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.