67. When connected in series, the rate at which electric energy dissipates is P
s
=
E
2
/(R
1
+ R
2
). When
connected in parallel, the corresponding rate is P
p
=
E
2
(R
1
+ R
2
)/R
1
R
2
. Letting P
p
/P
s
= 5, we get
(R
1
+ R
2
)
2
/R
1
R
2
= 5, where R
1
= 100 Ω. We solve for R
2
: R
2
= 38 Ω or 260 Ω.