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18.7
Circuit Analysis Using Kirchhoff’s Rules
657
We can replace the three capacitors with a single equivalent capacitor. In order for it to
store charge of magnitude Q for a potential difference , Q = Ceq. Therefore, Ceq = C1
+ C2 + C3. Once again, this result can be extended to the general case for any number of
capacitors connected in parallel.
For N capacitors connected in parallel,
Ceq = Σ Ci = C1 + C2 + • • • + CN
18.7
(18-18)
CIRCUIT ANALYSIS USING KIRCHHOFF’S RULES
Sometimes a circuit cannot be simplified by replacing parallel and series combinations
alone. In such cases, we apply Kirchhoff’s rules directly and solve the resulting equations simultaneously.
Problem-Solving Strategy: Using Kirchhoff ’s Rules to Analyze a Circuit
1. Replace any series or parallel combinations with their equivalents.
2. Assign variables to the currents in each branch of the circuit (I1, I2, . . .) and choose
directions for each current. Draw the circuit with the current directions indicated
by arrows. It does not matter whether or not you choose the correct direction.
3. Apply Kirchhoff’s junction rule to all but one of the junctions in the circuit.
(Applying it to every junction produces one redundant equation.) Remember
that current into a junction is positive; current out of a junction is negative.
4. Apply Kirchhoff’s loop rule to enough loops so that, together with the junction
equations, you have the same number of equations as unknown quantities. For
each loop, choose a starting point and a direction to go around the loop. Be careful with signs. For a resistor, if your path through a resistor goes with the current
(“downstream”), there is a potential drop; if your path goes against the current
(“upstream”), the potential rises. For an emf, the potential drops or rises depending on whether you move from the positive terminal to the negative or vice
versa; the direction of the current is irrelevant. A helpful method is to write “+”
and “–” signs on the ends of each resistor and emf to indicate which end is at the
higher potential and which is at the lower potential.
5. Solve the loop and junction equations simultaneously. If a current comes out
negative, the direction of the current is opposite to the direction you chose.
6. Check your result using one or more loops or junctions. A good choice is a loop
that you did not use in the solution.
Example 18.8
A Two-Loop Circuit
B
Find the currents through each branch of the circuit of
Fig. 18.29.
Strategy First we look for series and parallel combinations. R1 and 1 are in series, but since one is a resistor
and one an emf we cannot replace them with a single
equivalent circuit element. No pair of resistors is either in
series or in parallel. R1 and R2 might look like they’re in
parallel, but the emf 1 keeps points A and F at different
potentials, so they are not. The two emfs might look like
they’re in series, but the junction at point F means that
C
D
R1 = 4.0 Ω
R1
R2 = 6.0 Ω
R3
R2
R3 = 3.0 Ω
A
1 = 1.5 V
F
2 = 3.0 V
E
Figure 18.29
Circuit to be analyzed using Kirchhoff’s rules.
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Chapter 18
658
Electric Current and Circuits
Example 18.8 Continued
the current through the two is not the same. Since there
are no series or parallel combinations to simplify, we proceed to apply Kirchhoff’s rules directly.
Solution First we assign the currents variable names
and directions on the circuit diagram:
C
B
–
R1
+
A
Loop
I1 ABCFA
+
R2
I2
–
+ –
1
D
F
–
Loop
FCDEF R3
+
+ –
2
I3
From F to C, we move against the current I2
(“upstream”). The potential rises: ∆VF→C = +I2R2. From
C to D, the potential does not change. From D to E, we
again move upstream, so ∆VD→E = +I3R3. From E to F,
we move through a source of emf from the negative to the
positive terminal. The potential increases: ∆VE→F = +2.
Then the loop rule gives
+I2R2 + I3R3 + 2 = 0
(3)
Now we have three equations and three unknowns
(the three currents). To solve them simultaneously, we
first substitute known numerical values:
E
Points C and F are junctions between the three branches
of the circuit. We choose current I1 for branch FABC, current I3 for branch FEDC, and current I2 for branch CF.
Now we can apply the junction rule. There are two
junctions; we can choose either one. For point C, I1 and I3
flow into the junction and I2 flows out of the junction.
The resulting equation is
I1 + I3 – I2 = 0
(1)
Before applying the loop rule, we write “+” and “–”
signs on each resistor and emf to show which side is at
the higher potential and which at the lower, given the
directions assumed for the currents. In a resistor, current
flows from higher to lower potential. The emf symbol
uses the longer line for the positive terminal and the
shorter line for the negative terminal.
Now we choose a closed loop and add up the potential rises and drops as we travel around the loop.
Suppose we start at point A and travel around loop
ABCFA. The starting point and direction to go around
the loop are arbitrary choices, but once made, we stick
with it regardless of the directions of the currents. From
A to B, we move in the same direction as the current I1.
The current through a resistor travels from higher to
lower potential, so going from A to B is a potential drop:
∆VA→B = –I1R1.
From B to C, since the wire is assumed to have negligible resistance, there is no potential rise or drop. From C
to F, we move with current I2, so there is another potential drop: ∆VC→F = –I2R2.
Finally, from F to A, we move from the negative terminal of an emf to the positive terminal. The potential
rises: ∆VF→A = +1. A was the starting point, so the loop
is complete. The loop rule says that the sum of the potential changes is equal to zero:
–I1R1 – I2R2 + 1 = 0
(2)
We must choose another loop since we have not yet
gone through resistor R3 or emf 2. There are two choices
possible: the right-hand loop (such as FCDEF) or the
outer loop (ABCDEFA). Let’s choose FCDEF.
I1 + I3 – I2 = 0
–(4.0 Ω)I1 – (6.0 Ω)I2 + 1.5 V = 0
(6.0 Ω)I2 + (3.0 Ω)I3 + 3.0 V = 0
(1)
(2)
(3)
To solve simultaneous equations, we can solve one
equation for one variable and substitute into the other
equations, thus eliminating one variable. Solving Eq. (1)
for I1 yields I1 = –I3 + I2. Substituting in Eq. (2):
–(4.0 Ω)(–I3 + I2) – (6.0 Ω)I2 + 1.5 V = 0
Simplifying,
4.0I3 – 10.0I2 = –1.5 V/Ω = –1.5 A
(2a)
Eqs. (2a) and (3) now have only two unknowns. We
can eliminate I3 if we multiply Eq. (2a) by 3 and Eq. (3)
by 4 so that I3 has the same coefficient.
12.0I3 – 30.0I2 = –4.5 A 3 × Eq. (2a)
12.0I3 + 24.0I2 = –12.0 A 4 × Eq. (3)
Subtracting one from the other,
54.0I2 = –7.5 A
Now we can solve for I2:
7.5
I2 = – A = –0.139 A
54.0
Substituting the value of I2 into Eq. (2a) enables us to
solve for I3:
4I3 + 10 × 0.139 A = –1.5 A
–1.5 – 1.39
I3 = A = –0.723 A
4
Equation (1) now gives I1:
I1 = –I3 + I2 = +0.723 A – 0.139 A = +0.584 A
Rounding to two significant figures, the currents are I1 =
+0.58 A, I3 = – 0.72 A, and I2 = – 0.14 A. Since I3 and I2
came out negative, the actual directions of the currents in
those branches are opposite to the ones we arbitrarily
chose.
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18.8
Power and Energy in Circuits
659
Example 18.8 Continued
B
C
–
R1
A
+
–
0.58 A
+
D
R2
+
+ –
1
R3
0.14 A
F
0.72 A
–
+ –
E
2
Discussion Note that it did not matter that we chose
some of the current directions wrong. It also doesn’t matter which loops we choose (as long as we cover every
branch of the circuit), which starting point we use for a
loop, or which direction we go around a loop.
18.8
The hardest thing about applying Kirchhoff’s rules
is getting the signs correct. It is also easy to make an
algebraic mistake when solving simultaneous equations. Therefore, it is a good idea to check the answer. A
good way to check is to write down a loop equation for
a loop that was not used in the solution (see Practice
Problem 18.8).
Practice Problem 18.8
with the Loop Rule
Verifying the Solution
Apply Kirchhoff’s loop rule to loop CBAFEDC to verify
the solution of Example 18.8.
POWER AND ENERGY IN CIRCUITS
From the definition of electric potential, if a charge q moves through a potential difference ∆V, the change in electric potential energy is
∆UE = q ∆V
(17-7)
From energy conservation, a change in electric potential energy means that conversion
between two forms of energy takes place. For example, a battery converts stored chemical energy into electric potential energy. A resistor converts electric potential energy
into internal energy. The rate at which the energy conversion takes place is the power P.
Since current is the rate of flow of charge, I = q/∆t and
Power
∆UE q
P = = ∆t ∆V = I ∆V
∆t
(18-19)
Thus, the power for any circuit element is the product of current and potential difference. We can verify that current times voltage comes out in the correct units for power
by substituting coulombs per second for amperes and joules per coulomb for volts:
C J J
A × V = s × = s = W
C
According to the definition of emf, if the amount of charge pumped by an ideal
source of constant emf is q, then the work done by the battery is
W = q
The power supplied by the emf is the rate at which it does work:
q
∆W
P=
∆
t = ∆t = I
(18-2)
(18-20)
Since ∆V = for an ideal emf, Eqs. (18-20) and (18-19) are equivalent.
Power Dissipated by a Resistor
If an emf causes current to flow through a resistor, what happens to the energy supplied
by the emf? Why must the emf continue supplying energy to maintain the current?
Current flows in a metal wire when an emf gives rise to a potential difference
between one end and the other. The electric field makes the conduction electrons drift in
the direction of lower electric potential energy (higher potential). If there were no
The term power dissipated
means the rate at which
energy is dissipated. “Power” is
not dissipated in a resistor;
energy is.