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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
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| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /644/CH1/EX1.3/p3.sce | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '644/CH1/EX1.3/p3.sce')
| -rw-r--r-- | 644/CH1/EX1.3/p3.sce | 15 |
1 files changed, 15 insertions, 0 deletions
diff --git a/644/CH1/EX1.3/p3.sce b/644/CH1/EX1.3/p3.sce new file mode 100644 index 000000000..d8309331f --- /dev/null +++ b/644/CH1/EX1.3/p3.sce @@ -0,0 +1,15 @@ +// Example1.3. A coil consists of 2000 turns of copper wire having a cross sectional area of 0.8 mm^2. The mean length per turn is 80 cm and resistivity of copper is 0.02 micro-ohm- metre. Find the resistance of the coil and power absorbed by the coil when connected across 110 V d.c. supply.
+//1 millimetre = 10^-3 metre
+// 1 micro-ohm = 10^-6 ohms
+N = 2000; // number of turns
+A = 0.8*10^-6;// crossectional area (m^2)
+l = 80*10^-2;// mean length(m)
+p = 0.02*10^-6;// resistivity (ohm-m)
+V = 110; // supply voltage(V)
+// Let R ohms be the resistance of the coil and P watts be the power absorbed
+// FORMULAE: R=p*L/A , where L is the length of the coil ; P= V^2/R
+L= prod([l,N]);// length of the coil(m)
+R =prod([p,L])/A;// calculation of resisrance (ohms)
+P = (V^2)/R;// power absorbed by the coil (Watts)
+disp(R,"resistance of the coil (in ohms)= ")
+disp(P,"power absorbed by the coil(in watts)=")
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