initial commit / add all books

29 files changed, 549 insertions, 0 deletions

diff --git a/3507/CH9/EX19.14/Ex19_14.sce b/3507/CH9/EX19.14/Ex19_14.sce
new file mode 100644
index 000000000..be01ac88b
--- /dev/null
+++ b/3507/CH9/EX19.14/Ex19_14.sce
@@ -0,0 +1,23 @@
+//chapter19

+//example19.14

+//page424

+

+Vc=5 // V

+V_lower=2.5 // V

+V_upper=2.5 // V

+R=2 // kilo ohm

+

+// figure given in book is just for understanding purpose.It is not a part of solution.

+// however, the figure has been made in xcos and screenshot has been attached for reference

+

+// since power=(rms voltage)^2/R we get

+

+Pc=(0.707*Vc)^2/R

+P_lower=(0.707*V_lower)^2/R

+P_upper=(0.707*V_upper)^2/R

+Pt=Pc+P_lower+P_upper

+

+printf("power delivered by carrier = %.3f mW \n",Pc)

+printf("power delivered by lower sideband = %.3f mW \n",P_lower)

+printf("power delivered by upper sideband = %.3f mW \n",P_upper)

+printf("total power delivered by AM wave = %.3f mW \n",Pt)

diff --git a/3507/CH9/EX9.1/Ex9_1.sce b/3507/CH9/EX9.1/Ex9_1.sce
new file mode 100644
index 000000000..6a59df5cf
--- /dev/null
+++ b/3507/CH9/EX9.1/Ex9_1.sce
@@ -0,0 +1,8 @@
+//chapter9
+//example9.1
+//page142
+
+printf("in fig. (i), the conventional current coming out of battery flows in the \nbranch circuits. In diode D1,the conventional current flows in the \ndirection of arrowhead and hence this diode is forward biased. \nHowever in diode D2, the conventional current flows opposite \nto arrowhead and hence this diode is reverse biased.\n \n")
+printf("in fig. (ii), During the positive half cycle of input ac voltage, the \nconventional current flows in the direction of arrowhead and hence diode \nis forward biased. However, during the negative half cycle \nof input ac voltage, the diode is reverse biased.\n \n")
+printf("in fig. (iii), During the positive half cycle of input ac voltage, the \nconventional current flows in the direction of arrowhead in D1 but it flows \nopposite to arrowhead in D2. So during positive half cycle, \ndiode D1 is forward biased and diode D2 is reverse biased. \nHowever in the negative half cycle of the input ac voltage, diode D2 \nis forward biased and diode D1 is reverse biased.\n \n")
+printf("in fig. (iv),  During the positive half cycle of input ac voltage, \nboth diodes are reverse biased. However in the negative half cycle of the \ninput ac voltage, both diodes are forward biased.\n \n")
diff --git a/3507/CH9/EX9.10/Ex9_10.sce b/3507/CH9/EX9.10/Ex9_10.sce
new file mode 100644
index 000000000..10b306ed8
--- /dev/null
+++ b/3507/CH9/EX9.10/Ex9_10.sce
@@ -0,0 +1,21 @@
+//chapter9

+//example9.10

+//page152

+

+n=10

+Vp=230 // V

+

+Vpm=2^0.5*Vp

+Vsm=Vpm/n // since n=Vpm/Vsm=N1/N2

+

+// Idc=Im/%pi and Vdc=Idc*Rl so

+// Vdc=(Im/%pi)*Rl .Also Im*Rl=Vsm so

+Vdc=Vsm/%pi

+

+// in negative half cycle diode is reverse biased so maximum secondary voltage appears across diode.

+PIV=Vsm

+

+printf("output dc voltage = %.2f V \n",Vdc)

+printf("peak inverse voltage = %.2f V \n",PIV)

+

+// accurate answer for output dc voltage is 10.35 V not 10.36 V

diff --git a/3507/CH9/EX9.11/Ex9_11.sce b/3507/CH9/EX9.11/Ex9_11.sce
new file mode 100644
index 000000000..d6283d547
--- /dev/null
+++ b/3507/CH9/EX9.11/Ex9_11.sce
@@ -0,0 +1,23 @@
+//chapter9

+//example9.11

+//page152

+

+rf=20 // ohm

+Rl=800 // ohm

+Vm=50 // V

+

+Im=Vm/(rf+Rl) // in ampere

+Idc=Im/%pi // in ampere

+Irms=Im/2 // in ampere

+Pac=Irms^2*(rf+Rl)

+Pdc=Idc^2*Rl

+Vout=Idc*Rl

+efficiency=100*Pdc/Pac

+

+printf("Im = %.1f mA \n",Im*1000)

+printf("Idc = %.1f mA \n",Idc*1000)

+printf("Irms = %.1f mA \n \n",Irms*1000)

+printf("ac power input = %.3f W \n",Pac)

+printf("dc power output = %.3f W \n \n",Pdc)

+printf("dc output voltage = %.3f V \n \n",Vout)

+printf("efficiency = %.3f percent \n",efficiency)

diff --git a/3507/CH9/EX9.12/Ex9_12.sce b/3507/CH9/EX9.12/Ex9_12.sce
new file mode 100644
index 000000000..e3d154950
--- /dev/null
+++ b/3507/CH9/EX9.12/Ex9_12.sce
@@ -0,0 +1,17 @@
+//chapter9

+//example9.12

+//page153

+

+Vdc=50 // V

+rf=25 // ohm

+Rl=800 // ohm

+

+// Vdc=Idc*Rl and Idc=Im/%pi so

+// Vdc=Im*Rl/%pi

+// but Im=Vm/(rf+Rl) so 

+// Vdc=Vm*Rl/(%pi*(rf+Rl))

+// making Vm as subject we get

+

+Vm=Vdc*%pi*(rf+Rl)/Rl

+

+printf("ac voltage required = %.1f V \n",Vm)

diff --git a/3507/CH9/EX9.13/Ex9_13.sce b/3507/CH9/EX9.13/Ex9_13.sce
new file mode 100644
index 000000000..206331305
--- /dev/null
+++ b/3507/CH9/EX9.13/Ex9_13.sce
@@ -0,0 +1,15 @@
+//chapter9

+//example9.13

+//page157

+

+rf=20 // ohm

+Rl=980 // ohm

+Vs=50 // V

+

+Vm=Vs*2^0.5

+Im=Vm/(rf+Rl)

+Idc=2*Im/%pi // in ampere

+Irms=Im/2^0.5 // in ampere

+

+printf("mean load current  = %.3f mA \n",Idc*1000)

+printf("rms load current = %.3f mA \n",Irms*1000)

diff --git a/3507/CH9/EX9.14/Ex9_14.sce b/3507/CH9/EX9.14/Ex9_14.sce
new file mode 100644
index 000000000..3c3d9f6ee
--- /dev/null
+++ b/3507/CH9/EX9.14/Ex9_14.sce
@@ -0,0 +1,21 @@
+//chapter9

+//example9.14

+//page157

+

+rf=0

+n=5

+Vp=230 // V rms

+Rl=100 //ohm

+

+Vs=Vp/n // V rms

+Vsm=Vs*2^0.5 // maximum voltage across secondary 

+Vm=Vsm/2 // maximum voltage across half secondary winding

+

+Idc=2*Vm/(%pi*Rl)

+Vdc=Idc*Rl

+PIV=Vsm

+efficiency=100*0.812/(1+rf/Rl)

+

+printf("dc output voltage = %.3f V \n",Vdc)

+printf("PIV = %.3f V \n",PIV)

+printf("efficiency = %.3f percent \n",efficiency)

diff --git a/3507/CH9/EX9.15/Ex9_15.sce b/3507/CH9/EX9.15/Ex9_15.sce
new file mode 100644
index 000000000..faa0de2cc
--- /dev/null
+++ b/3507/CH9/EX9.15/Ex9_15.sce
@@ -0,0 +1,24 @@
+//chapter9

+//example9.15

+//page158

+

+n=4

+Rl=200 // ohm

+fin=50 // Hz

+Vp=230 // V rms

+

+Vs=Vp/n // V rms

+Vsm=Vs*2^0.5 // maximum voltage across secondary 

+

+Idc=2*Vsm/(%pi*Rl)

+Vdc=Idc*Rl

+PIV=Vsm

+

+// in full wave rectifier, output frequency is twice input frequency since there are two ouput pulses for each cycle of input

+fout=2*fin

+

+printf("dc output voltage = %.3f V \n",Vdc)

+printf("peak inverse voltage = %.3f V \n",PIV)

+printf("output frequency = %.3f Hz",fout)

+

+// the accurate answer for dc output voltage is 51.768 V but in book it is given as 52 V

diff --git a/3507/CH9/EX9.16/Ex9_16.sce b/3507/CH9/EX9.16/Ex9_16.sce
new file mode 100644
index 000000000..945120c2c
--- /dev/null
+++ b/3507/CH9/EX9.16/Ex9_16.sce
@@ -0,0 +1,45 @@
+//chapter9

+//example9.16

+//page158

+

+// for dc output

+    // for centre-tap circuit

+        n=5

+        Vp=230 // V rms

+        Rl=100 //ohm

+        Vs=Vp/n // V rms

+        Vsm=Vs*2^0.5 // maximum voltage across secondary 

+        Vm=Vsm/2 // maximum voltage across half secondary winding

+        Vdc=2*Vm/%pi // since Vdc=Idc*Rl and Idc=2*Vm/(%pi*Rl)

+

+    // for bridge circuit

+        n_dash=5

+        Vp_dash=230 // V rms

+        Rl_dash=100 //ohm

+        Vs_dash=Vp_dash/n_dash// V rms

+        Vsm_dash=Vs*2^0.5 // maximum voltage across secondary 

+        Vm_dash=Vsm_dash

+        Vdc_dash=2*Vm_dash/%pi // since Vdc=Idc*Rl and Idc=2*Vm/(%pi*Rl)

+

+

+// for same dc output Vm must be same for both circuits i.e. n=5 for centre-tap and n=10 for bridge

+    // for centre-tap circuit

+        n1=5

+        Vs1=Vp/n1 // V rms

+        Vsm1=Vs1*2^0.5 // maximum voltage across secondary 

+        Vm1=Vsm1/2

+        PIV1=2*Vm1

+    

+    // for bridge circuit

+        n2=5

+        Vs2=Vp/n2 // V rms

+        Vsm2=Vs2*2^0.5 // maximum voltage across secondary 

+        Vm2=Vsm2/2

+        PIV2=Vm2

+        

+printf("dc output voltage for centre-tap circuit = %.3f V \n",Vdc) 

+printf("dc output voltage for bridge circuit = %.3f V \n \n",Vdc_dash)  

+

+printf("for same output, PIV for centre-tap circuit = %.3f V and bridge circuit = %.3f V \n",PIV1,PIV2)     

+    

+// the figure of transformer is for reference only. Also it cannot be plotted in scilab since scilab does not have centre-tap transformer

diff --git a/3507/CH9/EX9.17/Ex9_17.sce b/3507/CH9/EX9.17/Ex9_17.sce
new file mode 100644
index 000000000..247da0d1b
--- /dev/null
+++ b/3507/CH9/EX9.17/Ex9_17.sce
@@ -0,0 +1,19 @@
+//chapter9

+//example9.17

+//page160

+

+Vin=240 // V rms

+Rl=480 // ohm

+rf=1 // ohm

+

+Vm=Vin*2^0.5

+// for bridge rectifier we know that

+Im=Vm/(2*rf+Rl)

+Idc=2*Im/%pi

+Irms=Im/2

+P=Irms^2*rf

+

+printf("mean load current = %.3f A \n",Idc)

+printf("power dissipated in each diode = %.3f W \n",P)

+

+// the accurate answers are mean load current = 0.448 A and power dissipated in each diode = 0.124 W 

diff --git a/3507/CH9/EX9.18/Ex9_18.sce b/3507/CH9/EX9.18/Ex9_18.sce
new file mode 100644
index 000000000..fffee6cdd
--- /dev/null
+++ b/3507/CH9/EX9.18/Ex9_18.sce
@@ -0,0 +1,19 @@
+//chapter9

+//example9.18

+//page162

+

+Vrms_A=0.5 // V

+Vdc_A=10 // V

+Vrms_B=1 // V

+Vdc_B=25 // V

+

+ripple_A=Vrms_A/Vdc_A

+ripple_B=Vrms_B/Vdc_B

+

+if ripple_A>ripple_B

+    printf("power supply B is better \n")

+elseif ripple_B>ripple_A

+   printf("power supply A is better \n") 

+else

+    printf("both are equal \n")

+end

diff --git a/3507/CH9/EX9.19/Ex9_19.sce b/3507/CH9/EX9.19/Ex9_19.sce
new file mode 100644
index 000000000..cd493b012
--- /dev/null
+++ b/3507/CH9/EX9.19/Ex9_19.sce
@@ -0,0 +1,16 @@
+//chapter9

+//example9.19

+//page165

+

+// the waveform given in book is for understanding only. It is not required to solve the problem. Also it cannot be plotted in scilab unless Vm and Vdc are given.

+

+R=25 // ohm

+Rl=750 // ohm

+Vm=25.7 // V

+

+Vdc_dash=2*Vm/%pi

+Vdc=Vdc_dash*Rl/(R+Rl)

+

+printf("voltage across load is %.3f V plus a small ripple \n",Vdc)

+

+// the accurate answer is 15.833 V but in book it is given as 15.9 V

diff --git a/3507/CH9/EX9.2/Ex9_2.sce b/3507/CH9/EX9.2/Ex9_2.sce
new file mode 100644
index 000000000..99a44add5
--- /dev/null
+++ b/3507/CH9/EX9.2/Ex9_2.sce
@@ -0,0 +1,23 @@
+//chapter9

+//example9.2

+//page145

+

+Vi_p=20 // V

+rf=10 // ohm

+Rl=500 // ohm

+Vo=0.7 // V

+Vin=20 // V

+

+// peak current through diode will occur when Vin=Vf so

+Vf=Vin

+// since Vf=Vo+If_peak(rf+Rl) making If_peak as subject we get

+If_peak1=(Vf-Vo)/(rf+Rl) // in ampere

+Vout_peak1=If_peak1*Rl

+

+// for ideal diode, Vo=0 and rf=0 so

+// Vf=If_peak*Rl so we get

+If_peak2=Vf/Rl  // in ampere

+Vout_peak2=If_peak2*Rl

+

+printf("peak current through given diode = %.3f mA and peak output voltage = %.3f V \n",If_peak1*1000,Vout_peak1)

+printf("peak current through ideal diode = %.3f mA and peak output voltage = %.3f V \n",If_peak2*1000,Vout_peak2)

diff --git a/3507/CH9/EX9.20/Ex9_20.sce b/3507/CH9/EX9.20/Ex9_20.sce
new file mode 100644
index 000000000..19c9f7120
--- /dev/null
+++ b/3507/CH9/EX9.20/Ex9_20.sce
@@ -0,0 +1,21 @@
+//chapter9

+//example9.20

+//page170

+

+R=5 // kilo ohm

+Rl=10 // kilo ohm

+Ei=120 // V

+Vz=50 // V

+

+V=Ei*Rl/(R+Rl) // voltage across open circuit if zener diode is removed

+Vo=Vz // output voltage

+V_R=Ei-Vz // drop across R

+Il=Vz/Rl // load current 

+I=V_R/R // current through R

+

+// by Kirchoff first law I=Iz+Il

+Iz=I-Il

+

+printf("output voltage = %.3f V \n",Vo)

+printf("voltage drop across series resistance = %.3f V \n",V_R)

+printf("current through Zener diode = %.3f mA \n",Iz)

diff --git a/3507/CH9/EX9.21/Ex9_21.sce b/3507/CH9/EX9.21/Ex9_21.sce
new file mode 100644
index 000000000..8d4c037ef
--- /dev/null
+++ b/3507/CH9/EX9.21/Ex9_21.sce
@@ -0,0 +1,28 @@
+//chapter9
+//example9.21
+//page171
+
+Vmax=120 // V
+Vmin=80 // V
+Vz=50 // V
+R_L=10 // kilo ohm
+R1=5 // kilo ohm
+
+// zener diode is on for Vmax and Vmin both since they are > Vz
+
+// for max Iz
+    V_R1=Vmax-Vz
+    I=V_R1/R1 // current through R1
+    I_L=Vz/R_L // current through load
+    // by Kirchoff first law I=I_L+Iz so applying it we get
+    Iz_max=I-I_L
+    
+// for min Iz
+    V_R1_dash=Vmin-Vz
+    I_dash=V_R1_dash/R1// current through R1
+    I_L_dash=Vz/R_L // current through load
+    // by Kirchoff first law I=I_L+Iz so we get
+    Iz_min=I_dash-I_L_dash
+    
+printf("maximum zener current = %.3f mA \n",Iz_max)
+printf("minimum zener current = %.3f mA \n",Iz_min)
diff --git a/3507/CH9/EX9.22/Ex9_22.sce b/3507/CH9/EX9.22/Ex9_22.sce
new file mode 100644
index 000000000..7562ad518
--- /dev/null
+++ b/3507/CH9/EX9.22/Ex9_22.sce
@@ -0,0 +1,18 @@
+//chapter9

+//example9.22

+//page172

+

+Ei=12 // V

+Vz=7.2 // V

+Eo=Vz

+Iz_min=10d-3 // A

+Il_max=100d-3 // A

+

+// we see that R=(Ei-Eo)/(Iz-Il) and minimum Iz occurs when Il is maximum so

+R=(Ei-Eo)/(Iz_min+Il_max)

+

+printf("required series resistance = %.3f ohm \n",R)

+

+// on inserting this series resistance the output voltage will remain constant at 7.2 V

+

+// the accurate answer is 43.636 ohm but in book it is given as 43.5 ohm

diff --git a/3507/CH9/EX9.23/Ex9_23.sce b/3507/CH9/EX9.23/Ex9_23.sce
new file mode 100644
index 000000000..3bdcdc0e0
--- /dev/null
+++ b/3507/CH9/EX9.23/Ex9_23.sce
@@ -0,0 +1,21 @@
+//chapter9

+//example9.23

+//page172

+

+Ei=22 // V     

+Vz=18 // V

+Rl=18 // ohm

+Eo=Vz

+Iz_min=200d-3 // A

+

+// Zener current will be min when input voltage is min

+

+// load current is

+Il_max=Vz/Rl

+

+// we see that R=(Ei-Eo)/(Iz-Il) and minimum Iz occurs when Il is maximum so

+R=(Ei-Eo)/(Iz_min+Il_max)

+

+printf("required series resistance = %.3f ohm \n",R)

+

+// on inserting this series resistance the output voltage will remain constant at 18 V

diff --git a/3507/CH9/EX9.24/Ex9_24.sce b/3507/CH9/EX9.24/Ex9_24.sce
new file mode 100644
index 000000000..88ca8041f
--- /dev/null
+++ b/3507/CH9/EX9.24/Ex9_24.sce
@@ -0,0 +1,16 @@
+//chapter9

+//example9.24

+//page172

+

+Ei=13 // V     

+Vz=10 // V

+Eo=Vz

+Iz_min=15d-3 // A

+Il_max=85d-3 // A

+

+// Zener current will be min when input voltage is min

+

+// we see that R=(Ei-Eo)/(Iz-Il) and minimum Iz occurs when Il is maximum so

+R=(Ei-Eo)/(Iz_min+Il_max)

+

+printf("required series resistance = %.3f ohm \n",R)

diff --git a/3507/CH9/EX9.25/Ex9_25.sce b/3507/CH9/EX9.25/Ex9_25.sce
new file mode 100644
index 000000000..07a5e64c6
--- /dev/null
+++ b/3507/CH9/EX9.25/Ex9_25.sce
@@ -0,0 +1,15 @@
+//chapter9

+//example9.25

+//page173

+

+Ei=45 // V     

+Vz1=15 // V

+Vz2=15 // V

+Iz=200d-3 // current rating for each zener in ampere

+

+Eo=Vz1+Vz2

+

+R=(Ei-Eo)/Iz

+

+printf("regulated output voltage = %.3f V \n",Eo)

+printf("required series resistance = %.3f ohm \n",R)

diff --git a/3507/CH9/EX9.26/Ex9_26.sce b/3507/CH9/EX9.26/Ex9_26.sce
new file mode 100644
index 000000000..2c56fe3c9
--- /dev/null
+++ b/3507/CH9/EX9.26/Ex9_26.sce
@@ -0,0 +1,17 @@
+//chapter9

+//example9.26

+//page173

+

+Ei=45 // V     

+Vz1=10 // V

+Vz2=10 // V

+Vz3=10 // V

+Iz=1000d-3 // current rating for each zener in ampere

+

+Eo=Vz1+Vz2+Vz3

+

+R=(Ei-Eo)/Iz

+

+printf("required series resistance = %.3f ohm \n",R)

+

+// since zener diode is not available in xcos, simple diodes are used to represent zener diode in the circuit made in xcos

diff --git a/3507/CH9/EX9.27/Ex9_27.sce b/3507/CH9/EX9.27/Ex9_27.sce
new file mode 100644
index 000000000..bb5c3e7bf
--- /dev/null
+++ b/3507/CH9/EX9.27/Ex9_27.sce
@@ -0,0 +1,22 @@
+//chapter9

+//example9.27

+//page174

+

+R=200 // ohm

+Rl=2000 // ohm

+Eo=30 // V

+

+// for minimum input voltage i.e. Iz=0

+Il=Eo/Rl

+I=Il // since Iz=0

+Vin_min=Eo+I*R

+

+// for maximum input voltage i.e. Iz=25 mA

+Iz=25d-3 // A

+Il_dash=Eo/Rl

+I_dash=Il_dash+Iz

+Vin_max=Eo+I_dash*R

+

+printf("minimum input voltage = %.3f V \n",Vin_min)

+printf("maximum input voltage = %.3f V \n",Vin_max)

+printf("thus range of input = %.3f to %.3f V \n",Vin_min,Vin_max)

diff --git a/3507/CH9/EX9.28/Ex9_28.sce b/3507/CH9/EX9.28/Ex9_28.sce
new file mode 100644
index 000000000..c47a0026c
--- /dev/null
+++ b/3507/CH9/EX9.28/Ex9_28.sce
@@ -0,0 +1,21 @@
+//chapter9

+//example9.28

+//page174

+

+Ei=16 // V     

+Vz=12 // V    since we want to ragulate at 12 V

+Eo=Vz

+Iz_min=0 // A

+Il_max=200d-3 // A

+

+// Zener current will be min when input voltage is min

+

+// we see that R=(Ei-Eo)/(Iz-Il) and minimum Iz occurs when Il is maximum so

+R=(Ei-Eo)/(Iz_min+Il_max)

+

+Izm=Il_max

+Pzm=Vz*Izm

+

+printf("Zener voltage = %.3f V \n",Vz)

+printf("required series resistance = %.3f ohm \n",R)

+printf("maximum power rating of zener diode = %.3f W \n",Pzm)

diff --git a/3507/CH9/EX9.3/Ex9_3.sce b/3507/CH9/EX9.3/Ex9_3.sce
new file mode 100644
index 000000000..2a8cfcef5
--- /dev/null
+++ b/3507/CH9/EX9.3/Ex9_3.sce
@@ -0,0 +1,13 @@
+//chapter9

+//example9.3

+//page146

+

+R1=50 // ohm

+R2=5 // ohm

+V=10 // V

+

+Eo=V*R2/(R1+R2) // thevenin voltage

+Ro=R1*R2/(R1+R2) // thevenin resistance

+I_D=Eo/Ro // current through diode in ampere

+

+printf("current through diode = %.3f mA \n",I_D*1000)

diff --git a/3507/CH9/EX9.4/Ex9_4.sce b/3507/CH9/EX9.4/Ex9_4.sce
new file mode 100644
index 000000000..2352c5c8c
--- /dev/null
+++ b/3507/CH9/EX9.4/Ex9_4.sce
@@ -0,0 +1,17 @@
+//chapter9

+//example9.4

+//page146

+

+V=10 // V

+V_D1=0.7 // V

+V_D2=0.7 // V

+R=48 // ohm

+R_D1=1 // ohm

+R_D2=1 // ohm

+

+// D1 and D3 are forward biased while D2 and D4 are reverse biased thus

+V_net=V-V_D1-V_D2

+R_t=R_D1+R+R_D2

+I=V_net/R_t

+

+printf("circuit current = %.3f mA \n",I*1000)

diff --git a/3507/CH9/EX9.5/Ex9_5.sce b/3507/CH9/EX9.5/Ex9_5.sce
new file mode 100644
index 000000000..c9fc15b5d
--- /dev/null
+++ b/3507/CH9/EX9.5/Ex9_5.sce
@@ -0,0 +1,13 @@
+//chapter9

+//example9.5

+//page147

+

+E1=24 // V

+E2=4 // V

+Vo=0.7 // V

+R=2 // kilo ohm

+

+// diode D1 is forward biased and diode D2 is reverse biased so

+I=(E1-E2-Vo)/R

+

+printf("current in the circuit = %.3f mA \n",I)

diff --git a/3507/CH9/EX9.6/Ex9_6.sce b/3507/CH9/EX9.6/Ex9_6.sce
new file mode 100644
index 000000000..2a6c9bc34
--- /dev/null
+++ b/3507/CH9/EX9.6/Ex9_6.sce
@@ -0,0 +1,11 @@
+//chapter9

+//example9.6

+//page147

+

+V=20 // V

+V_D_Ge=0.3 // V

+

+// when voltage is applied, Ge diode turns on first and 0.3 V is maintained across circuit so Si diode never turns on. So

+V_A=V-V_D_Ge

+

+printf("voltage V_A at point A = %.3f V \n",V_A)

diff --git a/3507/CH9/EX9.7/Ex9_7.sce b/3507/CH9/EX9.7/Ex9_7.sce
new file mode 100644
index 000000000..b5e10407a
--- /dev/null
+++ b/3507/CH9/EX9.7/Ex9_7.sce
@@ -0,0 +1,16 @@
+//chapter9

+//example9.7

+//page148

+

+V=10 // V

+V_D=0.7 // V

+R_BC=2 // kilo ohm

+R=2 // kilo ohm

+

+// by Kirchoff voltage law we get

+// -V_D-I_D*R_BC-2*I_D*R+V=0 thus making I_D as subject we get

+I_D=(V-V_D)/(R_BC+2*R)

+V_Q=2*I_D*R

+

+printf("I_D = %.3f mA \n",I_D)

+printf("V_Q = %.3f V \n",V_Q)

diff --git a/3507/CH9/EX9.8/Ex9_8.sce b/3507/CH9/EX9.8/Ex9_8.sce
new file mode 100644
index 000000000..3567f74b6
--- /dev/null
+++ b/3507/CH9/EX9.8/Ex9_8.sce
@@ -0,0 +1,15 @@
+//chapter9

+//example9.8

+//page148

+

+V=15 // V

+R=0.5 // kilo ohm

+V_D=0.7 // V

+

+// both diodes are forward biased

+

+I1=(V-V_D)/R

+I_D1=I1/2

+I_D2=I_D1

+

+printf("current through diode D1 = %.3f mA and diode D2 = %.3f mA \n",I_D1,I_D2)

diff --git a/3507/CH9/EX9.9/Ex9_9.sce b/3507/CH9/EX9.9/Ex9_9.sce
new file mode 100644
index 000000000..6b33fcb1e
--- /dev/null
+++ b/3507/CH9/EX9.9/Ex9_9.sce
@@ -0,0 +1,11 @@
+//chapter9

+//example9.9

+//page151

+

+P_dc=40 // W

+P_ac=100 // W

+

+efficiency=100*P_dc/P_ac

+

+printf("rectification efficiency = %.3f percent \n \n",efficiency)

+printf("remaining 60 watts are not lost. Crystal diode consumes only a \nlittle power due to its small internal resistance. \nActualy 100 W ac power is contained as 50 W in positive half \ncycle and 50 W in negative half cycle.\nThe 50 W of negative half cycle are not supplied at all. \nThe 50 W of positive half cycle are converted to 40 W \n")