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CHAPTE R 27 CI RCU ITS
We can now redraw the circuit as in Fig. 27-11c; note that
the current through R 23 must be i 1 because charge that
moves through R 1 and R 4 must also move through R 23 . For
this simple one-loop circuit, the loop rule (applied clockwise
from point a as in Fig. 27-11d) yields
ϩᏱ Ϫ i 1 R 1 Ϫ i 1 R 23 Ϫ i 1 R 4 ϭ 0.
Substituting the given data, we find
12 V Ϫ i 1 (20 ⍀) Ϫ i 1 (12 ⍀) Ϫ i 1 (8.0 ⍀) ϭ 0,
which gives us
(Answer)
(b) What is the current i 2 through R 2 ?
KEY IDEAS
(1) we must now work backward from the equivalent circuit
of Fig. 27-11d, where R 23 has replaced R 2 and R 3 . (2) Because
R 2 and R 3 are in parallel, they both have the same potential
difference across them as R 23 .
Working backward: We know that the current through R 23
is i 1 ϭ 0.30 A. Thus, we can use Eq. 26-8 (R ϭ V/i) and
Fig. 27-11e to find the potential difference V 23 across R 23 .
Setting R 23 ϭ 12 ⍀ from (a), we write Eq. 26-8 as
V 23 ϭ i 1 R 23 ϭ (0.30 A)(12 ⍀) ϭ 3.6 V.
The potential difference across R 2 is thus also 3.6 V
(Fig. 27-11f), so the current i 2 in R 2 must be, by Eq. 26-8 and
Fig. 27-11g,
(Answer)
(c) What is the current i 3 through R 3 ?
KEY IDEAS
We can answer by using either of two techniques: (1) Apply
Eq. 26-8 as we just did. (2) Use the junction rule, which tells
us that at point b in Fig. 27-11b, the incoming current i 1 and
the outgoing currents i 2 and i 3 are related by
i 1 ϭ i 2 ϩ i 3 .
Calculation: Rearranging this junction-rule result yields
the result displayed in Fig. 27-11g:
i 3 ϭ i 1 Ϫ i 2 ϭ 0.30 A Ϫ 0.18 A
ϭ 0.12 A.
(Answer)
i 2 ϭ
V 2
R 2
ϭ
3.6 V
20 ⍀
ϭ 0.18 A.
i 1 ϭ
12 V
40 ⍀
ϭ 0.30 A.
Sample Problem 27.02 Resistors in parallel and in series
Figure 27-11a shows a multiloop circuit containing one ideal
battery and four resistances with the following values:
(a) What is the current through the battery?
KEY IDEA
Noting that the current through the battery must also be
the current through R 1 , we see that we might find the
current by applying the loop rule to a loop that includes R 1
because the current would be included in the potential
difference across R 1 .
Incorrect method: Either the left-hand loop or the big loop
should do. Noting that the emf arrow of the battery points
upward, so the current the battery supplies is clockwise, we
might apply the loop rule to the left-hand loop, clockwise
from point a. With i being the current through the battery,
we would get
ϩᏱ Ϫ iR 1 Ϫ iR 2 Ϫ iR 4 ϭ 0 (incorrect).
However, this equation is incorrect because it assumes
that R 1 , R 2 , and R 4 all have the same current i. Resistances
R 1 and R 4 do have the same current, because the current
passing through R 4 must pass through the battery and then
through R 1 with no change in value. However, that current
splits at junction point b — only part passes through R 2 , the
rest through R 3 .
Dead-end method: To distinguish the several currents in
the circuit, we must label them individually as in Fig. 27-11b.
Then, circling clockwise from a, we can write the loop rule
for the left-hand loop as
ϩᏱ Ϫ i 1 R 1 Ϫ i 2 R 2 Ϫ i 1 R 4 ϭ 0.
Unfortunately, this equation contains two unknowns, i 1 and
i 2 ; we would need at least one more equation to find them.
Successful method: A much easier option is to simplify
the circuit of Fig. 27-11b by finding equivalent resistances.
Note carefully that R 1 and R 2 are not in series and thus
cannot be replaced with an equivalent resistance.
However, R 2 and R 3 are in parallel, so we can use either
Eq. 27-24 or Eq. 27-25 to find their equivalent resistance
R 23 . From the latter,
R 23 ϭ
R 2 R 3
R 2 ϩ R 3
ϭ
(20 ⍀)(30 ⍀)
50 ⍀
ϭ 12 ⍀.
R 3 ϭ 30 ⍀, R 4 ϭ 8.0 ⍀.
R 1 ϭ 20 ⍀, R 2 ϭ 20 ⍀, Ᏹ ϭ 12 V,
Additional examples, video, and practice available at WileyPLUS
CHAPTE R 27 CI RCU ITS
We can now redraw the circuit as in Fig. 27-11c; note that
the current through R 23 must be i 1 because charge that
moves through R 1 and R 4 must also move through R 23 . For
this simple one-loop circuit, the loop rule (applied clockwise
from point a as in Fig. 27-11d) yields
ϩᏱ Ϫ i 1 R 1 Ϫ i 1 R 23 Ϫ i 1 R 4 ϭ 0.
Substituting the given data, we find
12 V Ϫ i 1 (20 ⍀) Ϫ i 1 (12 ⍀) Ϫ i 1 (8.0 ⍀) ϭ 0,
which gives us
(Answer)
(b) What is the current i 2 through R 2 ?
KEY IDEAS
(1) we must now work backward from the equivalent circuit
of Fig. 27-11d, where R 23 has replaced R 2 and R 3 . (2) Because
R 2 and R 3 are in parallel, they both have the same potential
difference across them as R 23 .
Working backward: We know that the current through R 23
is i 1 ϭ 0.30 A. Thus, we can use Eq. 26-8 (R ϭ V/i) and
Fig. 27-11e to find the potential difference V 23 across R 23 .
Setting R 23 ϭ 12 ⍀ from (a), we write Eq. 26-8 as
V 23 ϭ i 1 R 23 ϭ (0.30 A)(12 ⍀) ϭ 3.6 V.
The potential difference across R 2 is thus also 3.6 V
(Fig. 27-11f), so the current i 2 in R 2 must be, by Eq. 26-8 and
Fig. 27-11g,
(Answer)
(c) What is the current i 3 through R 3 ?
KEY IDEAS
We can answer by using either of two techniques: (1) Apply
Eq. 26-8 as we just did. (2) Use the junction rule, which tells
us that at point b in Fig. 27-11b, the incoming current i 1 and
the outgoing currents i 2 and i 3 are related by
i 1 ϭ i 2 ϩ i 3 .
Calculation: Rearranging this junction-rule result yields
the result displayed in Fig. 27-11g:
i 3 ϭ i 1 Ϫ i 2 ϭ 0.30 A Ϫ 0.18 A
ϭ 0.12 A.
(Answer)
i 2 ϭ
V 2
R 2
ϭ
3.6 V
20 ⍀
ϭ 0.18 A.
i 1 ϭ
12 V
40 ⍀
ϭ 0.30 A.
Sample Problem 27.02 Resistors in parallel and in series
Figure 27-11a shows a multiloop circuit containing one ideal
battery and four resistances with the following values:
(a) What is the current through the battery?
KEY IDEA
Noting that the current through the battery must also be
the current through R 1 , we see that we might find the
current by applying the loop rule to a loop that includes R 1
because the current would be included in the potential
difference across R 1 .
Incorrect method: Either the left-hand loop or the big loop
should do. Noting that the emf arrow of the battery points
upward, so the current the battery supplies is clockwise, we
might apply the loop rule to the left-hand loop, clockwise
from point a. With i being the current through the battery,
we would get
ϩᏱ Ϫ iR 1 Ϫ iR 2 Ϫ iR 4 ϭ 0 (incorrect).
However, this equation is incorrect because it assumes
that R 1 , R 2 , and R 4 all have the same current i. Resistances
R 1 and R 4 do have the same current, because the current
passing through R 4 must pass through the battery and then
through R 1 with no change in value. However, that current
splits at junction point b — only part passes through R 2 , the
rest through R 3 .
Dead-end method: To distinguish the several currents in
the circuit, we must label them individually as in Fig. 27-11b.
Then, circling clockwise from a, we can write the loop rule
for the left-hand loop as
ϩᏱ Ϫ i 1 R 1 Ϫ i 2 R 2 Ϫ i 1 R 4 ϭ 0.
Unfortunately, this equation contains two unknowns, i 1 and
i 2 ; we would need at least one more equation to find them.
Successful method: A much easier option is to simplify
the circuit of Fig. 27-11b by finding equivalent resistances.
Note carefully that R 1 and R 2 are not in series and thus
cannot be replaced with an equivalent resistance.
However, R 2 and R 3 are in parallel, so we can use either
Eq. 27-24 or Eq. 27-25 to find their equivalent resistance
R 23 . From the latter,
R 23 ϭ
R 2 R 3
R 2 ϩ R 3
ϭ
(20 ⍀)(30 ⍀)
50 ⍀
ϭ 12 ⍀.
R 3 ϭ 30 ⍀, R 4 ϭ 8.0 ⍀.
R 1 ϭ 20 ⍀, R 2 ϭ 20 ⍀, Ᏹ ϭ 12 V,
Additional examples, video, and practice available at WileyPLUS