76. For t < 0, no current goes through L

2

, so i

2

= 0 and i

1

=

E/R. As the switch is opened there will be a

very brief sparking across the gap. i

1

drops while i

2

increases, both very quickly. The loop rule can be

written as

E − i

1

R

− L

1

di

1

dt

− i

2

R

− L

2

di

2

dt

= 0 ,

where the initial value of i

1

at t = 0 is given by

E/R and that of i

2

at t = 0 is 0. We consider the situation

shortly after t = 0. Since the sparking is very brief, we can reasonably assume that both i

1

and i

2

get

equalized quickly, before they can change appreciably from their respective initial values. Here, the loop
rule requires that L

1

(di

1

/dt), which is large and negative, must roughly cancel L

2

(di

2

/dt), which is large

and positive:

L

1

di

1

dt

≈ −L

2

di

2

dt

.

Let the common value reached by i

1

and i

2

be i, then

di

1

dt

≈

∆i

1

∆t

=

i

− E/R

∆t

and

di

2

dt

≈

∆i

2

∆t

=

i

− 0

∆t

.

The equations above yield

L

1

i

−

E

R



=

−L

2

(i

− 0) =⇒ i =

EL

1

L

2

R

1

+ L

1

R

2

=

L

1

L

1

+ L

2

E

R

.