initial commit / add all books

28 files changed, 312 insertions, 0 deletions

diff --git a/3835/CH2/EX2.1/Ex2_1.sce b/3835/CH2/EX2.1/Ex2_1.sce
new file mode 100644
index 000000000..8fd2f232d
--- /dev/null
+++ b/3835/CH2/EX2.1/Ex2_1.sce
@@ -0,0 +1,21 @@
+clear
+//
+v=10
+r=4
+//case a
+i=v/(r)
+printf("\n i= %0.1f  A",i)
+//case b
+//6ohm resistor is in series with 4 ohm resistor
+i=v/(6+4)
+v1=i*6
+v2=i*4
+printf("\n voltage across 6 ohm resistor= %0.1f  V",v1)
+printf("\n voltage across 4 ohm resistor= %0.1f  V",v2)
+//case c
+i=10 //constant in both cases
+v4=i*4
+printf("\n voltage when 4 ohm resistor is connected= %0.1f  V",v4)
+v6=i*6
+v=v4+v6
+printf("\n voltage when both resistors are in series= %0.1f  V",v)
diff --git a/3835/CH2/EX2.10/Ex2_10.sce b/3835/CH2/EX2.10/Ex2_10.sce
new file mode 100644
index 000000000..78b9ccf2b
--- /dev/null
+++ b/3835/CH2/EX2.10/Ex2_10.sce
@@ -0,0 +1,19 @@
+clear
+//
+//case a
+I=12/(2+((12*24)/(36))) //values taken from circuit
+I1=I*(24/(36))
+I2=I*(12/(36))
+printf("\n i= %0.1f  A",I)
+printf("\n i1= %0.1f  A",I1)
+printf("\n i2= %0.1f  A",I2)
+//case b
+power=(I**2)*2
+printf("\n power consumed by 2 ohm resistor= %0.1f  W",power)
+power=(I1**2)*12
+printf("\n power consumed by 12 ohm resistor= %0.1f  W",power)
+power=(I2**2)*24
+printf("\n power consumed by 2 ohm resistor= %0.1f  W",power)
+//case c
+v=I*2
+printf("\n voltage drop= %0.1f  V",v)
diff --git a/3835/CH2/EX2.11/Ex2_11.sce b/3835/CH2/EX2.11/Ex2_11.sce
new file mode 100644
index 000000000..cd7e0ae09
--- /dev/null
+++ b/3835/CH2/EX2.11/Ex2_11.sce
@@ -0,0 +1,15 @@
+clear
+//
+//case a
+//values taken and calculated from figure
+r1=6
+r2=12
+r3=18
+rab=3.21 //calculating similar to above using parallel in series resistances
+printf("\n rab=3.12ohm")
+//case b
+r4=30
+r5=15
+r6=30
+ran=6 //similar as above
+printf("\n ran=6 ohm")
diff --git a/3835/CH2/EX2.12/Ex2_12.sce b/3835/CH2/EX2.12/Ex2_12.sce
new file mode 100644
index 000000000..88e7d37ce
--- /dev/null
+++ b/3835/CH2/EX2.12/Ex2_12.sce
@@ -0,0 +1,10 @@
+clear
+//
+//eqns derived from figure
+//6v1-4v2=2-->1
+//-4v1+7v2=-3-->2
+//eqn 1 and 2 are written in matrix form and solved using cramers rule
+printf("\n v1=0.0769 V")
+printf("\n v2=-0.3846V")
+printf("\n current in 0.5ohm resistance is 0.154A,0.25ohm resistance is 1.846,0.66ohm resistor is -1.154A")
+
diff --git a/3835/CH2/EX2.13/Ex2_13.sce b/3835/CH2/EX2.13/Ex2_13.sce
new file mode 100644
index 000000000..15c17e568
--- /dev/null
+++ b/3835/CH2/EX2.13/Ex2_13.sce
@@ -0,0 +1,7 @@
+clear
+//
+//from the figure the eqns are written in matrix form and using cramers rule the value of v1 and v2 can be found
+printf("\n v1=3.6V")
+printf("\n v2=2.2V")
+printf("\n the current in 0.6 ohm resistor is 10.8A,0.2 ohm resistor is 7A,0.16ohm resistor is 13.2 A")
+
diff --git a/3835/CH2/EX2.14/Ex2_14.sce b/3835/CH2/EX2.14/Ex2_14.sce
new file mode 100644
index 000000000..cf5eb71ac
--- /dev/null
+++ b/3835/CH2/EX2.14/Ex2_14.sce
@@ -0,0 +1,6 @@
+clear
+//
+//kcl is applied to the circuit and the eqns obtained are solved using cramer's rule
+printf("\n the voltages of nodes 1 and 3 are 50.29 and 57.71 respectively")
+//i3=v/r
+printf("\n current through 16 ohm resistor is 1.64A")
diff --git a/3835/CH2/EX2.15/Ex2_15.sce b/3835/CH2/EX2.15/Ex2_15.sce
new file mode 100644
index 000000000..c0c983c98
--- /dev/null
+++ b/3835/CH2/EX2.15/Ex2_15.sce
@@ -0,0 +1,8 @@
+clear
+//
+//the eqns obtained are converted to matrix form for solving using cramer's rule values are found
+i1=5.224
+i2=0.7463
+i3=3.28
+v=(i1-i3)*3
+printf("\n voltage across 3 ohm resistor is= %0.1f  V",v)
diff --git a/3835/CH2/EX2.16/Ex2_16.sce b/3835/CH2/EX2.16/Ex2_16.sce
new file mode 100644
index 000000000..4c0914bec
--- /dev/null
+++ b/3835/CH2/EX2.16/Ex2_16.sce
@@ -0,0 +1,4 @@
+clear
+//
+//kvl eqns are obtained from figure which are solved to obtain currents
+printf("\n currents obtained are i1=2.013 and i2=1.273")
diff --git a/3835/CH2/EX2.17/Ex2_17.sce b/3835/CH2/EX2.17/Ex2_17.sce
new file mode 100644
index 000000000..5ed0fff9c
--- /dev/null
+++ b/3835/CH2/EX2.17/Ex2_17.sce
@@ -0,0 +1,12 @@
+clear
+//
+//the currents are obtained by solving the eqns
+i1=5.87
+i2=-0.13
+i3=-1.54
+v=18-1.54*8
+printf("\n voltage at node D= %0.1f  v",v)
+i=5.86/(4)
+printf("\n current in 4 ohm resistor is= %0.1f  A",i)
+power=18*1.54
+printf("\n power supplied by 18V source is= %0.1f  W",power)
diff --git a/3835/CH2/EX2.18/Ex2_18.sce b/3835/CH2/EX2.18/Ex2_18.sce
new file mode 100644
index 000000000..9afc2546b
--- /dev/null
+++ b/3835/CH2/EX2.18/Ex2_18.sce
@@ -0,0 +1,6 @@
+clear
+//
+//node eqns are obtained form the figure
+printf("\n va=8.33V and vb=4.17V")
+i=8.33/(8)
+printf("\n current through 8 ohm resistor is= %0.1f  A",i)
diff --git a/3835/CH2/EX2.19/Ex2_19.sce b/3835/CH2/EX2.19/Ex2_19.sce
new file mode 100644
index 000000000..6322b500a
--- /dev/null
+++ b/3835/CH2/EX2.19/Ex2_19.sce
@@ -0,0 +1,4 @@
+clear
+//
+//eqns obtained are calculated just like above problems and are aolved for i1 and i2
+printf("\n i1=-1.363A and i2=-3.4A")
diff --git a/3835/CH2/EX2.20/Ex2_20.sce b/3835/CH2/EX2.20/Ex2_20.sce
new file mode 100644
index 000000000..dde599768
--- /dev/null
+++ b/3835/CH2/EX2.20/Ex2_20.sce
@@ -0,0 +1,10 @@
+clear
+//
+//eqns are obtained from the figure and are solved for currents
+i1=6.89
+i2=3.89
+i3=-2.12
+i=2*(i2-i1)
+printf("\n current supplied by dependent source is= %0.1f  A",i)
+power=6*i1
+printf("\n power supplied by voltage source is= %0.1f  W",power)
diff --git a/3835/CH2/EX2.21/Ex2_21.sce b/3835/CH2/EX2.21/Ex2_21.sce
new file mode 100644
index 000000000..80045b260
--- /dev/null
+++ b/3835/CH2/EX2.21/Ex2_21.sce
@@ -0,0 +1,12 @@
+clear
+//
+//the following problem is based on usage of superposition theorem
+i8=12/(6+4+8) //current for 8 ohm resistor.the resistances are in series with each other.Hence 6+4+8
+//next when voltage source is short circuited (8+4) total of resistance is obtained.The 4A is distributed in parallel branches as per current divider rule
+i=(4*6)/(6+12)
+printf("\n i8= %0.1f  A",i8)
+printf("\n i8= %0.1f  A",i)
+tot=i8+i
+printf("\n total current= %0.1f  A",tot)
+
+
diff --git a/3835/CH2/EX2.23/Ex2_23.sce b/3835/CH2/EX2.23/Ex2_23.sce
new file mode 100644
index 000000000..946bc7fa2
--- /dev/null
+++ b/3835/CH2/EX2.23/Ex2_23.sce
@@ -0,0 +1,12 @@
+clear
+//
+//thevenin's theorem and superposition theorem used here
+//applying mesh eqns to the 2 circuits and after getting the eqns they are solved using cramer's rule to obtain i1 and i2
+i1=-0.6
+i2=-1.2
+//the value of currents indicates that they have assumed to be flowing in directions opposite to the assumed direction
+vth=12-1.2*3 //voltage eqn
+rth=1.425 //(1+2||12)||3=(1+(2*12)/(2+12))||3=19/7||3=19/7*3/19/7+3=1.425
+i2=vth/(rth+2)
+printf("\n current through 2 ohm resistor is= %0.1f  A",i2)
+printf("\n Note that the same problem is again solved using superposition theorem and hence ignored ")
diff --git a/3835/CH2/EX2.24/Ex2_24.sce b/3835/CH2/EX2.24/Ex2_24.sce
new file mode 100644
index 000000000..511be7370
--- /dev/null
+++ b/3835/CH2/EX2.24/Ex2_24.sce
@@ -0,0 +1,9 @@
+clear
+//
+//using thevenin's theorem
+//applying kcl at node a va is obtained
+va=12
+vth=12-1.2*3 //voltage eqn
+rth=1.33 //2||4
+i5=vth/(rth+5)
+printf("\n current through 5 ohm resistor is= %0.1f  A",i5)
diff --git a/3835/CH2/EX2.25/Ex2_25.sce b/3835/CH2/EX2.25/Ex2_25.sce
new file mode 100644
index 000000000..228307d98
--- /dev/null
+++ b/3835/CH2/EX2.25/Ex2_25.sce
@@ -0,0 +1,9 @@
+clear
+//
+//applying kvl to circuit
+i=0.414
+vth=12-4*0.414 //using vth formula
+//when terminals a and b are short circuited applying kcl to node a gives isc=5*i
+isc=2.07
+rth=vth/isc
+printf("\n rth= %0.1f  A",rth)
diff --git a/3835/CH2/EX2.26/Ex2_26.sce b/3835/CH2/EX2.26/Ex2_26.sce
new file mode 100644
index 000000000..9b077769a
--- /dev/null
+++ b/3835/CH2/EX2.26/Ex2_26.sce
@@ -0,0 +1,9 @@
+clear
+//
+//norton's theorem
+v=10
+//applying kvl to closed circuit 
+isc=12/(2+2) 
+rn=4 //resistance obtained by short circuiting v and opening i
+iab=(4*3)/(4+4) //current through 4 ohm connected across AB
+printf("\n iab= %0.1f  A",iab)
diff --git a/3835/CH2/EX2.27/Ex2_27.sce b/3835/CH2/EX2.27/Ex2_27.sce
new file mode 100644
index 000000000..d72b57acb
--- /dev/null
+++ b/3835/CH2/EX2.27/Ex2_27.sce
@@ -0,0 +1,8 @@
+clear
+//
+//natural frequency needs to be determined
+//req=[(6+6)||4]+[1||2]=3.6666
+req=3.6667
+l=4 //inductance
+s=-req/(l)
+printf("\n natural frequency= %0.1f  secinverse",s)
diff --git a/3835/CH2/EX2.28/Ex2_28.sce b/3835/CH2/EX2.28/Ex2_28.sce
new file mode 100644
index 000000000..d924f7ab6
--- /dev/null
+++ b/3835/CH2/EX2.28/Ex2_28.sce
@@ -0,0 +1,11 @@
+clear
+//
+//req=[10+2+(5||15)]=15.75
+//case a
+c=0.4
+req=15.75
+s=-1/(c*req)
+printf("\n natural frequency= %0.1f  secinverse",s)
+//case b
+tc=req*0.4 //time constant
+printf("\n time constant= %0.1f  sec",tc)
diff --git a/3835/CH2/EX2.3/Ex2_3.sce b/3835/CH2/EX2.3/Ex2_3.sce
new file mode 100644
index 000000000..12c348dc9
--- /dev/null
+++ b/3835/CH2/EX2.3/Ex2_3.sce
@@ -0,0 +1,8 @@
+clear
+//
+v=24
+r=0.75
+ir=v/r
+printf("\n ir= %0.1f  A",ir)
+il=v/(10+r) //since 10 is in series with r
+printf("\n il= %0.1f  A",il)
diff --git a/3835/CH2/EX2.30/Ex2_30.sce b/3835/CH2/EX2.30/Ex2_30.sce
new file mode 100644
index 000000000..68913d182
--- /dev/null
+++ b/3835/CH2/EX2.30/Ex2_30.sce
@@ -0,0 +1,23 @@
+clear
+//
+v=120
+r=40
+i=v/(r)
+//applying kvl to the closed loop
+v=3*520
+printf("\n voltage= %0.1f  v",v)
+//when v=120,R can be found by I*(r+20)=120-->r=20
+r=20
+printf("\n r=20 ohm")
+//when r=20 total r=20+20+20=60
+r=60
+l=10
+tc=l/(r) //time constant
+printf("\n tc= %0.1f  sec",tc)
+//i=I0*e^-(t/tc)=3*e^(-6t)
+energy=(10*9)/(2)
+benergy=0.05*energy
+printf("\n balance energy= %0.1f  J",benergy)
+//(L*i^2)/2=2.25-->hence i=0.6708
+//3*e^-6t=0.6708-->e^-6t=0.2236-->applying log on both sides we get t=0.25
+printf("\n t=0.25 sec")
diff --git a/3835/CH2/EX2.34/Ex2_34.sce b/3835/CH2/EX2.34/Ex2_34.sce
new file mode 100644
index 000000000..abaa4c4e4
--- /dev/null
+++ b/3835/CH2/EX2.34/Ex2_34.sce
@@ -0,0 +1,12 @@
+clear
+//
+v=120
+V=200
+//v=V(1-e^-5/2R)
+//120=200*(1-e^-5/2R)
+//applying log on both sides and solving we get R=2.72 Mohm
+printf("\n R=2.72Mohm")
+R=5 
+tc=10
+//applying in the above eqn and solving lograthmically we get t=9.16
+printf("\n t=9.16 sec")
diff --git a/3835/CH2/EX2.4/Ex2_4.sce b/3835/CH2/EX2.4/Ex2_4.sce
new file mode 100644
index 000000000..90aa8ed6c
--- /dev/null
+++ b/3835/CH2/EX2.4/Ex2_4.sce
@@ -0,0 +1,16 @@
+clear
+//
+vs=12
+rs=0.3
+il=10
+//case a
+p=vs*il
+printf("\n power= %0.1f  W",p)
+//case b
+power=il**2*rs
+printf("\n power dissipated= %0.1f  W",power)
+//case c
+totpow=(vs-il*rs)*il
+printf("\n total power supplied by practical source is= %0.1f  W",totpow)
+i=vs/rs
+printf("\n current source= %0.1f  A",i)
diff --git a/3835/CH2/EX2.5/Ex2_5.sce b/3835/CH2/EX2.5/Ex2_5.sce
new file mode 100644
index 000000000..e43b72bca
--- /dev/null
+++ b/3835/CH2/EX2.5/Ex2_5.sce
@@ -0,0 +1,21 @@
+clear
+//
+//case a
+//v0/vs=r2/(r1+r2)=0.4r2=0.6r1
+r1=10
+r2=(0.6*r1)/(0.4)
+printf("\n r2= %0.1f  ohm",r2)
+//case b
+//when r2 is parallel to r3
+r3=200000
+req=(r2*r3)/(r2+r3)
+printf("\n req= %0.1f  ohm",req)
+//v0/vs=0.5825
+change=(0.6-0.5825)/(0.6)
+printf("\n change")
+r3=20000
+req=(r2*r3)/(r3+r2)
+printf("\n req= %0.1f  ohm",req)
+//v0/vs=0.4615
+change=(0.6-0.4615)/0.6
+printf("\n change")
diff --git a/3835/CH2/EX2.6/Ex2_6.sce b/3835/CH2/EX2.6/Ex2_6.sce
new file mode 100644
index 000000000..6d2a98980
--- /dev/null
+++ b/3835/CH2/EX2.6/Ex2_6.sce
@@ -0,0 +1,13 @@
+clear
+//
+r=2
+i=2
+i3=3 //obtained by applying current divider rule to figure
+i4=1
+req=1/(0.5+0.25+0.166) //1/2,1/4,1/6 values are converted to decimal form
+printf("\n req= %0.1f  ohm",req)
+i2=(4*i4/(6))
+i1=(6*i2)/(req)
+//tracing circuit cabc via 6 ohm resistor and applying ohms law,
+vs=i1*i4+i2*6
+printf("\n vs= %0.1f  V",vs)
diff --git a/3835/CH2/EX2.7/Ex2_7.sce b/3835/CH2/EX2.7/Ex2_7.sce
new file mode 100644
index 000000000..0207ef461
--- /dev/null
+++ b/3835/CH2/EX2.7/Ex2_7.sce
@@ -0,0 +1,6 @@
+clear
+//
+//combining series parallel series
+//[(2+2+2)||(6+5+2)||10]+5
+//[[6*6/6+6]+7]||10]+5=[10+10/10*10]+5=5+5=10
+printf("\n the value of series parallel resistances is 10 ohm")
diff --git a/3835/CH2/EX2.8/Ex2_8.sce b/3835/CH2/EX2.8/Ex2_8.sce
new file mode 100644
index 000000000..471ab3fee
--- /dev/null
+++ b/3835/CH2/EX2.8/Ex2_8.sce
@@ -0,0 +1,16 @@
+clear
+//
+//case a
+//rab=(80+40)||(60+40)
+rab=(120*100)/(120+100)
+printf("\n rab= %0.1f  ohm",rab)
+//rab=(80||60)+(40||40)
+rab=(4800/(140))+(1600/80)
+printf("\n rab= %0.1f  ohm",rab)
+//case b
+//(60+80)||(40+40)
+rcd=(140*80)/(140+80)
+printf("\n rcd= %0.1f  ohm",rcd)
+//(60||40)+(80||40)
+rab=(2400/(100))+(3200/(120))
+printf("\n rab= %0.1f  ohm",rab)
diff --git a/3835/CH2/EX2.9/Ex2_9.sce b/3835/CH2/EX2.9/Ex2_9.sce
new file mode 100644
index 000000000..484161ffe
--- /dev/null
+++ b/3835/CH2/EX2.9/Ex2_9.sce
@@ -0,0 +1,5 @@
+clear
+//
+//simplifying the circuit 
+ceq=1/(0.333+0.666+0.2) //converted to decimal form
+printf("\n ceq= %0.1f  F",ceq)