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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
commitb1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch)
treeab291cffc65280e58ac82470ba63fbcca7805165 /1823/CH10
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-rwxr-xr-x1823/CH10/EX10.1/SolEx10_1.sce10
-rwxr-xr-x1823/CH10/EX10.2/SolEx10_2.sce9
-rwxr-xr-x1823/CH10/EX10.4/SolEx10_4.sce13
-rwxr-xr-x1823/CH10/EX10.7/SolEx10_7.sce14
-rwxr-xr-x1823/CH10/EX10.9/SolEx10_10.sce14
5 files changed, 60 insertions, 0 deletions
diff --git a/1823/CH10/EX10.1/SolEx10_1.sce b/1823/CH10/EX10.1/SolEx10_1.sce
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index 000000000..2b558c50b
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+++ b/1823/CH10/EX10.1/SolEx10_1.sce
@@ -0,0 +1,10 @@
+//Find the average values of (a) input voltage and (b) input current.
+//Example 10.1 page no 296
+clear
+clc
+V2=12 //load
+D=0.8 //duty cycle
+V1=V2/D //V
+P0=20 //average power
+I1=P0/V1
+printf("\n The value of I1=%0.3f A",I1)
diff --git a/1823/CH10/EX10.2/SolEx10_2.sce b/1823/CH10/EX10.2/SolEx10_2.sce
new file mode 100755
index 000000000..00e76bc25
--- /dev/null
+++ b/1823/CH10/EX10.2/SolEx10_2.sce
@@ -0,0 +1,9 @@
+//Determine the smallest value of duty cycle possible
+//Example 10.2 page no 296
+clear
+clc
+fs=30*10^3 //kHz.
+Lc=50*10^-6 // Inductor H
+Rl=7 //Load Ω
+D=1-((2*fs*Lc)/Rl)
+printf("\n The value of D=%0.3f ",D)
diff --git a/1823/CH10/EX10.4/SolEx10_4.sce b/1823/CH10/EX10.4/SolEx10_4.sce
new file mode 100755
index 000000000..691248742
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+++ b/1823/CH10/EX10.4/SolEx10_4.sce
@@ -0,0 +1,13 @@
+//Determine (a) the duty cycle and (b) the output power.
+//Ts for the buck converter.
+//Example 10.4 page no 296
+clear
+clc
+V2=5 //V
+V1=12 //V
+D=V2/V1
+Rl=5 //Ω
+V2=5 //V
+p0=V2^2/Rl
+printf("\n The value of D=%0.3f ",D)
+printf("\n The value of p0=%0.3f ",p0)
diff --git a/1823/CH10/EX10.7/SolEx10_7.sce b/1823/CH10/EX10.7/SolEx10_7.sce
new file mode 100755
index 000000000..5c7e7847d
--- /dev/null
+++ b/1823/CH10/EX10.7/SolEx10_7.sce
@@ -0,0 +1,14 @@
+//Find the maximum and minimum values of the inductor current
+//Example 10.7 page no 297
+clear
+clc
+D=0.6 //Duty cycle
+V1=24 //V
+Rl=7
+fs=30*10^3
+L=50*10^-6
+V2=D*V1
+Imax=V2/Rl+((V1-V2)*D)/(2*fs*L) //maximum values of the inductor current
+Imin=V2/Rl-((V1-V2)*D)/(2*fs*L) // minimum values of the inductor current
+printf("\n The value of Imax=%0.3f A ",Imax)
+printf("\n The value of Imin=%0.3f A",Imin)
diff --git a/1823/CH10/EX10.9/SolEx10_10.sce b/1823/CH10/EX10.9/SolEx10_10.sce
new file mode 100755
index 000000000..2d0d8a216
--- /dev/null
+++ b/1823/CH10/EX10.9/SolEx10_10.sce
@@ -0,0 +1,14 @@
+//Determine (a) the output voltage, (b) the load resistance, and (c) the load current.
+//Example 10.9
+//page no 298
+clear
+clc
+V1=12
+D=0.6
+V2=V1/(1-D) //output voltage
+P0=60 //w Supplying power
+Rl=V2^2/P0 //load resistance
+I2=V2/Rl //load current
+printf("\n The value of V2=%0.3f V ",V2)
+printf("\n The value of Rl=%0.3f ohm",Rl)
+printf("\n The value of I2=%0.3f A ",I2)