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(c) The power dissipated in
R
3
is
()
(
)
2
2
33
3
0.263A
5.00
0.346W.
Pi
R
==
−
Ω
=
(d) The power supplied by
ε
1
is
i
3
1
= (0.421 A)(3.00 V) = 1.26 W.
(e) The power
“supplied”
by
2
is
i
2
2
= (–0.158 A)(1.00 V) = –0.158 W. The negative
sign indicates that
2
is actually absorbing energy from the circuit.
43. (a) We first find the currents. Let
i
1
be the current in
R
1
and take it to be positive if it
is to the right. Let
i
2
be the current in
R
2
and take it to be positive if it is to the left. Let
i
3
be the current in
R
3
and take it to be positive if it is upward. The junction rule produces
iii
123
0
+
+
=
.
The loop rule applied to the lefthand loop produces
11
13
3
0
iR iR
−
+=
and applied to the righthand loop produces
22
23
3
0.
−
We substitute
i
3
= –
i
2
–
i
1
, from the first equation, into the other two to obtain
12
31
3
0
iR iR iR
−
−−=
and
3
−
Solving the above equations yield
3
2
3
1
(3.00 V)(2.00
5.00
) (1.00 V)(5.00
)
0.421 A.
(4.00
)(2.00
) (4.00
)(5.00 ) (2.00
)(5.00
)
RR
R
i
+−
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 Spring '08
 Any
 Physics, Current

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