Chapter 18
Direct Current Circuits
Read and take notes on
pages 548-558 in
Conceptual Physics Text
Or lesson 4 B Two Types of Connections
on Physics Classroom
Bright Storm on
Electric Circuits
(Start to minute 1:50)
Read and take notes on
pages 594-597 in
College Physics Text
Electric Circuits
• Electric circuits control the flow of electricity and
the energy associated with it.
• Circuits are used in many applications.
• Kirchhoff’s Rules will simplify the analysis of
simple circuits.
• Some circuits will be in steady state.
– The currents are constant in magnitude and direction.
• In circuits containing resistors and capacitors, the
current varies in time.
Introduction
Khan Academy on
Circuits and Ohm’s Law
Sources of emf
• The source that maintains the current in a closed
circuit is called a source of emf.
– Any devices that increase the potential energy of charges
circulating in circuits are sources of emf.
– Examples include batteries and generators
• SI units are Volts
– The emf is the work done per unit charge.
Section 18.1
emf and Internal Resistance
• A real battery has some
internal resistance.
• Therefore, the terminal
voltage is not equal to
the emf.
Section 18.1
More About Internal Resistance
• The schematic shows the
internal resistance, r
• The terminal voltage is ΔV =
Vb-Va
• ΔV = ε – Ir
• For the entire circuit,
ε = IR + Ir
Section 18.1
Internal Resistance and emf, Cont.
• ε is equal to the terminal voltage when the
current is zero.
– Also called the open-circuit voltage
• R is called the load resistance.
• The current depends on both the resistance
external to the battery and the internal
resistance.
Section 18.1
Internal Resistance and emf, Final
• When R >> r, r can be ignored.
– Generally assumed in problems
• Power relationship
– I e = I2 R + I 2 r
• When R >> r, most of the power delivered
by the battery is transferred to the load
resistor.
Section 18.1
Batteries and emf
• The current in a circuit depends on the
resistance of the battery.
– The battery cannot be considered a source of
constant current.
• The terminal voltage of battery cannot be
considered constant since the internal
resistance may change.
• The battery is a source of constant emf.
Section 18.1
Bright Storm on
Resistors in Series
Resistors in Series
• When two or more resistors are connected end-toend, they are said to be in series.
• The current is the same in all resistors because any
charge that flows through one resistor flows through
the other.
• The sum of the potential differences across the
resistors is equal to the total potential difference
across the combination.
Khan Academy on Resistors in
Series Circuits
(A must watch, explains much more than how to add
resistors)
Resistors in Series, Cont.
• Potentials add
– ΔV = IR1 + IR2 = I (R1+R2)
– Consequence of Conservation of Energy
• The equivalent resistance has the effect on the circuit
as the original combination of resistors.
Section 18.2
Equivalent Resistance – Series
• Req = R1 + R2 + R3 + …
• The equivalent resistance of a series
combination of resistors is the algebraic sum
of the individual resistances and is always
greater than any of the individual resistors.
• If one element in the series circuit fails, the
circuit would no longer be complete and none
of the elements would work.
Section 18.2
Equivalent Resistance – Series
An Example
• Four resistors are replaced with their equivalent
resistance.
Section 18.3
EXAMPLE 18.1 Four Resistors in Series
(a) Four resistors connected in series. (b) The equivalent
resistance of the circuit in (a).
Goal Analyze several resistors
connected in series.
Problem Four resistors are arranged as
shown in figure a. Find (a) the equivalent resistance of the circuit and (b) the
current in the circuit if the emf of the battery is 6.0 V.
Strategy Because the resistors are connected in series, summing their resistances
gives the equivalent resistance. Ohm's law can then be used to find the current.
SOLUTION
(a) Find the equivalent resistance of the circuit.
Sum the resistances to find the equivalent resistance.
Req = R1 + R2 + R3 + R4
Req = 2.0 Ω + 4.0 Ω + 5.0 Ω + 7.0 Ω
Req = 18.0 Ω
(b) Find the current in the circuit.
Apply Ohm's law to the equivalent resistor in figure b, solving for the current.
I=
ΔV
Req
=
6.0 V
18.0 Ω
= 1/3 A
LEARN MORE
Remarks A common misconception is that the current is "used up" and steadily
declines as it progresses through a series of resistors. That would be a violation of
conservation of charge.
Question The internal resistance of the battery is neglected in this example. How
would it affect the solution for the final current, if taken into account? (Select all
that apply.)
It would not affect the current.
resistance of the circuit.
circuit.
It would increase the equivalent
It would not affect the equivalent resistance of the
It would decrease the current.
resistance of the circuit.
It would decrease the equivalent
It would increase the current.
Bright Storm on
Resistors in Parallel
(Start to minute 3:30)
Read and take notes on
pages 598-599 in
College Physics Text
Resistors in Parallel
• The potential difference across each resistor is the
same because each is connected directly across the
battery terminals.
• The current, I, that enters a point must be equal to
the total current leaving that point.
– I = I1 + I2
– The currents are generally not the same.
– Consequence of Conservation of Charge
Section 18.3
Equivalent Resistance – Parallel
An Example
• Equivalent resistance replaces the two original resistances.
• Household circuits are wired so the electrical devices are
connected in parallel.
– Circuit breakers may be used in series with other circuit elements for
safety purposes.
Section 18.3
Khan Academy on
Resistors in Parallel
(Another must watch!)
Equivalent Resistance – Parallel
• Equivalent Resistance
• The inverse of the
equivalent resistance of two
or more resistors connected
in parallel is the algebraic
sum of the inverses of the
individual resistance.
– The equivalent is always less
than the smallest resistor in
the group.
Section 18.3
EXAMPLE 18.2 Three Resistors in Parallel
Three resistors connected in parallel. The voltage across each resistor is 18 V.
Goal Analyze a circuit having resistors connected in
parallel.
Problem Three resistors are connected in parallel as in
the figure. A potential difference of 18 V is maintained
between points a and b. (a) Find the current in each
resistor. (b) Calculate the power delivered to each
resistor and the total power. (c) Find the equivalent resistance of the circuit. (d)
Find the total power delivered to the equivalent resistance.
Strategy To get the current in each resistor we can use Ohm's law and the fact
that the voltage drops across parallel resistors are all the same. The rest of the
problem just requires substitution into the equation for power delivered to a
resistor, = I2R, and the reciprocal-sum law for parallel resistors.
SOLUTION
(a) Find the current in each resistor.
Apply Ohm's law, solved for the current I delivered by the battery to find the
current in each resistor.
ΔV 18 V
=
= 6.0 A
I1 =
R1 3.0 Ω
ΔV 18 V
=
= 3.0 A
I2 =
R2 6.0 Ω
I3 =
ΔV
R3
=
18 V
9.0 Ω
= 2.0 A
(b) Calculate the power delivered to each resistor and the total power.
Apply = I2R to each resistor, substituting the results from part (a).
3 Ω:
1
= I12R1 = (6.0 A)2(3.0 Ω) = 110 W
6 Ω:
2
= I22R2 = (3.0 A)2(6.0 Ω) = 54 W
9 Ω:
3
= I32R3 = (2.0 A)2(9.0 Ω) = 36 W
Sum to get the total power.
tot
= 110 W + 54 W + 36 W = 2.0 102 W
(c) Find the equivalent resistance of the circuit.
Apply the reciprocal-sum rule.
1
=
1
+
1
+
Req R1 R2
1
Req
=
1
3.0 Ω
+
1
R3
1
6.0 Ω
+
1
9.0 Ω
=
11
18 Ω
18
Req = Ω = 1.6 Ω
11
(d) Compute the power dissipated by the equivalent resistance.
Use the alternate power equation.
=
(ΔV)2
Req
=
(18 V)2
1.6 Ω
= 2.0 102 W
LEARN MORE
Remarks There's something important to notice in part (a): the smallest 3.0 Ω
resistor carries the largest current, whereas the other, larger resistors of 6.0 Ω and
9.0 Ω carry smaller currents. The largest current is always found in the path of
least resistance. In part (b) the power could also be found with = (ΔV )2/R. Note
that 1 = 108 W, but is rounded to 110 W because there are only two significant
figures. Finally, notice that the total power dissipated in the equivalent resistor is
the same as the sum of the power dissipated in the individual resistors, as it should
be.
Question If a fourth resistor were added in parallel to the other three, how would
the equivalent resistance change?
It would be larger.
It would be smaller.
needed to determine the effect.
More information is
Bright Storm on Solving
Complex Resistor Problems
(It is long but VERY effective!!!)
If you didn’t read and take notes on
pages 548-558 in
Conceptual Physics Text then read and take notes on the
following link
lesson 4 E Combination Circuits
on Physics Classroom
Read and take notes on
pages 600-602 in
College Physics Text
Problem-Solving Strategy, 1
• Combine all resistors in series.
– They carry the same current.
– The potential differences across them are not the
same.
– The resistors add directly to give the equivalent
resistance of the series combination: Req = R1 + R2
+…
– Draw the simplified circuit diagram.
Section 18.3
Problem-Solving Strategy, 2
• Combine all resistors in parallel.
– The potential differences across them are the same.
– The currents through them are not the same.
– The equivalent resistance of a parallel combination is
found through reciprocal addition:
– Remember to invert the answer after summing the
reciprocals.
– Draw the simplified circuit diagram.
Section 18.3
Problem-Solving Strategy, 3
• Repeat the first two steps as necessary.
– A complicated circuit consisting of several resistors and
batteries can often be reduced to a simple circuit with only
one resistor.
– Replace any resistors in series or in parallel using steps 1 or
2.
– Sketch the new circuit after these changes have been
made.
– Continue to replace any series or parallel combinations.
– Continue until one equivalent resistance is found.
Section 18.3
Problem-Solving Strategy, 4
• Use Ohm’s Law.
– Use ΔV = I R to determine the current in the
equivalent resistor.
– Start with the final circuit found in step 3 and
gradually work back through the circuits, applying
the useful facts from steps 1 and 2 to find the
current in the other resistors.
Section 18.3
Khan Academy Example Problem
(Great Practice Problem)
Example
• Complex circuit
reduction
– Combine the resistors in
series and parallel.
• Redraw the circuit
with the equivalents
of each set.
– Combine the resulting
resistors in series.
– Determine the final
equivalent resistance.
Section 18.3
EXAMPLE 18.3 Equivalent Resistance
Figure (a) The four resistors shown above can be reduced in steps to an
equivalent 14 Ω resistor.
Goal Solve a problem involving both series and parallel
resistors.
Problem Four resistors are connected as shown in
figure a. (a) Find the equivalent resistance between
points a and c. (b) What is the current in each resistor if
a 42-V battery is connected between a and c?
Strategy Reduce the circuit in steps, as shown in
figures b and c, using the sum rule for resistors in series
and the reciprocal-sum rule for resistors in parallel.
Finding the currents is a matter of applying Ohm's law while working backwards
through the diagrams.
SOLUTION
(a) Find the equivalent resistance of the circuit.
The 8.0-Ω and 4.0-Ω resistors are in series, so use the sum rule to find the
equivalent resistance between a and b:
Req = R1 + R2 = 8.0 Ω + 4.0 Ω = 12.0 Ω
The 6.0-Ω and 3.0-Ω resistors are in parallel, so use the reciprocal-sum rule to
find the equivalent resistance between b and c (don't forget to invert!):
1
1 1
1
1
1
= + =
+
=
Req R1 R2 6.0 Ω 3.0 Ω 2.0 Ω
Req = 2.0 Ω
In the new diagram, figure b, there are now two resistors in series. Combine them
with the sum rule to find the equivalent resistance of the circuit.
Req = R1 + R2 = 12.0 Ω + 2.0 Ω = 14.0 Ω
(b) Find the current in each resistor if a 42-V battery is connected between points
a and c.
Find the current in the equivalent resistor in figure c, which is the total current.
Resistors in series all carry the same current, so the value is the current in the 12Ω resistor in figure b and also in the 8.0-Ω and 4.0-Ω resistors in figure a:
I=
ΔVac
Req
=
42 V
= 3.0 A
14 Ω
Apply the junction rule to point b:
(1)
I = I1 + I2
The 6.0-Ω and 3.0-Ω resistors are in parallel, so the voltage drops across them are
the same:
ΔV6Ω = ΔV3Ω → (6.0 Ω)I1 = (3.0 Ω)I2 → 2.0 I1 = I2
Substitute this result into Equation (1), with I = 3.0 A.
3.0 A = I1 + 2I1 = 3I1 → I1 = 1.0 A I2 = 2.0 A
LEARN MORE
Remarks As a final check, note that ΔVbc = (6.0 Ω)I1 = (3.0 Ω)I2 = 6.0 V and
ΔVab = (12 Ω)I1 = 36 V; therefore, ΔVac + ΔVab + ΔVbc = 42 V, as expected.
Question Which of the original resistors dissipates energy at the greatest rate?
the 4.0 Ω resistor
Ω resistor
the 6.0 Ω resistor
the 8.0 Ω resistor
the 3.0
Gustav Kirchhoff
• 1824 – 1887
• Invented spectroscopy
with Robert Bunsen
• Formulated rules about
radiation
Section 18.4
Read and take notes on
pages 603-604 in
College Physics Text
Kirchhoff’s Rules
• There are ways in which resistors can be
connected so that the circuits formed cannot
be reduced to a single equivalent resistor.
• Two rules, called Kirchhoff’s Rules, can be
used instead.
Section 18.4
Statement of Kirchhoff’s Rules
• Junction Rule
– The sum of the currents entering any junction
must equal the sum of the currents leaving that
junction.
• A statement of Conservation of Charge
• Loop Rule
– The sum of the potential differences across all the
elements around any closed circuit loop must be
zero.
• A statement of Conservation of Energy
Section 18.4
More About the Junction Rule
• I1 = I 2 + I3
• From Conservation of
Charge
• Diagram b shows a
mechanical analog.
Section 18.4
Loop Rule
• A statement of Conservation of Energy
• To apply Kirchhoff’s Rules,
– Assign symbols and directions to the currents in all
branches of the circuit.
• If the direction of a current is incorrect, the answer will
be negative, but have the correct magnitude.
– Choose a direction to transverse the loops.
• Record voltage rises and drops.
Section 18.4
More About the Loop Rule
• Traveling around the loop
from a to b
• In a, the resistor is
transversed in the direction
of the current, the potential
across the resistor is –IR.
• In b, the resistor is
transversed in the direction
opposite of the current, the
potential across the resistor
is +IR.
Section 18.4
Loop Rule, Final
• In c, the source of emf is
transversed in the direction
of the emf (from – to +), the
change in the electric
potential is +ε
• In d, the source of emf is
transversed in the direction
opposite of the emf (from +
to -), the change in the
electric potential is -ε
Section 18.4
Junction Equations from Kirchhoff’s Rules
• Use the junction rule as often as needed, so
long as, each time you write an equation, you
include in it a current that has not been used
in a previous junction rule equation.
– In general, the number of times the junction rule
can be used is one fewer than the number of
junction points in the circuit.
Section 18.4
Loop Equations from Kirchhoff’s Rules
• The loop rule can be used as often as needed
so long as a new circuit element (resistor or
battery) or a new current appears in each new
equation.
• You need as many independent equations as
you have unknowns.
Section 18.4
Problem-Solving Strategy – Kirchhoff’s Rules
• Draw the circuit diagram and assign labels and
symbols to all known and unknown quantities.
• Assign directions to the currents.
• Apply the junction rule to any junction in the circuit.
• Apply the loop rule to as many loops as are needed
to solve for the unknowns.
• Solve the equations simultaneously for the unknown
quantities.
• Check your answers.
Section 18.4
EXAMPLE 18.4 Applying Kirchhoff's Rules
Goal Use Kirchhoff's rules to find currents in a circuit
with three currents and one battery.
Problem Find the currents in the circuit shown in the
figure by using Kirchhoff's rules.
Strategy There are three unknown currents in this
circuit, so we must obtain three independent equations,
which then can be solved by substitution. We can find
the equations with one application of the junction rule and two applications of the
loop rule. We choose junction c. (Junction d gives the same equation.) For the
loops, we choose the bottom loop and the top loop, both shown by blue arrows,
which indicate the direction we are going to traverse the circuit mathematically
(not necessarily the direction of the current). The third loop gives an equation that
can be obtained by a linear combination of the other two, so it provides no
additional information and isn't used.
SOLUTION
Apply the junction rule to point c. I1 is directed into the junction, I2 and I3 are
directed out of the junction.
I1 = I2 + I3
Select the bottom loop and traverse it clockwise starting at point a, generating an
equation with the loop rule.
ΣΔV = ΔVbat + ΔV4.0 Ω + ΔV9.0 Ω = 0
6.0 V - (4.0 Ω)I1 - (9.0 Ω)I3 = 0
Select the top loop and traverse it clockwise from point c. Notice the gain across
the 9.0 Ω resistor because it is traversed against the direction of the current!
ΣΔV = ΔV5.0 Ω + ΔV9.0 Ω = 0
-(5.0 Ω)I2 + (9.0 Ω)I3 = 0
Rewrite the three equations, rearranging terms and dropping units for the moment,
for convenience.
(1)
I1 = I2 + I3
(2) 4.0I1 + 9.0I3 = 6.0
(3) -5.0I2 + 9.0I3 = 0
Solve Equation (3) for I2 and substitute into Equation (1).
I2 = 1.8I3
I1 = I2 + I3 = 1.8I3 + I3 = 2.8I3
Substitute the latter expression into Equation (2) and solve for I3.
4.0 (2.8I3) + 9.0I3 = 6.0 → I3= 0.30 A
Substitute I3 back into Equation (3) to get I2.
-5.0I2 + 9.0(0.30 A) = 0 → I2 = 0.54 A
Substitute I3 into Equation (2) to get I1.
4.0I1 + 9.0(0.30 A) = 6.0 → I1 = 0.83 A
LEARN MORE
Remarks Substituting these values back into the original equations verifies that
they are correct, with any small discrepancies due to rounding. The problem can
also be solved by first combining resistors.
Question How would the answers change if the assumed directions of the
currents used in solving the problem were all reversed in the figure above, but you
followed the circuit paths in the same way to write down the equations based on
Kirchhoff's rules? (Select all that apply.)
The sign of each current-times-resistance term would reverse.
used for the voltage across the battery would reverse.
resistances would reverse.
reverse.
The sign
The sign of the
The signs of the currents after solving would
EXAMPLE 18.5 Another Application of Kirchhoff's Rules
Goal Find the currents in a circuit
with three currents and two
batteries when some current
directions are chosen wrongly.
Problem Find I1, I2, and I3 in figure
a.
Strategy Use Kirchhoff's two rules,
the junction rule once and the loop rule twice, to develop three equations for the
three unknown currents. Solve the equations simultaneously.
SOLUTION
Apply Kirchhoff's junction rule to junction c. Because of the chosen current
directions, I1 and I2 are directed into the junction and I3 is directed out of the
junction.
(1) I3 = I1 + I2
Apply Kirchhoff's loop rule to the loops abcda and befcb. (Loop aefda gives no
new information.) In loop befcb, a positive sign is obtained when the 6.0 Ω
resistor is traversed because the direction of the path is opposite the direction of
the current I1.
(2) Loop abcda: 10 V - (6.0 Ω)I1 - (2.0 Ω)I3 = 0
(3) Loop befcb: -(4.0 Ω)I2 - 14 V + (6.0 Ω)I1 - 10 V = 0
Using Equation (1), eliminate I3 from Equation (2) (ignore units for the moment):
10 - 6I1 - 2.0 (I1 + I2) = 0
(4) 10 = 8.0I1 + 2.0I2
Divide each term in Equation (3) by 2 and rearrange the equation so that the
currents are on the right side:
(5) -12 = -3.0I1 + 2.0I2
Subtracting Equation (5) from Equation (4) eliminates I2 and gives I1:
22 = 11I1 → I1 = 2.0 A
Substituting this value of I1 into Equation (5) gives I2:
2.0I2 = 3.0I1 - 12 = 3.0(2.0) - 12 = -6.0 A
I2 = -3.0 A
Finally, substitute the values found for I1 and I2 into Equation (1) to obtain I3:
I3 = I1 + I2 = 2.0 A - 3.0 A = -1.0 A
LEARN MORE
Remarks The fact that I2 and I3 are both negative indicates that the wrong
directions were chosen for these currents. Nonetheless, the magnitudes are
correct. Choosing the right directions of the currents at the outset is unimportant
because the equations are linear, and wrong choices result only in a minus sign in
the answer.
Question Is it possible for the current through a battery to be directed from the
positive terminal toward the negative terminal? (Select all that apply.)
Yes. It happens inside a rechargeable flashlight battery or a car battery
during recharge.
is greater than its emf.
reversible.
Yes. It can happen when the voltage applied to a battery
No. The charge carriers inside a battery is never
Household Circuits
• The utility company
distributes electric power to
individual houses with a
pair of wires.
• Electrical devices in the
house are connected in
parallel with those wires.
• The potential difference
between the wires is about
120V.
Section 18.6
Household Circuits, Cont.
• A meter and a circuit breaker are connected in series
with the wire entering the house.
• Wires and circuit breakers are selected to meet the
demands of the circuit.
• If the current exceeds the rating of the circuit
breaker, the breaker acts as a switch and opens the
circuit.
• Household circuits actually use alternating current
and voltage.
Section 18.6
Circuit Breaker Details
• Current passes through
a bimetallic strip.
– The top bends to the left
when excessive current
heats it.
– Bar drops enough to
open the circuit
• Many circuit breakers
use electromagnets
instead.
Section 18.6
240-V Connections
• Heavy-duty appliances
may require 240 V to
operate.
• The power company
provides another wire
at 120 V below ground
potential.
Section 18.6
Electrical Safety
• Electric shock can result in fatal burns.
• Electric shock can cause the muscles of vital
organs (such as the heart) to malfunction.
• The degree of damage depends on
– The magnitude of the current
– The length of time it acts
– The part of the body through which it passes
Section 18.7
Effects of Various Currents
• 5 mA or less
– Can cause a sensation of shock
– Generally little or no damage
• 10 mA
– Hand muscles contract
– May be unable to let go of a live wire
• 100 mA
– If passes through the body for just a few seconds, can be
fatal
Section 18.7
Ground Wire
• Electrical equipment
manufacturers use
electrical cords that
have a third wire, called
a case ground.
• Prevents shocks
Section 18.7
Ground Fault Interrupts (GFI)
• Special power outlets
• Used in hazardous areas
• Designed to protect people from electrical
shock
• Senses currents (of about 5 mA or greater)
leaking to ground
• Shuts off the current when above this level
Section 18.7
Electrical Signals in Neurons
• Specialized cells in the body, called neurons, form
a complex network that receives, processes, and
transmits information from one part of the body
to another.
• Three classes of neurons
– Sensory neurons
• Receive stimuli from sensory organs that monitor the
external and internal environment of the body
– Motor neurons
• Carry messages that control the muscle cells
– Interneurons
• Transmit information from one neuron to another
Section 18.8
Diagram of a Neuron
Section 18.8
Link to Homework Questions on
Web Assign
Unit 3 C Circuits
# 1-8
Link to Discussion Questions on
Web Assign
Unit 3 C Circuits
Grading Rubric for Unit 3C Circuits
Name: ______________________
Conceptual notes Lesson 4 B Two Types of Connections Physics Classroom --------------------------____
Lesson 4 E Combination Circuits Physics Classroom --------------------------------____
Conceptual Physics Text Pgs 548-558 ----------------------------------------------------____
Advanced notes from text book:
Pgs 594-597 --------------------------------------------------------------------------------------------_____
Pgs 598-599 --------------------------------------------------------------------------------------------_____
Pgs 600-602 --------------------------------------------------------------------------------------------_____
Pgs 603-604 --------------------------------------------------------------------------------------------_____
Example Problems and Questions:
18.1 a,b--------------------------------------------------------------------------------------------------_____
18.2 a-d--------------------------------------------------------------------------------------------------_____
18.3 a,b--------------------------------------------------------------------------------------------------_____
18.4 three currents------------------------------------------------------------------------------------------_____
18.5 three currents------------------------------------------------------------------------------------------_____
Web Assign 1, 2, 3, 4, 5, 6, 7, 8 -------------------------------------------------------------------------------------____