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I s
1
=
V s
1
R s
1
=
6.00 V
4.00
Æ
=
1.50 A
Next, notice that I s
1 is also the current in they are in series. (In Fig. 18.6b, I s
1
=
I p
1
R p
1
=
I
5 and R
=
5
, because
1.50 A
The resistors’ individual voltages are therefore
.)
V p
1
=
I s
1
R p
1
=
1
1.50 A
21
1.50
Æ
2
=
2.25 V and
V
5 =
I s
1
R
5 =
1
1.50 A
21
2.50
Æ
2
=
3.75 V
(As a check, note that the voltages do, in fact, add to 6.00 V.)
Finally, the voltage across R
(why?), and
3 and R
4 is the same as V p
1
V p
1
=
V
3 =
V
4 =
2.25 V
With these voltages and known resistances, the last two currents, I
3 and I
4
, are and
I
3 =
V
3
R
3
=
2.25 V
6.00
Æ
=
0.375 A
I
4 =
V
4
R
4
=
2.25 V
2.00
Æ
=
1.13 A
The current
1
I s
1
2 is expected to divide at the
Thus, a double-check is available: I
3 +
I
4
R
3
R
4 junction.
does, in fact, equal
I s
1
, within rounding errors.
F O L L O W - U P E X E R C I S E .
In this Example, verify that the total power delivered to all of the resistors is the same as the power output of the battery. Why must this be true?
6 3 1
D I D Y O U L E A R N ?
➥ In series, all circuit elements carry the same current.
➥ For resistors in series, the resistor with the most resistance has the largest voltage across it.
➥ For resistors in parallel, the resistor with the least resistance dissipates the most power.
R
1
R
2
V
2
R
4
R
5
18.2
L E A R N I N G P A T H Q U E S T I O N S
➥ Kirchhoff’s junction rule amounts to applying conservation of what quantity?
➥ Kirchhoff’s loop rule amounts to applying conservation of what quantity?
➥ If a a resistor is traversed in the direction opposite to its current direction, what is the sign of the voltage across it?
Series–parallel circuits with a single voltage source can be reduced to a single loop, as seen in Example 18.3. However, circuits may contain several loops, each one having several voltage sources, resistances, or both. In many cases, resistors may not be in series or in parallel. As an example of this situation, a multiloop circuit, which does not lend itself to the methods of Section 18.1, is shown in 䉴 Fig. 18.7a. Even though some groups of resistors may be replaced by their equivalent resistances (Fig. 18.7b), this circuit can be reduced only so far by using parallel and series procedures.
Analyzing these types of circuits requires a more general approach—that is, the application of Kirchhoff’s rules.* These rules embody conservation of charge and energy. (Although not stated specifically, Kirchhoff’s rules were applied to the parallel and series arrangements in Section 18.1.) First, it is useful to introduce some terminology that will help us describe more complex circuits:
■
■
A point where three or more wires are joined is called a junction or node—for example, point A in Fig. 18.7b.
A path connecting two junctions is called a branch. A branch may contain one or more circuit elements, and there may be more than two branches between two junctions.
*Gustav Robert Kirchhoff (1824–1887) was a German scientist who made important contributions to electrical circuit theory and light spectroscopy.
V
1
R
3
(a)
I
1
V
1
R s
I
1
A
I
3
I
2
V
2
R p
R
3
I
1
(b)
B
I
3
I
2
䉱
F I G U R E 1 8 . 7 Multiloop circuit
In general, a circuit that contains voltage sources in more than one loop cannot be completely reduced by series and parallel methods alone. However, some reductions within each loop may be possible, such as from part (a) to part (b). At a junction the current divides or comes together, as at junctions A and B in part (b), respectively.
Any path between two junctions is called a branch. In part (b), there are three branches—that is, there are three different ways to get from junction A to junction B.

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K I R C H H O F F ’ S J U N C T I O N T H E O R E M
Kirchhoff’s first rule , or junction theorem, states that the algebraic sum of the currents at any junction is zero: g
I i =
0
(sum of currents
at a junction)
(18.4)
This means that the sum of the currents entering a junction (taken as positive) and the currents leaving the junction (taken as negative) is zero. This rule is just a statement of charge conservation—no charge can pile up at a junction (why?). For the junction at point A in Fig. 18.7b, for example, using the sign conventions the algebraic sum of the currents is
I
1 -
I
2 -
I
3 =
0 or equivalently
I
1 =
I
2 +
I
3
current in
= current out
(This rule was applied in analyzing parallel resistances in Section 18.1.)
P R O B L E M - S O L V I N G H I N T
Sometimes it is not evident whether a particular current is directed into or out of a junction just from looking at a circuit diagram. In this case, a direction is simply assumed.
Then the currents are calculated, without worry about their directions. If some of the assumed directions turn out to be opposite to the actual directions, then negative answers for these currents will result. This outcome means that the directions of these currents are opposite to the directions initially chosen (or guessed).
–
V > 0
+
V < 0
Across battery
(a)
V < 0
R
I
V > 0
Across resistor
(b)
䉱
F I G U R E 1 8 . 8 Sign convention for Kirchhoff ’ s loop theorem (a) The battery voltage is taken as positive if it is traversed from the negative to the positive terminal. It is assigned a negative value if traversed from the positive to the negative. (b) The voltage across a resistor is taken as negative if the resistor is traversed in the direction of the assigned current (“downstream”). It is taken as positive if the resistance is traversed in the direction opposite that of the assigned branch current
(“upstream”).
K I R C H H O F F ’ S L O O P T H E O R E M
Kirchhoff’s second rule , or loop theorem, states that the algebraic sum of the potential differences (voltages) across all of the elements of any closed loop is zero: g
V i =
0
(sum of voltages
around a closed loop)
(18.5)
This expression means that the sum of the voltage rises (an increase in potential) equals the sum of the voltage drops (a decrease in potential) around a closed loop, which must be true if energy is conserved. (This rule was used in analyzing series resistances in Section 18.1.)
Notice that traversing a circuit loop in different directions will yield either a voltage rise or a voltage drop across each circuit element. Thus, it is important to establish a sign convention for voltages. The sign conventions used in this book are illustrated in 䉳 Fig. 18.8. The voltage across a battery is taken as positive (a voltage rise) if it is traversed from the negative terminal to the positive (Fig.
18.8a) and negative if it is traversed from positive to negative. (Note that the direction of the current through the battery has nothing to do with the sign of the battery voltage. The sign of this voltage depends only on the direction chosen to cross the battery.)
The voltage across a resistor is taken as negative (a decrease) if the resistor is traversed in the same direction as the assigned current, in essence going “downhill” potential-wise (Fig. 18.8b). The voltage will be positive (an increase in potential) if the resistor is traversed in the direction opposite to the current. Used together, these sign conventions allow the summation of the voltages around a closed loop, regardless of the direction chosen to do that sum. It should be clear that Eq. 18.5 is the same in either case. To see this, note that reversing the chosen loop direction simply amounts to multiplying Eq. 18.5 (from the original direction) by
-
1 . This operation, of course, does not change the equation or the physics.

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P R O B L E M - S O L V I N G H I N T
In applying Kirchhoff’s loop theorem, the sign of a voltage across a resistor is determined by the direction of the current in that resistor. However, there can be situations in which the current direction is not obvious. How do you handle the voltage signs in such cases? The answer is simple: After assuming a direction for the current, follow the voltage sign convention based on this assumed direction. This guarantees that the signs of all the voltage drops are mathematically consistent. Thus, if it turns out that the actual current direction is opposite your choice, the voltage drops will automatically reflect that.
A graphical interpretation of Kirchhoff’s loop theorem is presented in Learn by
Drawing 18.1, Kirchhoff Plots: A Graphical Inerpretation of Kirchhoff’s Loop Theorem. Integrated Example 18.4, in which a simple parallel circuit is reexamined using
Kirchhoff’s rules, shows that our previous series–parallel considerations were consistent with these circuit rules. In this Example, take care to notice how important it is draw a correct circuit diagram—it can guide you as to how to proceed.
I
6 3 3
I N T E G R A T E D E X A M P L E 1 8 . 4
Two resistors R
1 and R
2 are connected in parallel. This combination is in series with a third resistor R
3
, which has the largest resistance of the three. A battery completes the circuit, with one electrode connected to the beginning and the other to the end of this network. (a) Which resistor will carry the most current: (1) R
1
, (2) R
2
, or (3) R
3
? Explain.
(b) In the actual circuit, assume R
1 =
6.0
Æ
, R
2 =
3.0
Æ
, R
3 =
10.0
Æ
, and the battery’s terminal voltage is 12.0 V. Apply Kirchhoff’s rules to determine the current in each resistor and the voltage across each resistor.
( A ) C O N C E P T U A L R E A S O N I N G .
It is best to first look at a schematic circuit diagram based on the word description of the network ( 䉴 Fig. 18.9). It might appear that the resistor with the least resistance would carry the most current. But be careful; this holds only if all the resistors are in parallel. This is not true here. The two parallel resistors each carry only a portion of the total current. However, because the total of their two currents is in R
3
, that resistor carries the total, and therefore the most, current. Thus, the correct answer is (3).
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N .
Given: R
1
R
2
R
3
=
=
=
6.0
3.0
Æ
Æ
10.0
Æ
V
=
12.0 V
Find:
I
1
, I
2
, and I
3
(current in each resistor) and
V
1
, V
2
, and V
3
(voltage across each resistor)
There are three unknown currents: the total current (I) and the currents in each of the parallel resistors (I
1 and I
2
). Since there is only one battery, the current must be clockwise (shown in the figure). Applying Kirchhoff’s junction theorem to the first junction
(J in Fig. 18.9a) g
I i =
0 or I
-
I
1 -
I
2 =
0 (1)
Using the loop theorem in the clockwise direction in Fig. 18.9b, the battery is crossed from the negative to the positive terminal and then R
1 and R
3 are traversed to complete the loop. The resulting equation (showing the voltage signs explicitly) is g
V i =
0 or
+
V
+
1
-
I
1
R
1
2
+
1
-
IR
3
2
=
0 (2)
A third equation can be obtained by applying the loop theorem but this time going through R
2 instead of R
1 g
V i
(Fig. 18.9c). This yields
=
0 or
+
V
+
1
-
I
2
R
2
2
+
1
-
IR
3
2
=
0 (3)
Putting in the battery voltage (in volts) and resistances (in ohms) and rearranging gives three equations and three unknowns (the currents):
(continued on next page)
+
–
V
V =
12 V
V =
12 V
R
2
3.0
=
Ω
R
2
I
2
(a)
R
3
=
10.0
Ω
(b)
J
I
1
R
1
R
3
R
1
=
6.0
Ω
R
3
=
10.0
Ω
(c)
䉱
F I G U R E 1 8 . 9 Sketching circuit diagrams and Kirchhoff ’ s rules
(a) The circuit diagram resulting from the written description in Integrated Example 18.4. (b) and (c) The two loops used in the analysis of
Integrated Example 18.4.

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I
2
R
2
= 9.0
Ω
12
-
6 I
1
I
=
I
1
-
10 I
=
0
+
I
2 or 6
-
3 I
1 -
5 I
=
0
(1a)
(2a)
12
-
3 I
2 -
10 I
=
0
Adding Eqs. (2a) and (3a) yields
I
=
I
1 +
I
2
18
-
3
1
I
1 +
. Therefore, this equation becomes
I
2
2
-
15 I
=
0 . However, from Eq. (1a),
18
-
3 I
-
15 I
=
0 or 18 I
=
18
(3a) and solving for the total current yields I
=
1.00 A .
Eqs. (3a) and (1a) can then be solved for the remaining currents:
I
2 =
2
3
A and I
1 =
1
3
A
These answers are consistent with our circuit diagram reasoning in part (a).
Because the currents are now known, the voltages can be obtained from Ohm’s law,
V
=
IR . Thus,
V
1
V
2
V
3
=
I
1
R
1
=
I
2
R
2
=
I
3
R
3
=
=
A
A 2
3
1
3
A
A
B
B 1
1
6.0
3.0
Æ
Æ
2
2
=
=
2.0 V
2.0 V
=
1
1.0 A
21
10.0
Æ
2
=
10.0 V
Take a quick look at the results to see if they are reasonable. As expected, the voltage drops across the parallel resistors are equal. Because of that, two-thirds of the total current is in the resistor with the least resistance. Also, the total voltage across the network is 12.0 V, as it must be.
F O L L O W - U P E X E R C I S E .
(a) In this Example, predict what will happen to each of the currents if R
2 changing R
2 is increased. Explain your reasoning. (b) Rework part (b) of this Example, to 8.0
Æ
, and see if your reasoning is correct.
I
2
I
3
Loop 2
I
2
P R O B L E M - S O L V I N G H I N T
Note that answers in Example 18.4 were in amperes and volts because amps, volts, and ohms were used consistently throughout. By staying within this SI system (that is, expressing electrical quantities in volts, amps and ohms), carrying units is not necessary; the answers will automatically be in these units. (Of course, it is always a good idea to check units.)
I
1
V
2
=
12 V
R
3
=
2.0
Ω
R
1
=
6.0
Ω
Loop 1
V
1
= 6.0 V
I
1
Loop 3
䉱
F I G U R E 1 8 . 1 0 Application of
Kirchhoff ’ s rules To analyze a circuit such as the one shown for Example
18.5, assign a current and its direction for each branch in the circuit
(most conveniently done at junctions). Identify each loop and the direction of traversal. Then write current equations for each independent junction (using Kirchhoff’s junction theorem). Also write voltage equations for as many loops as needed to include every branch
(using Kirchhoff’s loop theorem). Be careful to observe sign conventions.
A P P L I C A T I O N O F K I R C H H O F F ’ S R U L E S
As indicated by its whimsical title, Integrated Example 18.4 could have been worked as a relatively simple series–parallel combination. However, more complicated, multiloop circuits (where resistors are neither in parallel nor in series) require a more structured approach using basic principles. In this book, the following general steps will be used when applying these principles (Kirchhoff’s rules):
1. Assign a current and direction of current for each branch in the circuit. This assignment is done most conveniently at junctions.
2. Indicate the loops and the directions in which they are to be traversed
(
䉳
Fig. 18.10). Every branch must be in at least one loop.
3. Apply Kirchhoff’s first rule (junction rule) at each junction that gives a unique equation. (This step gives a set of equations that includes all currents, but there may be redundant equations from two different junctions.)
4. Traverse the number of loops necessary to include all branches. In traversing a loop, apply Kirchhoff’s second rule, the loop theorem (using V
=
IR each resistor), and write the equations, using the proper sign conventions.
for
If this procedure is applied properly, Steps 3 and 4 give a set of N equations if there are N unknown currents. These equations may then be solved for the currents. If more loops are traversed than necessary, one (or more) redundant loop equation(s) might appear. Only the number of loops that includes each branch once is needed.

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This procedure may seem complicated, but it’s generally straightforward, as
Integrated Example 18.5 shows.
I N T E G R A T E D E X A M P L E 1 8 . 5
Consider the circuit diagrammed in Fig. 18.10. (a) What can you say about the magnitude of the electric potential change between the two nodes if the path taken is through R
2 compared to the magnitude if instead the path went through R
3
(1) the change is greater if the path goes through R
2
, (2) the change is greater if the path goes through the battery, (3) the change is the same for both paths? Explain. (b) Find the current in each branch and determine the two changes in part (a).
(c) Determine the power dissipated in each resistor and compare the total to that rate at which energy is lost (or gained) by the two batteries.
(A) CONCEPTUAL REASONING.
The electric potential difference between the nodes must be the same. If not, the nodes would have different potential values, which cannot be the case. Hence the correct answer is (3), the change is the same for both paths.
( B ) A N D ( C ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N .
The solution is begun by assigning current directions (“best guesses”) in each loop, and then using the junction theorem and the loop theorem (twice—once for each inner loop) to generate three equations, because there are three currents.
From the currents and resistance/battery values, potential changes can be determined. Lastly, these quantities can be used to find power levels in each of the circuit elements.
Given: Values in
Fig. 18.10
Find: (b) I
1
, I
2
, and I
3
(current in each of the three branches) and
V (voltage drop between the two nodes via two different paths)
(c) P
1
, P
2
, and P
3
(power dissipated in each resistor) and
P b
1
and P b
2
(rate of energy lost or gained by each battery)
(b) The chosen current directions and loop traversal directions are shown in the figure. (Remember, these directions are not unique; choose them, and check the final current signs to see if they were correct.) There is a current in every branch, and every branch is in at least one loop. (Some branches are in more than one loop, which is acceptable.)
Applying Kirchhoff’s first rule at the left-hand junction
I
1 -
I
2 -
I
3 =
0 or I
1
(1)
Going around loop 1 as in Fig. 18.10 and applying Kirchhoff’s loop theorem with the sign conventions gives g
V i = +
V
1 +
1
-
I
1
R
1
2
+
1
-
V
2
2
+
1
-
I
3
R
3
2
=
0 (2)
Putting in the numerical values yields
=
I
2 +
I
3
+
6
-
6 I
1 -
12
-
2 I
3 =
0
Rearranging this equation and dividing both sides by 2
3 I
1 +
I
3 = -
3
For convenience, units are omitted (they are all in amps and ohms, and are self-consistent).
For loop 2, the loop theorem yields g
V i = +
V
2 +
1
-
I
2
R
2
2
+
1
+
I
3
R
3
2
=
0 (3)
After substituting in the values and rearranging
9 I
2 -
2 I
3 =
12 (3a)
Equations (1), (2a), and (3a) form a set of three equations with three unknowns. The currents can be found in many ways.
One is to substitute Eq. (1) into Eq. (2a) and eliminate I
1
:
3
1
I
2 +
I
3
2
+
I
3 = -
3
After rearranging and dividing by 3, this simplifies to
I
2 = -
1
-
4
3
I
3
Then, substituting Eq. (4) into Eq. (3a) eliminates I
2
:
9
A
-
1
-
4
3
I
3
B
-
2 I
3 =
12
Finishing the algebra and solving for I
3
,
-
14 I
3 =
21 or I
3 = -
1.5 A
(4)
The minus sign means the wrong direction was assumed for I
3
.
Putting the value of I
3 into Eq. (4) gives I
2
:
I
2 = -
1
-
4
3
1
-
1.5 A
2
=
1.0 A
Then, from Eq. (1),
I
1 =
I
2 +
I
3 =
1.0 A
-
1.5 A
= -
0.5 A
Again, the sign means the direction of I
1 was wrong.
The electric potential change, from the left to the right node through battery #2 , is determined by the sign conventions. The current through R
3 is actually to the left, so its potential change is positive; however, the battery is traversed from anode to cathode, and this change is negative:
V
= -
12 V
+ ƒ
I
3 ƒ
R
3 = -
12 V
+ ƒ -
1.5 A
ƒ
1
2.0
Æ
2
= -
9.0 V
The electric potential change, going from the left node to the right node through R
2
, is negative, because the path is taken in the direction of the current. Its value is
V
= -
I
2
R
2 = -
1
1.0 A
21
9.0
Æ
2
= -
9.0 V in agreement with the conceptual part (a).
(c) The resistors all dissipate energy as follows:
P
R
1
P
R
2
P
R
3
=
I
2
1
R
1
=
I
2
2
R
2
=
I
2
3
R
3
=
1
0.50 A
2 2 1
6.0
Æ
2
=
1.5 W
=
1
1.0 A
2 2 1
9.0
Æ
2
=
9.0 W
=
1
1.50 A
2 2 1
2.0
Æ
2
=
4.5 W for a total dissipation rate of 15 W.
The batteries can gain or lose energy, depending upon the direction of their current. Battery #2 is losing energy because its current leaves its anode. Thus
P b
2
=
I
3
V b
2
=
1
1.5 A
21
12 V
2
=
18 W (a loss)
Battery #1 increasing its stored energy (being “recharged”) because its current enters the anode.
P b
1
=
I
1
V b
1
=
1
-
0.5 A
21
6.0 V
2
= -
3.0 W (a gain)
The end energy rate (power) result shows conservation of energy, since the net battery rate equals the total resistor loss rate of 15 W.
F O L L O W - U P E X E R C I S E .
Find the currents in this Example by using the junction theorem and loops 3 and 1 instead of loops 1 and 2.

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L E A R N B Y D R A W I N G 1 8 . 1
The equation form of Kirchhoff’s loop theorem has a geometrical visualization that may help you develop better insight into its meaning. This graphical approach allows the visualization of how the potential changes in a circuit, either to anticipate the results of mathematical analysis or to qualitatively confirm the results. (Don’t forget that a complete analysis usually also includes the junction theorem—see Example 18.5.)
The idea is to make a three-dimensional plot based on the circuit diagram. The wires and elements of the circuit form the basis for the x–y plane, or the diagram’s “floor.” Plotted perpendicularly to this plane, along the z-axis, is the electric potential (V), with an appropriate choice for zero. Such a diagram is called a Kirchhoff plot (Fig. 1).
The rules for constructing a Kirchhoff plot are simple:
Start at a known potential value, and go around a complete loop, finishing where you started. Because you come back to the same location, the sum of all the rises in potential
(positive voltages) must be balanced by the sum of the drops (negative voltages). This requirement is the geometrical expression of energy conservation, embodied mathematically by Kirchhoff’s loop theorem.
Thus, if the potential increases (say, in traversing a battery from negative terminal to positive terminal), draw a rise in the
z-direction. In this instance, the rise represents the terminal voltage of the battery. Similarly, if the potential decreases (for example, in traversing a resistor in the direction of the current), make sure that the potential drops. If possible, try to draw the rises and drops (the voltages) to scale.
Potential
ε IR
ε I c
R d
V b
ε a
V = terminal voltage = f
r e
– Ir <
Ir
F I G U R E 1 Kirchhoff plots: A graphical problem-solving strategy The schematic of the circuit is laid out in the x–y plane, and the electric potential is plotted perpendicularly along the z-axis. Usually, the zero of the potential is taken to be the negative terminal of the battery. A direction for current is assigned, and the value of the potential is plotted around the circuit, following the rules for gains and losses.
This particular plot shows a rise in potential when the battery is traversed from cathode to anode, followed by a drop in potential across the external resistor, and a smaller drop in potential across the battery’s internal resistance.
For elaborate circuits, this graphical method may prove to be too complicated for practical use. Nevertheless, it is always good to keep this concept in mind, as it illustrates the fundamental physics behind the loop theorem.
As an example of the power of this method, consider the circuit in Fig. 1: a battery with internal resistance r wired to a single external resistor R. The direction of the current is from the anode to the cathode through the external resistor. The potential of the battery’s cathode is chosen as zero. Starting there and traversing the circuit in the direction of the current, there is a rise in potential from the battery cathode to the anode. From there the potential remains constant as the current travels through the wires to the external resistor. That is, no significant voltage drop should be indicated along connecting wires (why?).
At the resistor, there must be a drop in potential. However, it must not drop to zero, because there must be some potential difference left to produce current in the internal resistance. Thus it can be reasoned visually why the terminal voltage of the battery, V, must be less than its emf (the rise between a and b).
Figure 2 shows two resistors in series, and that combination in parallel with a third resistor. For simplicity, all three resistors have the same resistance (R) and the battery’s internal resistance is assumed to be zero. Starting at point a, there is a rise in potential corresponding to the battery voltage. Then, as the loop is traced, it takes a route through the single resistor, so
Following the loop that includes the two resistors, each must have half the total drop (why?). So each will carry half the current of the single resistor. Recall that in parallel circuits, the largest resistance carries the least current. Notice how nicely the geometrical approach helps develop intuition and allows anticipation of quantitative results.
As an exercise, try redrawing Fig. 2 if, instead, the series resistors had resistance values of R and 2R. Which of these two now has the largest voltage? How do the currents in the resistors compare to the previous situation? Lastly, analyze the circuit mathematically, to see whether your expectations are confirmed.
Potential
ε
ε b a
I c
R d e
R
R g
F I G U R E 2 Kirchhoff plot of a more complex circuit
Imagine how the plot would change if you were to vary the values of the three resistors. Then analyze the circuit mathematically to see whether your plot allowed you to anticipate the voltages and currents.