138
CHAPTER 18
Apply Kirchhoff’s loop rule around loop
abcfa
to find
(
)
(
)
3
6.00
6.00
0
I
I
−
Ω
+
Ω
=
or
I
3
I
=
(1)
Apply Kirchhoff’s loop rule around loop
abcda
to find
(
)
(
)
2
0
I
I
Ω
=
6.00
6.00 V
10.0
−
Ω
+
−
or
2
0.600 A
0.600
I
I
=
−
(2)
Apply Kirchhoff’s loop rule around loop
abcea
to find
(
)
(
)
1
0
I
I
Ω
=
6.00
4.50 V
5.00
−
Ω
+
−
or
1
0.900 A
1.20
I
I
=
−
(3)
Finally, apply Kirchhoff’s junction rule at either point
a
or point
c
to obtain
(4)
Substitute equations (1), (2), and (3) into equation (4) to obtain the current through the
ammeter. This gives
or
and
3
1
2
I
I
I
I
+
=
+
0.900 A
1.20
0.600
I
I
I
+
=
−
+
3.80
1.50 A
I
=
A
0
−
.600
I
1.50 A
3.80
0.395 A
I
=
=
18.64
In the figure given below, note that all bulbs have the same resistance,
.
R
(a)
In the series situation, Case 1, the same current
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 Fall '10
 STAFF
 Physics, Current, Resistor, Electrical resistance, Kirchhoff, bulbs, potential difference ∆v

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