61. Suppose that the switch had been in position a for a long time so that the current had reached the
steady-state value i
0
. The energy stored in the inductor is U
B
=
1
2
Li
2
0
. Now, the switch is thrown to
position b at time t = 0. Thereafter the current is given by
i = i
0
e
−t/τ
L
,
where τ
L
is the inductive time constant, given by τ
L
= L/R. The rate at which thermal energy is
generated in the resistor is given by
P = i
2
R = i
2
0
Re
−2t/τ
L
.
Over a long time period the energy dissipated is
∞
0
P dt = i
2
0
R
∞
0
e
−2t/τ
L
dt =
−
1
2
i
2
0
Rτ
L
e
−2t/τ
L
∞
0
=
1
2
i
2
0
Rτ
L
.
Upon substitution of τ
L
= L/R this becomes
1
2
Li
2
0
, the same as the total energy originally stored in the
inductor.