65. When all the batteries are connected in parallel, each supplies a current i; thus, i
R
= N i. Then from
E = ir + i
R
R = ir + N ir, we get i
R
= N
E/[(N + 1)r]. When all the batteries are connected in series,
i
r
= i
R
and
E
total
= N
E = Ni
r
r + i
R
R = N i
R
r + i
R
r, so i
R
= N
E/[(N + 1)r].