For the circuit in problem 3 above assume that S 120 V R 7

For the circuit in problem #3 above, assume that S = 12.0 V. R = 7.60 Ohm , and L = 5.00 H. The ideal battery is connected at time t = 0.  Determine the amount of energy delivered by the battery during the first 3.00 s. Determine how much of this energy is stored in the magnetic field of the  inductor. Determine how much of this energy is dissipated in the resistor.

Solution

Emf of the battery E = 12 volt

Resistance R = 7.6 ohm

Inductance L = 5 H

(a). The amount of energy deliveres by the battery during the first 3 s is U = (E ² /R ) t

     U = (12 ² / 7.6) 3

        = 18.94(3)

        = 56.84 J

(b). Current after time t is i = io [1-e ^{-(R/L) t}]

Where io = E/R = 12/7.6 = 1.578 A

Substitute values you get i = 1.5789 [1- e ^-(7.6/5)3]

                                        = 1.5789 [1-0.0104]

                                        = 1.562 A

(b).Energy stored in the magnetic field of the inductor U \' = (1/2)Li 2

       U \' = 0.5 x 5 x 1.562 ²

           = 6.1 J

(c).Energy dissipated in the resistor U \" = i ² Rt

                                                          = 1.562 ² (7.6) 3

                                                          = 55.62 J