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| author | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
| commit | 7bc77cb1ed33745c720952c92b3b2747c5cbf2df (patch) | |
| tree | 449d555969bfd7befe906877abab098c6e63a0e8 /3830/CH1/EX1.21/Ex1_21.sce | |
| parent | d1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff) | |
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Diffstat (limited to '3830/CH1/EX1.21/Ex1_21.sce')
| -rw-r--r-- | 3830/CH1/EX1.21/Ex1_21.sce | 24 |
1 files changed, 24 insertions, 0 deletions
diff --git a/3830/CH1/EX1.21/Ex1_21.sce b/3830/CH1/EX1.21/Ex1_21.sce new file mode 100644 index 000000000..3db36aa53 --- /dev/null +++ b/3830/CH1/EX1.21/Ex1_21.sce @@ -0,0 +1,24 @@ +// Exa 1.21
+
+clc;
+clear;
+
+// Given
+
+// A Volt box design
+Vs = 100; // Input voltage (V)
+V2 = 5; // Output voltage (V)
+Rs = 10*10^6; // Desired sum of resistance(R1+R2) Ohms
+
+
+// Solution
+
+// By voltage divider formula, we get
+// R2/(R1+R2) = V2/Vs ;
+// i.e, By simplifying
+R2 = Rs*V2/Vs;
+
+R1 = Rs - R2;
+printf(' The desired values of R1 and R2 to satisfy Volt box requirements are %.1f M ohms and %.2f M ohms respectively \n ',R1/10^6,R2/10^6);
+
+//The answer provided in the textbook is wrong
|