A current of 1.24A in an inductor L results in a stored energy of 0.195J. The current is then changed to 6.47A in the opposite direction. Calculate the change of stored energy.
Situation 1: MPE1 1/2*L*I1^2 Situation 2: MPE2 1/2*L*I2^2 L doesn't change, since it only depends on geometry and core material properties of the inductor. Solve for L in situation 1: L 2*MPE1/I1^2 Substitute: MPE2 MPE1 * I2^2/I1^2 We are interested in MPE2 - MPE1: MPE2 - MPE1 MPE1*(I2^2/I1^2 - 1) Data: MPE1:0.195 J; I1:1.24 A; I2:6.47 A; Result: MPE2 - MPE1 5.1138 Joules
Dec 6, 2017