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+//Example 23.9
+L=7.5*10^-3;//Inductance (H)
+R=3;//Resistance (ohm)
+tau=L/R;//Time constant (s)
+printf('a.Time constant tau = %0.2f ms',tau*1000)
+I_0=10;//Initial current (A)
+I=0.368*I_0;//Current decreases to 0.368 times the initial value in tau seconds (A)
+t=tau;//Time (s) 
+while t<5*10^-3 
+    I=0.368*I;//Current (A)
+    t=t+tau;//Time (s)
+end// To find decline in current with time
+printf('\nb.Current = %0.2f A',I)
+//Here we used two iterations as we know 5ms is twice the characteristic time tau. I=I_0*exp(-t/tau) can also be used to find the current at 5ms.
+//Openstax - College Physics
+//Download for free at http://cnx.org/content/col11406/latest
+\ No newline at end of file