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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /2240/CH8/EX7.1/EX7_1.sce | |
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initial commit / add all books
Diffstat (limited to '2240/CH8/EX7.1/EX7_1.sce')
-rwxr-xr-x | 2240/CH8/EX7.1/EX7_1.sce | 17 |
1 files changed, 17 insertions, 0 deletions
diff --git a/2240/CH8/EX7.1/EX7_1.sce b/2240/CH8/EX7.1/EX7_1.sce new file mode 100755 index 000000000..286136d0b --- /dev/null +++ b/2240/CH8/EX7.1/EX7_1.sce @@ -0,0 +1,17 @@ +// Grob's Basic Electronics 11e
+// Chapter No. 07
+// Example No. 7_1
+clc; clear;
+// Three 50 Ohms resistors R1, R2 and R3 are in series across an applied voltage of 180 V. How much is the IR voltage drop across each resistor?
+
+// Given data
+
+R1 = 50*10^3; // Resistor 1=50k Ohms
+R2 = 50*10^3; // Resistor 2=50k Ohms
+R3 = 50*10^3; // Resistor 3=50k Ohms
+Vt = 180; // Applied Voltage=180 Volts
+
+R = R1 // R = R1 = R2 = R3
+Rt = R1+R2+R3;
+V = Vt*(R/Rt);
+disp (V,'The Voltage Drop across each Resistor in Volts')
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