At time t0, emf of 500V is applied to a coil which has industance of 0.7H and resistance of 40ohm. What is the energy stored in the magnetic field in the inductor when the current reaches one third of its maximum value?OMG im stuck on this, i don't know which formula to use.
1kWh3.6*10^6 J SO 3kWh3*3.6*10^610.8*10^6 J Inductors save the flexibility in variety of magnetic field. power saved by using inductor0.5*L*I^2 10.8*10^60.5*L*I^2 21.6*10^6L*I^2 L21.6*10^6/(9*10^4) L2.4*10^2240 H (Unit of inductance is Henry)
The maximum current is 500 V / 40.0 ? 12.5 A. Two-thirds of that is 8.33 A. The formula you need to use is (1/2)LI?. At the two-thirds point, the energy stored in the inductor is U (1/2) x (0.700 H) x (8.33 A)? 24.3 J
