2. Identify the unknowns. You will need as manyindependent equations as there are unknowns. Youmay write down more equations than this, but youwill find that some of the equations will be redundant(that is, not be independent in the sense of providingnew information). You may usefor eachresistor, which sometimes will reduce the numberof unknowns.
3. Apply Kirchhoff’s junction rule
at one or more
junctions.
V
=
IR
I
1
,
I
2
,
I
3
h
a
b
c
d
g
f
e
30
Ω
40
Ω
20
Ω
r
=
1
Ω
e
1
=
80 V
r
=
1
Ω
e
2
=
45 V
I
3
I
1
I
2
FIGURE 19–13
Currents can be
calculated using Kirchhoff’s rules.
See Example 19
–
8.
Using Kirchhoff’s rules.
Calculate the currents
and
in the three branches of the circuit in Fig. 19
–
13 (which is the same as Fig. 19
–
11).
APPROACH and SOLUTION
1. Label the currents
and their directions. Figure 19
–
13 uses the labels
and
for the current in the three separate branches. Since (positive) current
tends to move away from the positive terminal of a battery, we choose
and
to have the directions shown in Fig. 19
–
13. The direction of
is not obvious
in advance, so we arbitrarily chose the direction indicated. If the current
actually flows in the opposite direction, our answer will have a negative sign.
2.Identify the unknowns. We have three unknowns and thereforewe need three equations, which we get by applying Kirchhoff’s junction andloop rules.
3.
Junction rule
: We apply Kirchhoff’s junction rule to the currents at point a,
where
enters and
and
leave:
(i)
This same equation holds at point d, so we get no new information by writing
an equation for point d.
I
3
=
I
1
+
I
2
.
I
1
I
2
I
3
(I
1
,
I
2
, and
I
3
)
I
1
I
3
I
2
I
3
I
1
,
I
2
,
I
3
I
1
,
I
2
,
EXAMPLE 19
;
8
P R O B L E M S O L V I N G
Choose current directions arbitrarily

SECTION 19
–
3
Kirchhoff’s Rules
535
P R O B L E M S O L V I N G
Be consistent
w
ith signs
w
hen
applying the loop rule
P R O B L E M S O L V I N G
I
1
is in the opposite direction
from that assumed in Fig. 19
–
13
4.
Loop rule
: We apply Kirchhoff’s loop rule to two different closed loops. First
we apply it to the upper loop ahdcba. We start (and end) at point a. From
a to h we have a potential decrease
From h to d there is no
change,
but from d to c the potential increases by 45 V:
that is,
From c to a the potential decreases through the two resistances by an amount
Thus we have
or
(ii)
where we have omitted the units (volts and amps) so we can more easily see
the algebra. For our second loop, we take the outer loop ahdefga. (We could
have chosen the lower loop abcdefga instead.) Again we start at point a, and
going to point h we have
Next,
But when we
take our positive test charge from d to e, it actually is going uphill, against the
current
—
or at least against the
assumed
direction of the current, which is
what counts in this calculation. Thus
has a
positi
v
e
sign.

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