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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3835/CH2/EX2.21/Ex2_21.sce | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3835/CH2/EX2.21/Ex2_21.sce')
| -rw-r--r-- | 3835/CH2/EX2.21/Ex2_21.sce | 12 |
1 files changed, 12 insertions, 0 deletions
diff --git a/3835/CH2/EX2.21/Ex2_21.sce b/3835/CH2/EX2.21/Ex2_21.sce new file mode 100644 index 000000000..80045b260 --- /dev/null +++ b/3835/CH2/EX2.21/Ex2_21.sce @@ -0,0 +1,12 @@ +clear +// +//the following problem is based on usage of superposition theorem +i8=12/(6+4+8) //current for 8 ohm resistor.the resistances are in series with each other.Hence 6+4+8 +//next when voltage source is short circuited (8+4) total of resistance is obtained.The 4A is distributed in parallel branches as per current divider rule +i=(4*6)/(6+12) +printf("\n i8= %0.1f A",i8) +printf("\n i8= %0.1f A",i) +tot=i8+i +printf("\n total current= %0.1f A",tot) + + |