+E=A*V*N//Total energy loss per hour in joules
+printf('Thus total energy loss per hour is %5.2f Joules',E)
diff --git a/2345/CH15/EX15.23/Ex15_23.sce b/2345/CH15/EX15.23/Ex15_23.sce new file mode 100755 index 000000000..46a3099a1 --- /dev/null +++ b/2345/CH15/EX15.23/Ex15_23.sce @@ -0,0 +1,15 @@ +//Finding inductance and energy
+//Example 15.23(pg. 407)
+clc
+clear
+l=0.5//length of coil in meters
+d=0.1//diameter of coil
+N=1500//no of turns of coil
+a=%pi*(d^2)/4//cross sectional area of coil in m^2
+Ur=1//relative permeability
+Uo=4*%pi*(10^-7)//permeability
+I=8//current in A
+L=((N^2)*a*Uo*Ur)/l//self inductance of coil in H
+E=(1/2)*L*(I^2)//Energy stored in Joules
+printf('Thus Self Inductance of coil is %2.3f H\n',L)
+printf('and Energy stored is %1.2f Joules',E)
diff --git a/2345/CH15/EX15.24/Ex15_24.sce b/2345/CH15/EX15.24/Ex15_24.sce new file mode 100755 index 000000000..10c99412c --- /dev/null +++ b/2345/CH15/EX15.24/Ex15_24.sce @@ -0,0 +1,15 @@ +//Finding flux and field strength
+//Example 15.24(pg. 408)
+clc
+clear
+N=600//number of turns on the coil
+I=2//current passing through solenoid in A
+l=0.6//length of solenoid in meter
+H=N*I/l//magnetic field at the centre in AT/m
+Ur=1//relative permeability
+Uo=4*%pi*(10^-7)//permeability
+d=0.025//diameter in meters
+a=%pi*(d^2)/4//cross sectional area of coil in m^2
+phi=Uo*Ur*H*a//flux in Wb
+printf('Thus Magenetic field at centre is %3.2f AT/m',H)
+printf('\n and Flux is %e Wb',phi)
diff --git a/2345/CH15/EX15.25/Ex15_25.sce b/2345/CH15/EX15.25/Ex15_25.sce new file mode 100755 index 000000000..92aeac097 --- /dev/null +++ b/2345/CH15/EX15.25/Ex15_25.sce @@ -0,0 +1,11 @@ +//Finding Ampere-Turns
+//Example 15.25(pg. 408)
+clc
+clear
+Ur=1//relative permeability
+B=1.257//flux density in Wb/m^2
+Uo=4*%pi*(10^-7)//permeability
+H=B/(Uo*Ur)//magnetising force in AT/m
+l=0.004//length of air gap in meter
+AT=H*l//AT required for the air gap
+printf('Thus AT required for the air gap is %3.1f ',AT)
diff --git a/2345/CH15/EX15.26/Ex15_26.sce b/2345/CH15/EX15.26/Ex15_26.sce new file mode 100755 index 000000000..8e9c29e99 --- /dev/null +++ b/2345/CH15/EX15.26/Ex15_26.sce @@ -0,0 +1,14 @@ +//Finding flux
+//Example 15.26(pg. 409)
+clc
+clear
+D=0.3//diameter of anchor ring in m
+l=%pi*D//length of iron ring in m
+N=400//number of turns on the iron ring
+a=0.0012//area of cross section of iron path in m^2
+Ur=1000//relative permeability
+Uo=4*%pi*(10^-7)//permeability
+I=2//current in A
+phi=(N*I)/(l/(Uo*Ur*a))//flux through iron path in WB
+phi1=phi/(10^-3)//flux in mWb
+printf('Thus flux through iron path is %2.2f mWb',phi1)
diff --git a/2345/CH15/EX15.27/Ex15_27.sce b/2345/CH15/EX15.27/Ex15_27.sce new file mode 100755 index 000000000..eb528372d --- /dev/null +++ b/2345/CH15/EX15.27/Ex15_27.sce @@ -0,0 +1,22 @@ +//Finding magnetising current
+//Example 15.27(pg. 409)
+clc
+clear
+a=0.01//crosssectional area of ring in m^2
+Uo=4*(%pi)*(10^-7)//absolute permeability
+lf=1.25//leakage factor
+Ur=400//permeability
+N=175//no of turns
+phig=0.8*(10^-3)//flux through air gap in Wb
+Bg=phig/a//Flux density in air gap in Wb/m^2
+Hg=Bg/Uo//magnetising force in air gap in AT/m
+Lg=0.004//length of air gap in m
+ATg=Hg*Lg//AT required for air gap in AT
+phii=phig*lf//flux through iron path in Wb
+Bi=phii/a//Flux density in iron path in Wb/m^2
+Hi=Bi/(Uo*Ur)//magnetising force in iron path in AT/m
+Li=1.5//length of iron path in m
+ATi=Hi*Li//At required for iron path in AT
+AT=ATi+ATg//total AT required
+I=ATg/N//Magnetising current required in A
+printf('Thus the magnetising current required is %2.2f Amps',I)
diff --git a/2345/CH15/EX15.28/Ex15_28.sce b/2345/CH15/EX15.28/Ex15_28.sce new file mode 100755 index 000000000..576f3fdfa --- /dev/null +++ b/2345/CH15/EX15.28/Ex15_28.sce @@ -0,0 +1,12 @@ +//Finding charge and capacity
+//Example 15.28(pg. 411)
+clc
+clear
+SI=0.2//steady current in A
+t=0.2//time in sec
+Q=SI*t//charge given to condenser in Coulomb
+V=220//PD across condenser in Volts
+C=Q/V//Capacitance of condenser in F
+C1=C*(10^6)//Capacitance in mircoF
+printf('Thus the Charge of condenser is %2.2f Coulomb\n',Q)
+printf('And the Capacitance of condenser is %3.2f microF',C1)
diff --git a/2345/CH15/EX15.29/Ex15_29.sce b/2345/CH15/EX15.29/Ex15_29.sce new file mode 100755 index 000000000..09702de7e --- /dev/null +++ b/2345/CH15/EX15.29/Ex15_29.sce @@ -0,0 +1,9 @@ +//Finding heat
+//Example 15.29(pg. 411)
+clc
+clear
+C=2*(10^-6)//capacitance of condenser in F
+V=10000//PD across condenser in Volts
+E=(1/2)*C*(V^2)//energy stored in condenser in Joules
+H=E/4.2//heat produced in the wire in calories
+printf('Thus heat produced in the wire is %2.2f calories',H)
diff --git a/2345/CH15/EX15.3/Ex15_3.sce b/2345/CH15/EX15.3/Ex15_3.sce new file mode 100755 index 000000000..7fca59d5e --- /dev/null +++ b/2345/CH15/EX15.3/Ex15_3.sce @@ -0,0 +1,15 @@ +//Finding current
+//Example 15.3(pg 393)
+clc
+clear
+M=250000//weight of water lifted per hr in kg
+h=50//height in metres
+g=9.81//gravitational const.
+WD=M*h*g//work done by pump per hr in watt-sec
+P=WD/3600//Power output of pump per sec in watts
+V=500//supply voltage in volts
+Ep=0.8//efficiency of pump
+Em=0.9//efficiency of motor
+E=Em*Ep//overall efficiency
+I=P/(V*E)//current in amperes
+printf('Current drawn by the motor is %3.2f Amperes',I)
diff --git a/2345/CH15/EX15.30/Ex15_30.sce b/2345/CH15/EX15.30/Ex15_30.sce new file mode 100755 index 000000000..c33ebde03 --- /dev/null +++ b/2345/CH15/EX15.30/Ex15_30.sce @@ -0,0 +1,14 @@ +//Finding K and flux density
+//Example 15.30(pg. 411)
+clc
+clear
+V=15*(10^3)//potential difference applied in V
+A=0.02//surface area of plate in m^2
+d=0.001//distance between plates in m
+C=4.5*(10^-10)//Capacitance of capacitor in F
+Ko=8.854*(10^-12)//constant
+K=(C*d)/(Ko*A)//dielectric constant
+q=C*V//charge on condenser in C
+D=q/A//Electric flux density in C/m^2
+printf('Thus the Charge of condenser is %e Coulomb\n',q)
+printf('And the electric flux density of condenser is %e microF',D)
diff --git a/2345/CH15/EX15.31/Ex15_31.sce b/2345/CH15/EX15.31/Ex15_31.sce new file mode 100755 index 000000000..c96145cbb --- /dev/null +++ b/2345/CH15/EX15.31/Ex15_31.sce @@ -0,0 +1,17 @@ +//Finding resistance
+//Example 15.31(pg. 412)
+clc
+clear
+m=0.6//mass of water in kgm
+S=4200//specific heat of water
+T1=100//temperature in degreeC
+T2=10//temperature in degreeC
+t=5*60//time in sec
+V=230//Supply voltage in Volts
+H=m*S*(T1-T2)//Heat required to raise the temp of water from 0 to 100 degree. in J
+e=0.78//efficiency of kettle
+Ei=H/e//Energy input in Joules
+Ei1=Ei/(100*3600)//Energy input in kWh
+W=Ei/t//Rating of kettle in watts
+R=(V*V)/W//Resistance of heating element in ohms
+printf('Thus Resistance of heating element is %2.1f ohms',R)
diff --git a/2345/CH15/EX15.32/Ex15_32.sce b/2345/CH15/EX15.32/Ex15_32.sce new file mode 100755 index 000000000..79e15dfcd --- /dev/null +++ b/2345/CH15/EX15.32/Ex15_32.sce @@ -0,0 +1,18 @@ +//Finding time
+//Example 15.32(pg. 413)
+clc
+clear
+m1=120//mass of water to be heated in kg
+m2=20//mass of copper tank in kg
+S1=1//specific heat of water
+S2=0.095//specific heat of copper
+T1=10//temp in degreeC
+T2=60//temp in degreeC
+H=(m1*S1*(T2-T1))+(m2*S2*(T2-T1))//heat required to raise the temp of water and tank in kcal
+H1=H*4200//heat required in Joules
+e=0.8//thermal efficiency
+E=H1/e//Energy input in joules
+E1=E/(1000*3600)//energy input in kWh
+r=3//rating of heater in kW
+t=E1/r//time taken in hours
+printf('Thus the time taken to raise the temp is %2.3f hours',t)
diff --git a/2345/CH15/EX15.33/Ex15_33.sce b/2345/CH15/EX15.33/Ex15_33.sce new file mode 100755 index 000000000..624932d3d --- /dev/null +++ b/2345/CH15/EX15.33/Ex15_33.sce @@ -0,0 +1,10 @@ +//Finding frequency
+//Example 15.33(pg. 414)
+clc
+clear
+rho=5*(10^-5)//specific resistance for steel in ohm-cm
+U=1//relative permeability
+d=0.15//depth of penetration in cm
+f=(rho*(10^9))/(U*d*d*4*(%pi^2))//frequency required in cycles per sec
+f1=f/1000//frquency in k.cycles/sec
+printf('Thus the frequency required is %3.3f k.cycles/sec',f1)
diff --git a/2345/CH15/EX15.34/Ex15_34.sce b/2345/CH15/EX15.34/Ex15_34.sce new file mode 100755 index 000000000..dd4721118 --- /dev/null +++ b/2345/CH15/EX15.34/Ex15_34.sce @@ -0,0 +1,30 @@ +//Finding current,voltage and power
+//Example 15.34(pg. 414)
+clc
+clear
+v=50*20*2//Volume of board to be heated in cm^3
+Mw=0.56//weight of wood in gm/cm^3
+m=Mw*v/1000//mass of wood in kgm
+S=0.35//specific heat of wood
+t=15/60//time in hrs
+f=30*(10^6)//frequency in cycles/sec
+t2=150,t1=30//temp in degreeC
+H=m*S*(t2-t1)//heat required to raise the temp in kcal
+Hw=H*1000/860//heat required in kW
+P=Hw/t//power required in Watts
+e=0.5//efficiency of dielectric heating process
+Pi=P/e//power input required in Watts
+Ko=8.854*(10^-12)//absolute permittivity
+K=5//relative permittivity
+A=0.5*0.2//area in m
+i=0.02
+C=Ko*K*A/i//capacitance of parallel plate capacitor in F
+Xc=1/(2*%pi*f*C)//capacitive reactance in ohms
+cosx=0.05
+tanx=19.97
+R=Xc*tanx//resistance
+V=sqrt(Pi*R)//voltage in volts
+Ic=V/Xc//current through the board in Amps
+printf('Thus the power required is %2.1f Watts\n',Pi)
+printf('And Voltage across the board is %3.2f volts\n',V)
+printf('And the current through the board is %2.3f Amps',Ic)
diff --git a/2345/CH15/EX15.35/Ex15_35.sce b/2345/CH15/EX15.35/Ex15_35.sce new file mode 100755 index 000000000..ee2dd5bc5 --- /dev/null +++ b/2345/CH15/EX15.35/Ex15_35.sce @@ -0,0 +1,14 @@ +//Finding efficiency
+//Example 15.35(pg. 416)
+clc
+clear
+m=2//quantity of aluminium to be melted in kg
+t1=15,t2=660//temp in degreeC
+S=0.212//specific heat of aluminium
+L=78.8//latent heat of aluminium in kcal/kg
+H=(m*S*(t2-t1))+(m*L)//total heat required to melt Al in kcal
+i=5//input to furnace in kW
+E=i*(1000*10*60)//Energy input to furnace in watt-sec
+E1=E/4180//energy input in kcal
+e=H*100/E1//efficiency of furnace
+printf('Thus the efficiency of furnace is %2.3f percent',e)
diff --git a/2345/CH15/EX15.36/Ex15_36.sce b/2345/CH15/EX15.36/Ex15_36.sce new file mode 100755 index 000000000..782d77e04 --- /dev/null +++ b/2345/CH15/EX15.36/Ex15_36.sce @@ -0,0 +1,12 @@ +//Finding cost
+//Example 15.36(pg. 417)
+clc
+clear
+O=5*735.5//output of motor in W
+e=0.85//efficiency of motor
+c=2//cost of energy per unit in Rs
+I=O/e//input of motor in Watts
+t=4//time in hrs
+E=I*t/1000//energy consumed in kWh
+C=c*E//cost of using the motor in Rs
+printf('Thus the cost of using the motor is %2.3f Rs',C)
diff --git a/2345/CH15/EX15.37/Ex15_37.sce b/2345/CH15/EX15.37/Ex15_37.sce new file mode 100755 index 000000000..6d6faf91f --- /dev/null +++ b/2345/CH15/EX15.37/Ex15_37.sce @@ -0,0 +1,10 @@ +//Finding no of electrons
+//Example 15.37(pg. 417)
+clc
+clear
+I=2.5*(10^-3)//current in Amp
+t=30*(10^-3)//time in sec
+Q=I*t//charge passing through the person in Coulumbs
+e=1.602*(10^-19)//charge of 1 electron in C
+N=Q/e//no of electrons passing through the person
+printf('Thus the no of electrons passing through the person is %e electrons',N)
diff --git a/2345/CH15/EX15.38/Ex15_38.sce b/2345/CH15/EX15.38/Ex15_38.sce new file mode 100755 index 000000000..b28bf1fb8 --- /dev/null +++ b/2345/CH15/EX15.38/Ex15_38.sce @@ -0,0 +1,15 @@ +//Finding resistance
+//Example 15.38(pg. 417)
+clc
+clear
+//(a)Finding resistance between 2 ends
+l=1//length in m
+a=2.5*(10^-2)*0.05*(10^-2)//area of cross section in m^2
+rho=1.724*(10^-8)//specific resistance of copper in ohm-m
+R=rho*l/a//resistance of the strip in ohm
+//(b) Finding resistance between 2 faces
+l1=0.05*(10^-2)//length in m
+a1=2.5*(10^-2)*1//area of cross section in m^2
+R1=rho*l1/a1//resistance in ohm
+printf('Thus the resistance of the strip is %e ohms\n ',R)
+printf('And the resistance between the faces is %e ohms',R1)
diff --git a/2345/CH15/EX15.39/Ex15_39.sce b/2345/CH15/EX15.39/Ex15_39.sce new file mode 100755 index 000000000..8a9fb6d82 --- /dev/null +++ b/2345/CH15/EX15.39/Ex15_39.sce @@ -0,0 +1,20 @@ +//Finding resistance and cost
+//Example 15.39(pg. 418)
+clc
+clear
+m=2//weight of water to be heated in kg
+t2=98,t1=15//temp in degreeC
+s=1//specific heat of water
+V=200//voltage in volts
+H=m*s*(t2-t1)//energy required to raise the temp of water in kcal
+H1=H*4200//energy in Watt-sec or Joules
+e=0.85//efficiency of kettle
+E=H1/e//energy input required in watt-sec
+E1=E/(1000*3600)//energy input in kWh
+c=35//cost per unit in paise
+C=c*E1//ocst of energy used in paise
+t=10/60//time in hrs
+W=E1*1000/t//wattage of kettle in watts
+R=V*V/W//resistance of heating element in ohms
+printf('Thus the resistance of heating element is %2.0f ohms\n',R)
+printf('And the cost of energy used is %2.0f paisa',C)
diff --git a/2345/CH15/EX15.4/Ex15_4.sce b/2345/CH15/EX15.4/Ex15_4.sce new file mode 100755 index 000000000..6001db4b3 --- /dev/null +++ b/2345/CH15/EX15.4/Ex15_4.sce @@ -0,0 +1,15 @@ +//Finding torque
+//Example 15.4(pg 394)
+clc
+clear
+P=10//Power developed by motor in H.P
+N=600//Speed of motor in rpm
+//1HP=735.5Nw-m/sec=75kgm/sec
+a=75
+b=735.5
+//Torque in kg-m
+Tkgm=(P*a*60)/(2*%pi*N)//since P=2*pi*NT/60
+//Torque in Nw-m
+TNwm=(P*b*60)/(2*%pi*N)//since P=2*pi*NT/60
+printf('(i)Torque in kg.meter is %3.2f kg-m \n',Tkgm)
+printf('(ii)Torque in Newton.meter is %3.2f Nw-m',TNwm)
diff --git a/2345/CH15/EX15.40/Ex15_40.sce b/2345/CH15/EX15.40/Ex15_40.sce new file mode 100755 index 000000000..cf87200c5 --- /dev/null +++ b/2345/CH15/EX15.40/Ex15_40.sce @@ -0,0 +1,18 @@ +//Finding current
+//Example 15.40(pg. 418)
+clc
+clear
+phi=70000/(10^8)//flux to be set up in Wb since 10^8lines =1Wb
+d=0.03//diameter in m
+a=%pi*d*d/4//area of cross section in m^2
+B=phi/a//flux density in Wb/m^2
+Lg=0.002//length of air gap in m
+Ls=(%pi*0.2)-Lg//length of steel path
+Uo=4*%pi*(10^-7)//absolute permitivity
+Ur=800//relative permitivity of steel
+Hg=B/Uo
+Hs=B/(Uo*Ur)
+AT=(Hg*Lg)+(Hs*Ls)//total ampere turns required
+N=500// no of turns
+I=AT/N//exciting current in amps
+printf('Thus the value of exciting current is %2.3f A',I)
diff --git a/2345/CH15/EX15.5/Ex15_5.sce b/2345/CH15/EX15.5/Ex15_5.sce new file mode 100755 index 000000000..38560dd52 --- /dev/null +++ b/2345/CH15/EX15.5/Ex15_5.sce @@ -0,0 +1,14 @@ +//finding mass and energy
+//Example 15.5(pg 395)
+clc
+clear
+P=25//Output of diesel engine in kW
+s=12500//calorific value of fuel oil in k-cal/kgm
+e=0.35//overall efficiency of diesel set
+P1=P/e//input energy required in 1 hour in kWh
+P2=P1*860//input energy in kcal
+m=P2/s//mass of oil needed per hr in kgm
+w=1000//weight of 1 ton of oil in kgm
+Eg=(P*w)/m//Energy generated by 1ton of oil in kWh
+printf('(i)Mass of oil required per hr is %3.3f kgm \n',m)
+printf('(ii)Eletrical energy generated per ton of fuel is %4.1f Kwh',Eg)
diff --git a/2345/CH2/EX2.1/Ex2_1.sce b/2345/CH2/EX2.1/Ex2_1.sce new file mode 100755 index 000000000..1274c2d91 --- /dev/null +++ b/2345/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,15 @@ +//Finding resistance
+//Example 2.1(pg. 21)
+clc
+clear
+l=300//in meters
+a=25*(10^-6)//in meter square
+d15=2.7//density at 15 degree C in ohm-meter
+R15=d15*(l/a)
+printf('The value of Resistance at 15 degree C is %3.2f.ohms \n',R15)
+k0=0.004//temp coefficient in ohm/degree C at 0 degree C
+t=15,T=50//in degree C
+k15=k0/(1+(k0*t))
+R50=R15*(1+k15*(T-t))
+printf('The value of Resistance at 50 degree C is %3.2f.ohms',R50)
+
diff --git a/2345/CH2/EX2.10/Ex2_10.sce b/2345/CH2/EX2.10/Ex2_10.sce new file mode 100755 index 000000000..041d0a76c --- /dev/null +++ b/2345/CH2/EX2.10/Ex2_10.sce @@ -0,0 +1,12 @@ +//Finding resistance and temperature
+//Example 2.10(pg. 25)
+clc
+clear
+R15=50,RT=58// resistance in ohms
+t=15// te mp in degree C
+k0=0.00425// temp coefficient at 0 degree C
+R0=R15/[1+(k0*t)]// resistance at 0 degree C in ohms
+T=[(RT/R0)-1]/k0// temp in degree C
+
+printf('The value of Resistance at 0 degree C is %3.1f ohms \n',R0)
+printf('The value of Temperature at 58 ohm resistance is %3.4f degree C',T)
diff --git a/2345/CH2/EX2.11/Ex2_11.sce b/2345/CH2/EX2.11/Ex2_11.sce new file mode 100755 index 000000000..c08774759 --- /dev/null +++ b/2345/CH2/EX2.11/Ex2_11.sce @@ -0,0 +1,9 @@ +//Finding temperature coefficient
+//Example 2.11(pg. 25)
+clc
+clear
+R25=50,R70=57.2// resistance in ohms
+t=25,T=70// temp in degree C
+//since Rt=R0(1+(k0*t))
+k0=(R70-R25)/[(R25*T)-(R70*t)]
+printf('The temp coefficient at 0 degree C is %3.3f',k0 )
diff --git a/2345/CH2/EX2.12/Ex2_12.sce b/2345/CH2/EX2.12/Ex2_12.sce new file mode 100755 index 000000000..0c618021f --- /dev/null +++ b/2345/CH2/EX2.12/Ex2_12.sce @@ -0,0 +1,11 @@ +//Finding resistance and conductivity
+//Example 2.12(pg. 26)
+clc
+clear
+R0=15.5// resistance in ohms
+t=16//in degree C
+k0=0.00428//temp coefficient
+R16=R0*[1+(k0*t)]
+G=(R0/R16)*100// since conductance=reciprocal of resistance
+printf('The value of Resistance at 16 degree C is %3.4f ohms \n',R16)
+printf('The value of percentage conductivity at 16 degree C is %3.2f percent',G)
diff --git a/2345/CH2/EX2.13/Ex2_13.sce b/2345/CH2/EX2.13/Ex2_13.sce new file mode 100755 index 000000000..9f1832bb5 --- /dev/null +++ b/2345/CH2/EX2.13/Ex2_13.sce @@ -0,0 +1,9 @@ +//finding temperature
+//Example 2.13(pg. 26)
+clc
+clear
+RT=144,R20=10// in ohms
+t=20// in degree C
+k20=5*(10^-3)//temp coefficient at 20 degree C
+T={[(RT/R20)-1]/k20}+t
+printf('The value of temp required for tungsten bulb is %4.2f degree C',T)
diff --git a/2345/CH2/EX2.14/Ex2_14.sce b/2345/CH2/EX2.14/Ex2_14.sce new file mode 100755 index 000000000..02826e042 --- /dev/null +++ b/2345/CH2/EX2.14/Ex2_14.sce @@ -0,0 +1,16 @@ +//Finding temperature
+//Example 2.14(pg. 27)
+clc
+clear
+V15=250,Vt=250//voltage in volts
+I15=5,It=4//current in amperes
+T=15//temp in degree C
+R15=V15/I15//resistance in ohms at 15 degreeC
+printf('Resistance at 15 degree C is %3.1f ohms \n',R15)
+Rt=Vt/It//resistance at t degreeC
+printf('Resistance at t degree C is %3.1f ohms \n',Rt)
+k0=0.0038
+R0=R15/[1+(k0*T)]
+printf('Resistance at 0 degree C is %3.2f ohms \n',R0)
+t=[(Rt/R0)-1]/k0
+printf('Temperature t is %3.2f degree C',t)
diff --git a/2345/CH2/EX2.15/Ex2_15.sce b/2345/CH2/EX2.15/Ex2_15.sce new file mode 100755 index 000000000..e7778043c --- /dev/null +++ b/2345/CH2/EX2.15/Ex2_15.sce @@ -0,0 +1,18 @@ +//Finding resistance
+//Example 2.15(pg. 28)
+clc
+clear
+n=100//no of slots
+c=12//conductors per slot
+Lm=300// mean length of turn in cm
+a=1.5*0.2//cross section of each conductor in cm^2
+s=1.72*(10^-6)//specific resistance of copper at 20 degreeC
+p=4// poles
+t=20,T=75//temp in degreeC
+k0=0.00427//temp coefficient of resistivity for copper
+L=n*c*Lm//total length of conductors
+Ls=L/p//length of conductors in each parallel path
+s0=s*(1-(k0*t))
+printf('Thus specific resistance at 0 degree C is %e ohm-cm \n',s0)
+RT=(s0*Ls)/a
+printf('Thus resistance at working temp of 75 degree C is %3.4f ohm',RT)
diff --git a/2345/CH2/EX2.16/Ex2_16.sce b/2345/CH2/EX2.16/Ex2_16.sce new file mode 100755 index 000000000..b0a5b5484 --- /dev/null +++ b/2345/CH2/EX2.16/Ex2_16.sce @@ -0,0 +1,12 @@ +//Finding resistance
+//Example 2.16(pg. 28)
+clc
+clear
+a=15//cross section area in cm^2
+l=100000//length in cm
+p0=7.6*(10^-6)//specific resistance at 0 degree C in ohm-cm
+k0=0.005//temp coefficient at 0 degree C
+t=50//temp in degree C
+p50=p0*[1+(t*k0)]//resistivity at 50 degree C
+R50=p50*(l/a)
+printf('Thus resistance at 50 degree C is %3.5f ohms \n',R50)
diff --git a/2345/CH2/EX2.17/Ex2_17.sce b/2345/CH2/EX2.17/Ex2_17.sce new file mode 100755 index 000000000..d1b59c95c --- /dev/null +++ b/2345/CH2/EX2.17/Ex2_17.sce @@ -0,0 +1,8 @@ +//Finding fusing current
+//Example 2.17(pg. 29)
+clc
+clear
+I2=27.5//current of No.25 wire in Amperes
+d=1/2//since I1/I2=1/2
+I1=I2*(d^(3/2))
+printf('Thus fusing current of No.33 wire is %3.3f amperes \n',I1)
diff --git a/2345/CH2/EX2.18/Ex2_18.sce b/2345/CH2/EX2.18/Ex2_18.sce new file mode 100755 index 000000000..7e95c93ba --- /dev/null +++ b/2345/CH2/EX2.18/Ex2_18.sce @@ -0,0 +1,15 @@ +//Finding ratios
+//Example 2.18(pg. 30)
+clc
+clear
+sAl=2.85*(10^-6),sCu=1.7*(10^-6)//specific resistance in ohm-cm
+gAl=2.71,gCu=8.89//specific gravity
+cAl=5000,cCu=10000//cost per tonne
+//P=V^2/R, power is same for both so resistance must also be same
+//so R=(p*l)/(pi*d^2)=(p*l)/(pi*d'^2)
+Kd=sqrt(sAl/sCu)//Kd=d/d'
+printf('Thus the ratio of diameters is %3.3f \n',Kd)
+Km=(Kd^2)*(gAl/gCu)
+printf('Thus the ratio of weights is %3.4f \n',Km)
+Kc=Km*(cAl/cCu)
+printf('Thus the ratio of costs is %3.4f',Kc)
diff --git a/2345/CH2/EX2.19/Ex2_19.sce b/2345/CH2/EX2.19/Ex2_19.sce new file mode 100755 index 000000000..aa0e60d7d --- /dev/null +++ b/2345/CH2/EX2.19/Ex2_19.sce @@ -0,0 +1,10 @@ +//Finding resistance
+//Example 2.19(pg. 33)
+clc
+clear
+R1=18.6//resistacne in ohms
+Kl=5//since l2=5*l1
+Ka=3// since a2=3*a1
+R2=R1*Kl/Ka
+// resistivity is same because wires are of same material
+printf('Thus the resistance of another conductor is %3.1f ohms',R2)
diff --git a/2345/CH2/EX2.2/Ex2_2.sce b/2345/CH2/EX2.2/Ex2_2.sce new file mode 100755 index 000000000..9fc2b6a95 --- /dev/null +++ b/2345/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,10 @@ +//Finding resistance
+//Example 2.2(pg. 21)
+clc
+clear
+R20=400// in ohms
+k0=0.0038
+t=20,T=80//degree C
+k1=k0/(1+(k0*t))
+R80=R20*{1+k1*(T-t)}
+printf('The value of Resistance at 80 degree C is %3.4f ohms',R80)
diff --git a/2345/CH2/EX2.20/Ex2_20.sce b/2345/CH2/EX2.20/Ex2_20.sce new file mode 100755 index 000000000..3e741fb5e --- /dev/null +++ b/2345/CH2/EX2.20/Ex2_20.sce @@ -0,0 +1,15 @@ +//Finding heat efficiency
+//Example 2.20(pg. 57)
+clc
+clear
+m=1//mass in kg
+S=4200//specific heat of water
+T2=100,T1=15// temp in degree C
+H=m*S*(T2-T1)//heat utilised in J
+printf('Heat utilised is %6.2f Joules \n',H)
+W=500//wattage rating of kettle in volts
+t=15*60// time in sec
+Hd=W*t//heat developed in J
+printf('Heat developed is %6.2f Joules \n',Hd)
+He=(H/Hd)*100//Heat efficiency
+printf('Thus heat efficiency is %3.2f percent',He)
diff --git a/2345/CH2/EX2.21/Ex2_21.sce b/2345/CH2/EX2.21/Ex2_21.sce new file mode 100755 index 000000000..9a6263ced --- /dev/null +++ b/2345/CH2/EX2.21/Ex2_21.sce @@ -0,0 +1,15 @@ +//Finding time
+//Example 2.21(pg. 58)
+clc
+clear
+m=3.6//mass in kg
+S=4200//specific heat of water
+T2=95,T1=15// temp in degree C
+H=m*S*(T2-T1)//heat utilised in J
+printf('Heat utilised is %7.2f Joules \n',H)
+e=0.84//efficiency of kettle
+Ei=H/e//Energy input in J
+printf('Energy input is %8.2f Joules \n',Ei)
+W=1000//rating of kettle in watts
+t=(Ei/W)/60//time taken in min
+printf('Thus time taken is %2.1f min \n',t)
diff --git a/2345/CH2/EX2.3/Ex2_3.sce b/2345/CH2/EX2.3/Ex2_3.sce new file mode 100755 index 000000000..8a2719976 --- /dev/null +++ b/2345/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,10 @@ +//Finding temperature
+//Example 2.3(pg. 22)
+clc
+clear
+t=15//degree C
+R15=250,RT=300//ohms
+k0=0.0038//ohm/degree C
+k1=k0/(1+(k0*t))
+T=[{(RT/R15)-1}/k1]+t//since RT=R15{1+k1*(T-t)}
+printf('The value of Temperature at 300 ohm resistance is %3.1f degree C',T)
diff --git a/2345/CH2/EX2.4/Ex2_4.sce b/2345/CH2/EX2.4/Ex2_4.sce new file mode 100755 index 000000000..c09a15210 --- /dev/null +++ b/2345/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,20 @@ +//Finding length
+//Example 2.4(pg. 22)
+clc
+clear
+//Part (a)
+d=0.4*(10^-3)//diameter in meter
+a=%pi*(d^2)/4//area in meter square
+p1=100*(10^-8)//resistivity of nichrome in ohm-meter
+R=40//resistance in ohms
+l1=R*a/p1
+printf('Thus the length of heater element with nichrome wire is %2.1f meter \n',l1)
+
+//Part(b)
+d=0.4*(10^-3)//diameter in meter
+a=12.6*(10^-8)//area in meter square
+p2=1.72*(10^-8)//resistance of copper wire in ohm-meter
+R=40//resistance in ohms
+l2=R*a/p2
+printf('Thus the length of heater element with copper wire is %2.1f meter',l2)
+
diff --git a/2345/CH2/EX2.5/Ex2_5.sce b/2345/CH2/EX2.5/Ex2_5.sce new file mode 100755 index 000000000..e780e415d --- /dev/null +++ b/2345/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,9 @@ +//Finding resistance
+//Example 2.5(pg. 23)
+clc
+clear
+R0=80//in ohms
+t=40// in degree C
+k0=0.0043
+R40=R0*(1+(k0*t))
+printf('The value of Resistance at 40 degree C is %3.2f ohms',R40)
diff --git a/2345/CH2/EX2.6/Ex2_6.sce b/2345/CH2/EX2.6/Ex2_6.sce new file mode 100755 index 000000000..72063bf9c --- /dev/null +++ b/2345/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,10 @@ +//Finding temperature coefficient
+//Example 2.6(pg. 23)
+clc
+clear
+R80=50,R28=40// resistance in ohms
+t=28,T=80// temp in degrees
+k28=[(R80/R28)-1]/(T-t)//since RT=Rt{1+k*(T-t)}
+printf('The value of Temperature coefficient at 28 degree C is %3.4f ohms per degree C \n',k28)
+k0=k28/(1-k28*t)// since k28=k0/(1+k0*t)
+printf('The value of Temperature coefficient at 0 degree C is %3.4f ohms per degree C',k0)
diff --git a/2345/CH2/EX2.7/Ex2_7.sce b/2345/CH2/EX2.7/Ex2_7.sce new file mode 100755 index 000000000..e71b45ed5 --- /dev/null +++ b/2345/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,10 @@ +//Finding resistance
+//Example 2.7(pg. 24)
+clc
+clear
+l=1000// length in meters
+d=0.09/100// diameter in meters
+p=1.724*(10^-8)// specific resistance in ohm meter
+a=%pi*(d^2)/4// area in meter square
+R=p*l/a//resistance in ohms
+printf('The value of Resistance is %3.2f ohms',R)
diff --git a/2345/CH2/EX2.8/Ex2_8.sce b/2345/CH2/EX2.8/Ex2_8.sce new file mode 100755 index 000000000..506db5849 --- /dev/null +++ b/2345/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,11 @@ +//Finding resistance
+//Example 2.8(pg. 24)
+clc
+clear
+R20=50// resistance in ohms
+T=60,t=20// temp in degree C
+k0=0.00427//temp coefficient at zero degreeC
+R0=R20/{1+(k0*t)}
+printf('The value of Resistance at 0 degree C is %3.2f ohms \n',R0)
+R60=R0*{1+(k0*T)}
+printf('The value of Resistance at 60 degree C is %3.2f ohms',R60)
diff --git a/2345/CH2/EX2.9/Ex2_9.sce b/2345/CH2/EX2.9/Ex2_9.sce new file mode 100755 index 000000000..b1b9ea4a1 --- /dev/null +++ b/2345/CH2/EX2.9/Ex2_9.sce @@ -0,0 +1,12 @@ +//Finding resistivity and temp coefficient
+//Example 2.9(pg. 24)
+clc
+clear
+k20=1/254.5// temperature coefficient at 20 degreeC
+p0=1.6*(10^-6)// resistivity at 0 degree C in ohm-cm
+t=20,T=50//temp in degree C
+k0=k20/(1-(t*k20))//temperature coefficient at 0 degreeC
+p50=p0*[1+(T*k0)]// resistivity at 50 degree C in ohm-cm
+k50=1/[T+(1/k0)]//temperature coefficient at 50 degreeC
+printf('Thus the temperature coefficient at 50 degree C is %3.4f \n',k0)
+printf('Thus the resistivity at 50 degree C is %e in ohm-cm',p50)
diff --git a/2345/CH4/EX4.1/Ex4_1.sce b/2345/CH4/EX4.1/Ex4_1.sce new file mode 100755 index 000000000..a1fb9414d --- /dev/null +++ b/2345/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,14 @@ +//Finding capacitance +//Example 4.1(pg 110) +clc +clear +// Let C1 and C2 be unknown capacities +//C1+C2=0.16 +//(C1*C2)/(C1 + C2)=0.03 +// from the above 2 equations we get the following polynomial +s=poly(0,"s"); +p=s^2 -0.16*s +0.0048 +[c1]=roots(p) +c2=0.16-c1 +printf('Thus the capacitance of condensers is %3.2f microF \n ',c1) + diff --git a/2345/CH4/EX4.2/Ex4_2.sce b/2345/CH4/EX4.2/Ex4_2.sce new file mode 100755 index 000000000..eb3f0efe0 --- /dev/null +++ b/2345/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,13 @@ +//Finding capacitance
+//Example 4.2(pg 110)
+clc
+clear
+n=9;
+Ko=8.854*10^-12;
+K=5;
+A=12*10^-4;
+d=2*10^-4;
+
+C=(n-1)*Ko*K*A/d
+printf('Thus the capacitance is %e F',C);
+//The Answer in the Textbook has a calculation error, hence it doesn't match the answer here.
diff --git a/2345/CH4/EX4.3/Ex4_3.sce b/2345/CH4/EX4.3/Ex4_3.sce new file mode 100755 index 000000000..217d9fc51 --- /dev/null +++ b/2345/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,13 @@ +//Finding heat
+//Example 4.3(pg 110)
+clc
+clear
+C=10^-6
+V=10000
+//here C is capacitance and V voltage
+E=1/2*C*V^2
+//E is the energy stored in the capacitor
+// when the capacitor is discharged all this energy is dissipated as heat in the wire
+H=E/4.2
+//H is heat produced in calories since 4.2 Joules=1 calorie
+printf('Thus the heat produced is %3.4f calories',H)
diff --git a/2345/CH4/EX4.4/Ex4_4.sce b/2345/CH4/EX4.4/Ex4_4.sce new file mode 100755 index 000000000..91148253a --- /dev/null +++ b/2345/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,16 @@ +//Finding electric flux density
+//Example 4.4(pg 111)
+clc
+clear
+
+A=0.02;//surface area of plates in meter square
+d=0.001;//distance between the plates in meter
+C=4.5*10^-10;//capacitance of the capacitor in farad
+//for paralel plate condenser C=KoKA/d
+Ko=8.854*10^-12;
+//dielectric constant K is given by
+K=(C*d)/(Ko*A)
+V=15000;//volatage in volts
+Q=C*V// charge on condenser in columb
+D=Q/A// electric flux density in columb per meter square
+printf('Thus the electric flux density is %e C/(m^2)',D)
diff --git a/2345/CH4/EX4.5/Ex4_5.sce b/2345/CH4/EX4.5/Ex4_5.sce new file mode 100755 index 000000000..7903a3ffb --- /dev/null +++ b/2345/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,19 @@ +//Finding relative permittivity
+//Example 4.5(pg 111)
+clc
+clear
+//before inserting the second sheet
+d=0.003;//distacne between plates in m^2
+K1=6;// relative permittivity of air
+Ko=8.854*10^-12;
+// capacitance C1=Ko*K1*A/d in Farad
+//after inserting the second sheet
+d1=0.003;//thickness of first sheet in meter
+d2=0.005;//thickness of second sheet in meter
+//K2 is unknown
+//C2=Ko*A/(d1/K1 + d2/K2)
+// but given that C2=(1/3)*C1
+//from equations 1,2,3
+K2= (d2*K1)/(3*d-d1)
+// since Ko*A/(d1/K1 + d2/K2)=Ko*K1*A/3*d
+printf('Thus K2 is %3.4f',K2)
diff --git a/2345/CH4/EX4.6/Ex4_6.sce b/2345/CH4/EX4.6/Ex4_6.sce new file mode 100755 index 000000000..80fa89917 --- /dev/null +++ b/2345/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,14 @@ +//Finding force
+//Example 4.6(pg 113)
+clc
+clear
+q1=1;// in coulomb
+q2=1;// in coulomb
+Eo=8.854*10^-12;// in Farad per meter
+Er=1;
+d=1// in meter
+pi=3.14;
+// F is the force between 2 charges in NEWTONS
+F=(q1*q2)/(4*pi*Eo*Er*d^2)
+
+printf('Thus the force between 2 charges is %e',F)
diff --git a/2345/CH4/EX4.7/Ex4_7.sce b/2345/CH4/EX4.7/Ex4_7.sce new file mode 100755 index 000000000..fbba474bd --- /dev/null +++ b/2345/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,11 @@ +//Finding charge
+//Example 4.7(pg 114)
+clc
+clear
+//q1=q2=q
+pi=3.14;
+d=0.2;// in meters
+K=9*10^9;// here K=1/4*pi*Eo*Er constant
+F=9.81*10^-1;// in newtons or 10^-1 kgm
+q=sqrt((F*(d^2))/K)
+printf('Thus charge is %e in coulomb',q)
diff --git a/2345/CH5/EX5.1/Ex5_1.sce b/2345/CH5/EX5.1/Ex5_1.sce new file mode 100755 index 000000000..4d27b152e --- /dev/null +++ b/2345/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,15 @@ +//Finding charge and capacitance
+//Example 5.1(pg 193)
+clc
+clear
+t=0.25//time in sec
+I=0.22//Current in A
+V=220//voltage in V
+Q=I*t//charge given to condenser
+C=Q/V//capacitance of condenser
+C1=C*(10^6)
+printf('Charge given to condenser is %3.3f Coulombs \n',Q)
+printf('Capacitance of condenser is %3.4f F',C)
+printf('or %3.0f microF',C1)
+
+
diff --git a/2345/CH5/EX5.2/Ex5_2.sce b/2345/CH5/EX5.2/Ex5_2.sce new file mode 100755 index 000000000..674acd347 --- /dev/null +++ b/2345/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,11 @@ +//Finding charge and potential gradient
+//Example 5.2(pg 193)
+clc
+clear
+C=0.0002*(10^-6)//capacitance in F
+V=20000//P.D across condenser in V
+t=2//thickness in mm
+Q=C*V//charge on each plate in coulomb
+g=(V/t)*(1/1000)// potential gradient in kV/mm
+printf('Charge given to condenser is %e Coulombs \n',Q)
+printf('Potential gradient of condenser is %3.0f kV/mm',g)
diff --git a/2345/CH5/EX5.3/Ex5_3.sce b/2345/CH5/EX5.3/Ex5_3.sce new file mode 100755 index 000000000..7cccf9cae --- /dev/null +++ b/2345/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,18 @@ +//Finding charge and energy
+//Example 5.3(pg 194)
+clc
+clear
+//Before immersion of oil
+C=0.005*(10^-6)
+V=500
+q=C*V
+E=(1/2)*(C*V*V)
+printf('Charge of condenser is %e coulomb \n',q)
+printf('Energy stored in condenser before immersion of oil is %e Joules \n',E)
+
+//After immersion of oil
+K=2.5
+q1=q// since no loss of charge
+C1=K*C//capacity of condenser
+E1=(q1^2)/(2*C1)// energy stored in condenser
+printf('Energy stored in condenser after immersion of oil is %e Joules',E1)
diff --git a/2345/CH5/EX5.4/Ex5_4.sce b/2345/CH5/EX5.4/Ex5_4.sce new file mode 100755 index 000000000..a8fad58a3 --- /dev/null +++ b/2345/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,14 @@ +//dielectric constant and flux density
+//Example 5.4(pg 194)
+clc
+clear
+A=0.02//surface area of plate in m^2
+d=0.001//distance between plates in m
+C=4.5*(10^-10)//capacitance in F
+V=15000//voltage in volts
+K0=8.854*(10^-12)
+K=(C*d)/(K0*A)
+q=C*V// charge on condenser in coulombs
+D=q/A//Electric flux density in Coulomb/m^2
+printf('Thus dielectric constant is %3.2f \n',K)
+printf('Thus Electric flux density is %e Coulombs/m^2',D)
diff --git a/2345/CH5/EX5.5/Ex5_5.sce b/2345/CH5/EX5.5/Ex5_5.sce new file mode 100755 index 000000000..f719d04d9 --- /dev/null +++ b/2345/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,10 @@ +//Finding capacitance
+//Example 5.5(pg 195)
+clc
+clear
+A=0.2//surface area of plate in m^2
+t=2.5*(10^-5)//thickness of dielectric in m
+K0=8.854*(10^-12)//permittivity of air in F/m
+K=5//relative permittivity of dielectric
+C=(K*K0*A*(10^6))/t//capacitance of condenser in microF
+printf('Thus the Capacitance of condenser is %3.3f microF',C)
diff --git a/2345/CH6/EX6.1/Ex6_1.sce b/2345/CH6/EX6.1/Ex6_1.sce new file mode 100755 index 000000000..65697b159 --- /dev/null +++ b/2345/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,12 @@ +//Finding current
+//Example 6.1(pg 212)
+clc
+clear
+f=0.01//flux in Wb
+l=1//mean circumference in m
+N=1000//tunrs
+Ur=1000//relative permeability
+Uo=4*%pi*(10^-7)//permeability of free space in H/m
+a=0.001// cross section area in m^2
+I=(f*l)/(N*Uo*Ur*a)// current in Amp. since f=A*T/(l/Uo*Ur*a)
+printf('Thus Current required is %3.3f Amp',I)
diff --git a/2345/CH6/EX6.2/Ex6_2.sce b/2345/CH6/EX6.2/Ex6_2.sce new file mode 100755 index 000000000..4ccc47a66 --- /dev/null +++ b/2345/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,12 @@ +//relative permeability
+//Example 6.2(pg 212)
+clc
+clear
+f=1.2*(10^-3)//flux in Wb
+l=1.4//mean circumference in m
+N=500//tunrs
+Uo=4*%pi*(10^-7)//permeability of free space in H/m
+a=0.0012// cross section area in m^2
+I=2//current in Amp
+Ur=(f*l)/(N*I*Uo*a)//relative permeability
+printf('Thus the relative permeability of iron is %3.2f ',Ur)
diff --git a/2345/CH6/EX6.3/Ex6_3.sce b/2345/CH6/EX6.3/Ex6_3.sce new file mode 100755 index 000000000..b873a470d --- /dev/null +++ b/2345/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,17 @@ +//Flux density, field intensity and permeability
+//Example 6.3(pg 213)
+clc
+clear
+l=0.4//mean circumference in m
+N=200//tunrs
+Uo=4*%pi*(10^-7)//permeability of free space in H/m
+a=5*(10^-4)// cross section area in m^2
+I=6.4//current in Amp
+f=0.8*(10^-3)//flux in Wb
+fd=f/a//flux density in Wb/m^2
+fi=I*N/l//Field intensity in AT/m
+Ur=(f*l)/(N*I*Uo*a)//relative permeability
+printf('(i) The Flux density is %3.2f Wb/m^2 \n',fd)
+printf('(ii) The Field intensity is %3.2f AT/m \n',fi)
+printf('(iii) The Relative permeability of steel is %3.2f ',Ur)
+//The answer to part(iii) has a calculation error in the textbook, hence it doesn't match the answer here.
diff --git a/2345/CH6/EX6.4/Ex6_4.sce b/2345/CH6/EX6.4/Ex6_4.sce new file mode 100755 index 000000000..e5d5ddfe0 --- /dev/null +++ b/2345/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,10 @@ +//Finding loss
+//Example 6.4(pg 214)
+clc
+clear
+Hl=250//Hysteresis loss per m^3 in J/cycle
+V=1/150//Volume of specimen in m^3
+N=50//No of cycles/sec
+E=Hl*V*N//Energy loss per sec in J
+Eh=(E*3600)/1000//Energy loss per hour in kWh
+printf('Thus Energy loss per hour is %3.2f kWh',Eh)
diff --git a/2345/CH6/EX6.5/Ex6_5.sce b/2345/CH6/EX6.5/Ex6_5.sce new file mode 100755 index 000000000..647dd2622 --- /dev/null +++ b/2345/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,14 @@ +//Finding loss and frequency
+//Example 6.5(pg 214)
+clc
+clear
+P=4//no of poles
+N=1600// Speed in rpm
+f=P*N/120//Frequency of magnetic reversal
+V=5400//volume
+d=7.5//density
+m=(V*d)/1000//Mass of armature in kg
+L=1.76//Loss in W/kg
+Cl=L*m//Core loss in Watts
+printf('Thus Frequency of magnetic reversal is %3.2f c/s',f)
+printf('\n and Core loss is %3.2f Watts',Cl)
diff --git a/2345/CH6/EX6.6/Ex6_6.sce b/2345/CH6/EX6.6/Ex6_6.sce new file mode 100755 index 000000000..0173f3834 --- /dev/null +++ b/2345/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,16 @@ +//Finding core loss
+//Example 6.6(pg 214)
+clc
+clear
+v=76300//volume in c.c
+P=8// no of poles
+N=375//rpm
+f=P*N/120//freqency in c/s
+Bmax=12000//max. flux density in lines/cm^2
+n=0.002//(assumed)
+d=7.8//densityin gm/c.c
+l=1.7//loss in watts per kg
+Hl=n*v*f*(Bmax^1.6)*(10^-7)//Hysteresis loss in Watts
+Al=v*d*l/1000//Additional loss under particular running conditions
+Tl=Hl+Al//total core loss
+printf('Thus the total core loss is %4.0f Watts',Tl)
diff --git a/2345/CH6/EX6.7/Ex6_7.sce b/2345/CH6/EX6.7/Ex6_7.sce new file mode 100755 index 000000000..902f661fc --- /dev/null +++ b/2345/CH6/EX6.7/Ex6_7.sce @@ -0,0 +1,12 @@ +//Finding energy loss
+//Example 6.7(pg 215)
+clc
+clear
+m=12000//mass in gm
+d=7.5//density of iron in gm/c.c
+Hl=3000//Hysteresis loss per cc in ergs/cycle
+N=50//No of cycles per sec
+v=m/d//volume of specimen
+E=v*Hl*N//Energy loss per cc in ergs
+Eh=E/(10^10)//Energy loss per hour in kWh
+printf('Thus the Loss in energy is %3.3f kWh',Eh)
diff --git a/2345/CH6/EX6.8/Ex6_8.sce b/2345/CH6/EX6.8/Ex6_8.sce new file mode 100755 index 000000000..8dba475cb --- /dev/null +++ b/2345/CH6/EX6.8/Ex6_8.sce @@ -0,0 +1,21 @@ +//Finding losses
+//Example 6.8(pg 215)
+clc
+clear
+m=10//mass in kg
+T1=20//total loss in watts
+f1=50//frequency in c/s
+T2=35//total loss in watts
+f2=75//frequency in c/s
+//both have same peak flux density
+//total loss=hysteresis loss+ Eddy current loss
+//all quantities except frequency are constant
+//so Total loss=Af+Bf^2
+//let c1 and c2 be constants such that total loss=c1*f + c2*f^2
+c2=[T2-(T1*f2/f1)]/(f2^2-f1*f2)
+c1=(T1-c2*f1^2)/f1
+k=c1/c2//hysteresis loss/eddy current loss
+H50=T1*k/101//hysteresis loss at 50 c/s
+E50=T1-H50//eddy current loss at 50 c/s
+printf('Thus hysteresis loss at 50 c/s is %3.1f Watts \n',H50)
+printf('And Eddy current loss at 50c/s is %3.1f Watts',E50)
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