P28_024.PDF

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24. The currents
i
1
,
i
2
and
i
3
are obtained from Eqs. 28-15 through 28-17:
i
1
=
=
i
2
=
=
i
3
=
E
1
(R
2
+
R
3
)
− E
2
R
3
(4.0 V)(10 Ω + 5.0 Ω)
−
(1.0 V)(5.0 Ω)
=
R
1
R
2
+
R
2
R
3
+
R
1
R
3
(10 Ω)(10 Ω) + (10 Ω)(5.0 Ω) + (10 Ω)(5.0 Ω)
0.275 A
,
E
1
R
3
− E
2
(R
1
+
R
2
)
(4.0 V)(5.0 Ω)
−
(1.0 V)(10 Ω + 5.0 Ω)
=
R
1
R
2
+
R
2
R
3
+
R
1
R
3
(10 Ω)(10 Ω) + (10 Ω)(5.0 Ω) + (10 Ω)(5.0 Ω)
0.025 A
,
i
2
−
i
1
= 0.025 A
−
0.275 A =
−0.250
A
.
V
d
−
V
c
can now be calculated by taking various paths. Two examples: from
V
d
−
i
2
R
2
=
V
c
we get
V
d
−
V
c
=
i
2
R
2
= (0.0250 A)(10 Ω) = +0.25 V; from
V
d
+
i
3
R
3
+
E
2
=
V
c
we get
V
d
−
V
c
=
−i
3
R
3
− E
2
=
−(−0.250
A)(5.0 Ω)
−
1.0 V = +0.25 V.

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