Question

A `3.56 H` inductor is placed in series with a `12.8Omega` resistor. An emf of `3.24 V` is then suddenly applied across the `RL` combination.
(a) At `0.278 s` after the emf is applied what is the rate at which energy is being delivered by the battery?
(b) At `0.278 s`, at what rate is energy appearing as thermal energy in the resistor? (c) At` 0.278 s`, at what rate is energy being stored in the magnetic field?

Answer 1

Correct Answer - A
`i_0=V/R=3.24/12.8=0.253A`
`tau_L=L/R=3.56/12.8=0.278s`
a. After one constant `(t=0.278s=tau_C)`
`i=(1/1-e)i_0`
`~~0.63i_0`
`=0.16A`
Power supplied by battery `=Ei`
`P=(3.24)(0.16)=0.518W`
b.`P_R=i^2R`
`=(0.16)^2(12.8)=0.328W`
c. `P_L=P_P_r=0.191W`