Question

A `5H` inductor is placed in series with a `10Omega` resistor. An emf of `5V` being suddenly applied to the combination. Using these values prove the principle of conservation of energy, for time equal to the time constant.

Answer 1

At any instant `t` current in `L-R` circuit is given as
`i=i_0(1-e^(-t//tau_L))`
Here, `i_0=E/R` and `tau_L=L/R`
After one time constant `(t=tau_L)` current in the circuit is
`i=E/R(1-1/e)=5/10(1-1/e)=0.316A`
The rate at which the energy is delivered by the battery is
`P_1=Ei=(5)(0.316)=1.58W`.............i
At this time rate by which energy is dissipated in the resistor is
`P_2=i^2R=(0.316)^2(10)=0.998w` ...........ii
The rate at which energy is stored in the inductor is
`P_3=d/(dt)(1/2Li^2)=Li((di)/(dt))`
Here, `(di)/(dt)=i_0/tau_Le^-1=E/(eL)` (after one time constant)
Substituting the values we get
`P_3=(L)(i)(E/(eL))=(Ei)/e`
`=(5xx0.316)/2.718=0.582W`............iii
From eqn i, ii and iii we have
`P_1=P_2+P_3`
It is the same as required by the principle of conservation of energy.