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7. (a) Let
i
be the current in the circuit and take it to be positive if it is to the left in
R
1
.
We use Kirchhoff’s loop rule:
ε
1
–
iR
2
–
iR
1
–
2
= 0. We solve for
i
:
i
RR
=
−
+
=
−
+
=
12
12
6 0
80
050
VV
4.0
A
.
.
..
ΩΩ
A positive value is obtained, so the current is counterclockwise around the circuit.
If
i
is the current in a resistor
R
, then the power dissipated by that resistor is given by
2
PiR
=
.
(b) For
R
1
,
P
1
=
2
1
iR
=
(0.50 A)
2
(4.0
Ω
) = 1.0 W,
(c) and for
R
2
,
P
2
=
2
2
=
(0.50 A)
2
(8.0
Ω
) = 2.0 W.
If
i
is the current in a battery with emf
, then the battery supplies energy at the rate
P =i
provided the current and emf are in the same direction. The battery absorbs energy at the
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 Spring '08
 Any
 Physics, Current

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