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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 /608/CH13/EX13.20 | |
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initial commit / add all books
Diffstat (limited to '608/CH13/EX13.20')
-rwxr-xr-x | 608/CH13/EX13.20/13_20.jpg | bin | 0 -> 21012 bytes | |||
-rwxr-xr-x | 608/CH13/EX13.20/13_20.sce | 13 |
2 files changed, 13 insertions, 0 deletions
diff --git a/608/CH13/EX13.20/13_20.jpg b/608/CH13/EX13.20/13_20.jpg Binary files differnew file mode 100755 index 000000000..a22bdd4f6 --- /dev/null +++ b/608/CH13/EX13.20/13_20.jpg diff --git a/608/CH13/EX13.20/13_20.sce b/608/CH13/EX13.20/13_20.sce new file mode 100755 index 000000000..c9542b71e --- /dev/null +++ b/608/CH13/EX13.20/13_20.sce @@ -0,0 +1,13 @@ +//Problem 13.20: The circuit diagram of Figure 13.48 shows dry cells of source e.m.f. 6 V, and internal resistance 2.5 ohm. If the load resistance RL is varied from 0 to 5 ohm in 0.5 ohm steps, calculate the power dissipated by the load in each case. Plot a graph of RL (horizontally) against power (vertically) and determine the maximum power dissipated.
+
+//initializing the variables:
+V = 6; // in volts
+r = 2.5; // in ohms
+
+//calculation:
+RL=(0:0.5:5)';
+function [y]=f(RL)
+ y = RL*(V/(r + RL))^2;
+endfunction
+fplot2d(RL, f)
+xtitle("graph of (RL) against power(P)", "RL(ohm)", "P(W)")
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