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+clear
+//
+v=120
+r=40
+i=v/(r)
+//applying kvl to the closed loop
+v=3*520
+printf("\n voltage= %0.1f v",v)
+//when v=120,R can be found by I*(r+20)=120-->r=20
+r=20
+printf("\n r=20 ohm")
+//when r=20 total r=20+20+20=60
+r=60
+l=10
+tc=l/(r) //time constant
+printf("\n tc= %0.1f sec",tc)
+//i=I0*e^-(t/tc)=3*e^(-6t)
+energy=(10*9)/(2)
+benergy=0.05*energy
+printf("\n balance energy= %0.1f J",benergy)
+//(L*i^2)/2=2.25-->hence i=0.6708
+//3*e^-6t=0.6708-->e^-6t=0.2236-->applying log on both sides we get t=0.25
+printf("\n t=0.25 sec")