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(a) The larger
i
is 3.41 A.
(b) We use
V =
ε
– ir
= 2.00 V –
i
(0.500
Ω
). We substitute value of
i
obtained in part (a)
into the above formula to get
V
= 0.293 V.
(c) The smaller
i
is 0.586 A.
(d) The corresponding
V
is 1.71 V.
110. The power delivered by the motor is
P
= (2.00 V)(0.500 m/s) = 1.00 W. From
P
=
i
2
R
motor

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