Consider the `RL` circuit in Fig. When the switch is closed in position `1` and opens in position `2`, electrical work must be performed on the inductor and on the resistor. The energy stored in the inductor is for the resistor energy appears as heat.
a. What is the ratio of `P_(L)//P_(R )` of the rate at which energy is stored in the inductor to the rate at which energy is dissipated in the resistor?
b. Express the ratio `P_(L)//P_(R )` as a function of time.
c. If the time constant of circuit is `t`, what is the time at which `P_(L) = P_(R )`?

Explanation

The anergy stored in the inductor is `U_(L) = (1)/(2)LI^(2)`
Power `P_(L) = (dU)/(dt) = (d)/(dt)((1)/(2)LI^(2)) = LI(dI)/(dt)`(i)
Power of resitor `P_(R) = I^(2)R` (ii)

So the ratio `(P_(L))/(P_(R)) = (LIdI//dt)/(I^(2)R) = (tau(dI//dt))/(I)` (iii)
b. As `I = (E)/(R ) [1 - e^(-(t//tau))]` (iv)
`(dI)/(dt) = (d)/(dt) [(E)/(R ) (1 - e^(-(t//tau)))] = (E)/(Rtau) e(-(t//tau)` (v)
Inserting the value of `dI//dt` and `I` in Eq.(iii), we have
`(P_(L))/(P_(R)) = tau = ((E)/(R tau) e^(-t//tau))/((E)/(R ) (1-e^(-t//tau))) =(e^(-(t//tau)))/(1-e^(-(t//tau))) = (1)/(e^((t//tau))-1)` (vi)
When `t = 0`, this ratio tends to infinity, due to large initial value of `dI//dt` and small initial value of `I`. When `t` tends to infinity, after many time constants, the current tends to zero. As a result, vanishes and the ratio goes to zero.
c. From Eq.(vi),`P_(L) = P_(R )` when `(1)/(e(t//tau) - 1) = 1` which on
simplification yields `e^((t//tau)) = 2, (t)/(tau) = In2 rArr t = 0.693 tau`