Chapter 18
Direct Current Circuits
DC Circuits
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
EMF and Terminal Voltage
This resistance behaves as though it were in
series with the emf.
emf and Internal
Resistance


A real battery has
some internal
resistance
Therefore, the
terminal voltage
is not equal to the
emf
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
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
Internal Resistance and
emf, final

When R >> r, r can be ignored


Generally assumed in problems
Power relationship

I
e = I2 R + I2 r

When R >> r, most of the
power delivered by the battery
is transferred to the load
resistor
Resistors in Series



When two or more resistors are
connected end-to-end, 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
A series connection has a single path from
the battery, through each circuit element in
turn, then back to the battery.
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
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
Equivalent Resistance –
Series: An Example

Four resistors are replaced with their
equivalent resistance
•The current through each resistor is the
same.
•The voltage depends on the resistance.
•The sum of the voltage drops across the
resistors equals the battery voltage.
1) 12 V
2) zero
3) 3 V
Assume that the voltage of the
4) 4 V
battery is 9 V and that the three
5) you need to know the
actual value of R
resistors are identical. What is
the potential difference across
each resistor?
9V
Assume that the voltage of the
battery is 9 V and that the three
1) 12 V
resistors are identical. What is
2) zero
the potential difference across
3) 3 V
each resistor?
4) 4 V
5) you need to know
the actual value of R
Since the resistors are all equal,
the voltage will drop evenly
across the 3 resistors, with 1/3 of
9 V across each one. So we get a
3 V drop across each.
9V
1) 12 V
2) zero
3) 6 V
In the circuit
below, what is
the voltage
across R1?
4) 8 V
5) 4 V
R1= 4 W
R2= 2 W
12 V
In the circuit below, what
1) 12 V
is the voltage across R1?
2) zero
3) 6 V
4) 8 V
5) 4 V
The voltage drop across R1 has
to be twice as big as the drop
across R2. This means that V1 =
R1= 4 W
R2= 2 W
8 V and V2 = 4 V. Or else you
could find the current I = V/R =
(12 V)/(6 W) = 2 A, then use
Ohm’s Law to get voltages.
12 V
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
A parallel connection splits the current; the
voltage across each resistor is the same:
Equivalent Resistance –
Parallel, 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
Equivalent Resistance –
Parallel

Equivalent Resistance
1
1
1
1




Req R1 R2 R3

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
1) 10 A
2) zero
3) 5 A
In the circuit to
the right, what
is the current
through R1?
4) 2 A
5) 7 A
R2= 2 W
R1= 5 W
10 V
In the circuit below, what is
1) 10 A
the current through R1?
2) zero
3) 5 A
4) 2 A
5) 7 A
The voltage is the same (10 V) across each
R2= 2 W
resistor because they are in parallel. Thus,
we can use Ohm’s Law, V1 = I1 R1 to find the
R1= 5 W
current I1 = 2 A.
10 V
1) increases
2) remains the same
3) decreases
4) drops to zero
Points P and Q are
connected to a battery of
fixed voltage. As more
resistors R are added to
the parallel circuit, what
happens to the total
current in the circuit?
Points P and Q are connected
to a battery of fixed voltage.
As more resistors R are added
to the parallel circuit, what
1) increases
2) remains the same
3) decreases
happens to the total current in 4) drops to zero
the circuit?
As we add parallel resistors, the overall
resistance of the circuit drops. Since V =
IR, and V is held constant by the battery,
when resistance decreases, the current
must increase.
1) all the current continues to flow through
the bulb
2) half the current flows through the wire,
the other half continues through the
bulb
3) all the current flows through the wire
Current flows
through a
lightbulb. If a
wire is now
connected across
the bulb, what
happens?
4) none of the above
1) all the current continues to flow through
the bulb
Current flows
through a lightbulb.
If a wire is now
connected across the
2) half the current flows through the wire,
the other half continues through the
bulb
3) all the current flows through the wire
4) none of the above
bulb, what happens?
The current divides based on the
ratio of the resistances. If one of the
resistances is zero, then ALL of the
current will flow through that path.
1) glow brighter than before
2) glow just the same as before
3) glow dimmer than before
4) go out completely
5) explode
Two lightbulbs A and B
are connected in series
to a constant voltage
source. When a wire
is connected across B,
bulb A will:
1) glow brighter than before
2) glow just the same as before
Two lightbulbs A and B
3) glow dimmer than before
are connected in series
to a constant voltage
source. When a wire
is connected across B,
bulb A will:
4) go out completely
5) explode
Since bulb B is bypassed by the wire,
the total resistance of the circuit
decreases. This means that the current
through bulb A increases.
1) circuit 1
2) circuit 2
3) both the same
The lightbulbs in the
circuit below are
identical with the same
resistance R. Which
circuit produces more
light? (brightness 
power)
4) it depends on R
The lightbulbs in the circuit
1) circuit 1
below are identical with the
2) circuit 2
same resistance R. Which
3) both the same
circuit produces more light?
4) it depends on R
(brightness  power)
In #1, the bulbs are in parallel,
lowering the total resistance of the
circuit. Thus, circuit #1 will draw
a higher current, which leads to
more light, because P = I V.
1) twice as much
2) the same
3) 1/2 as much
The three lightbulbs in the
4) 1/4 as much
circuit all have the same
5) 4 times as much
resistance of 1 W . By how
A
much is the brightness of
bulb B greater or smaller
than the brightness of bulb
A? (brightness  power)
C
B
10 V
1) twice as much
2) the same
The three light bulbs in the circuit all
3) 1/2 as much
have the same resistance of 1 W . By
4) 1/4 as much
how much is the brightness of bulb B
5) 4 times as much
greater or smaller than the brightness
of bulb A? (brightness  power)
We can use P = V2/R to compare the power:
A
C
B
PA = (VA)2/RA = (10 V) 2/1 W = 100 W
PB = (VB)2/RB = (5 V) 2/1 W = 25 W
10 V
1) increase
2) decrease
3) stay the same
What happens to the
R1
voltage across the
resistor R1 when the
switch is closed? The
voltage will:
S
R3
V
R2
1) increase
2) decrease
3) stay the same
What happens to the voltage
across the resistor R1 when
the switch is closed? The
R1
voltage will:
S
With the switch closed, the addition of
R2 to R3 decreases the equivalent
resistance, so the current from the
battery increases. This will cause an
increase in the voltage across R1 .
R3
V
R2
1) increases
2) decreases
What happens to the
3) stays the same
voltage across the
resistor R4 when the
switch is closed?
R1
S
R3
V
R2
R4
What happens to the voltage
across the resistor R4
when the switch is closed?
1) increases
2) decreases
3) stays the same
We just saw that closing the switch
causes an increase in the voltage
across R1 (which is VAB). The
voltage of the battery is constant,
so if VAB increases, then VBC must
decrease!
A
R1
B
S
R3
V
R2
C
R4
1) R1
2) both R1 and R2 equally
3) R3 and R4
Which resistor has
4) R5
the greatest current
5) all the same
going through it?
Assume that all the
resistors are equal.
V
Which resistor has the
1) R1
greatest current going
2) both R1 and R2 equally
through it? Assume that
all the resistors are
4) R5
equal.
The same current must flow
through left and right
combinations of resistors.
On the LEFT, the current
splits equally, so I1 = I2. On
the RIGHT, more current will
go through R5 than R3 + R4
since the branch containing
R5 has less resistance.
3) R3 and R4
5) all the same
V
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 + …
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:
1
1
1
1




Req R1 R2 R3
Problem-Solving Strategy,
3

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
Problem-Solving Strategy,
4

If the current in or the potential
difference across a resistor in the
complicated circuit is to be
identified, start with the final
circuit found in step 3 and
gradually work back through the
circuits

Use ΔV = I R and the procedures in
steps 1 and 2
Equivalent
Resistance –
Complex
Circuit
An analogy using
water may be helpful
in visualizing
parallel circuits:
Example 1
A 4.0-Ω resistor, an 8.0-Ω resistor, and a 12-Ω
resistor are connected in series with a 24-V
battery. What are (a) the equivalent
resistance and (b) the current in each
resistor? (c) Repeat for the case in which all
three resistors are connected in parallel
across the battery.
Example 2
A 9.0-Ω resistor and a 6.0-Ω resistor are connected
in series with a power supply. (a) The voltage drop
across the 6.0-Ω resistor is measured to be 12 V.
Find the voltage output of the power supply. (b)
The two resistors are connected in parallel across
a power supply, and the current through the 9.0-Ω
resistor is found to be 0.25 A. Find the voltage
setting of the power supply.
Example 3
(a) Find the equivalent resistance of the circuit in Figure
P18.8. (b) If the total power supplied to the circuit is 4.00
W, find the emf of the battery.
Example 4
Three 100-Ω resistors are connected as shown in Figure
P18.12. The maximum power that can safely be delivered to
any one resistor is 25.0 W. (a) What is the maximum voltage
that can be applied to the terminals a and b? (b) For the
voltage determined in part (a), what is the power delivered
to each resistor? What is the total power delivered?
Practice 1
What is the equivalent resistance of the combination of
resistors between points a and b in Figure P18.7? Note that
one end of the vertical resistor is left free.
Practice 2
(a) You need a 45-Ω resistor, but the stockroom has
only 20-Ω and 50-Ω resistors. How can the desired
resistance be achieved under these circumstances? (b)
What can you do if you need a 35-Ω resistor?
Some circuits cannot be broken down into
series and parallel connections.
Gustav Kirchhoff



1824 – 1887
Invented
spectroscopy with
Robert Bunsen
Formulated rules
about radiation
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
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
More About the Junction
Rule



I1 = I2 + I3
From
Conservation of
Charge
Diagram b shows
a mechanical
analog
1) 2 A
2) 3 A
3) 5 A
What is the current
in branch P?
4) 6 A
5) 10 A
5A
P
8A
2A
What is the current
in branch P?
1) 2 A
2) 3 A
3) 5 A
4) 6 A
5) 10 A
The current entering the junction
S
5A
in red is 8 A, so the current
leaving must also be 8 A. One
exiting branch has 2 A, so the
other branch (at P) must have 6 A.
P
8A
junction
2A
6A
Setting Up Kirchhoff’s
Rules

Assign symbols and directions to the
currents in all branches of the circuit


If a direction is chosen incorrectly, the
resulting answer will be negative, but the
magnitude will be correct
When applying the loop rule, choose a
direction for transversing the loop

Record voltage drops and rises as they
occur
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
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
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
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
EMFs in series in the same direction: total
voltage is the sum of the separate voltages
EMFs in series, opposite direction: total
voltage is the difference, but the lowervoltage battery is charged.
EMFs in parallel only make sense if the
voltages are the same; this arrangement can
produce more current than a single emf.
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
1) both bulbs go out
2) intensity of both bulbs increases
3) intensity of both bulbs decreases
4) A gets brighter and B gets dimmer
5) nothing changes
The lightbulbs in
the circuit are
identical. When
the switch is
closed, what
happens?
1) both bulbs go out
2) intensity of both bulbs increases
3) intensity of both bulbs decreases
The lightbulbs in the
4) A gets brighter and B gets dimmer
circuit are identical.
5) nothing changes
When the switch is
closed, what
happens?
When the switch is open, the point
between the bulbs is at 12 V. But so is
the point between the batteries. If
there is no potential difference, then
no current will flow once the switch is
closed!! Thus, nothing changes.
1) 2 – I1 – 2I2 = 0
2) 2 – 2I1 – 2I2 – 4I3 = 0
3) 2 – I1 – 4 – 2I2 = 0
4) I3 – 4 – 2I2 + 6 = 0
Which of the
5) 2 – I1 – 3I3 – 6 = 0
equations is valid for
1W
the circuit below?
I2
2W
6V
22 VV
4V
I1
1W
I3
3W
Which of the equations is
1) 2 – I1 – 2I2 = 0
valid for the circuit below?
2) 2 – 2I1 – 2I2 – 4I3 = 0
3) 2 – I1 – 4 – 2I2 = 0
4) I3 – 4 – 2I2 + 6 = 0
5) 2 – I1 – 3I3 – 6 = 0
Eqn. 3 is valid for the left loop:
The left battery gives +2V, then
there is a drop through a 1W
resistor with current I1 flowing.
Then we go through the middle
battery (but from + to – !), which
gives –4V. Finally, there is a
drop through a 2W resistor with
current I2.
1W
I2
2W
6V
22 VV
4V
I1
1W
I3
3W
Example 5
The ammeter shown in Figure P18.16 reads
2.00 A. Find I1, I2, and ε.
Example 6
Determine the potential difference Є1 for the circuit
in Figure P18.18.
Example 7
In the circuit of Figure P18.20, the current I1 is 3.0 A while
the values of ε and R are unknown. What are the currents
I2 and I3?
Example 8
Four resistors are connected to a battery with a terminal
voltage of 12 V, as shown in Figure P18.22. Determine the
power delivered to the 50-Ω resistor.
Practice 3
Calculate each of the unknown currents I1, I2,
and I3 for the circuit of Figure P18.25.
RC Circuits




A direct current circuit may contain
capacitors and resistors, the current will
vary with time
When the circuit is completed, the
capacitor starts to charge
The capacitor continues to charge until
it reaches its maximum charge (Q = Cε)
Once the capacitor is fully charged, the
current in the circuit is zero
When the switch is closed, the capacitor will
begin to charge.
The voltage across the capacitor increases
with time:
The charge follows a similar curve:
This curve has a characteristic time constant:
If an isolated charged capacitor is connected
across a resistor, it discharges:
Notes on Time Constant



In a circuit with a large time
constant, the capacitor charges
very slowly
The capacitor charges very quickly
if there is a small time constant
After t = 10 t, the capacitor is over
99.99% charged
Example 9
An uncharged capacitor and a resistor are
connected in series to a source of emf. If ε = 9.00
V, C = 20.0 μF, and R = 100 Ω, find (a) the time
constant of the circuit, (b) the maximum charge on
the capacitor, and (c) the charge on the capacitor
after one time constant.
Example 10
Consider a series RC circuit for which R = 1.0 MΩ,
C = 5.0 μF, and ε = 30 V. Find the charge on the
capacitor 10 s after the switch is closed.
Practice 4
A series RC circuit has a time constant of
0.960 s. The battery has an emf of 48.0 V, and
the maximum current in the circuit is 0.500
mA. What are (a) the value of the capacitance
and (b) the charge stored in the capacitor
1.92 s after the switch is closed?
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
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
A person receiving a
shock has become part
of a complete circuit.
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
Effects of Various Currents

5 mA or less



10 mA



Can cause a sensation of shock
Generally little or no damage
Hand muscles contract
May be unable to let go a of live wire
100 mA

If passes through the body for just a few
seconds, can be fatal
Ground Wire


Electrical
equipment
manufacturers
use electrical
cords that have a
third wire, called
a case ground
Prevents shocks
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
Example 11
An electric heater is rated at 1 300 W, a toaster at
1 000 W, and an electric grill at 1 500 W. The
three appliances are connected in parallel to a
common 120-V circuit. (a) How much current does
each appliance draw? (b) Is a 30.0-A circuit
breaker sufficient in this situation? Explain.
Example 12
A lamp (R = 150 Ω), an electric heater (R = 25
Ω), and a fan (R = 50 Ω) are connected in
parallel across a 120-V line. (a) What total
current is supplied to the circuit? (b) What is the
voltage across the fan? (c) What is the current in
the lamp? (d) What power is expended in the
heater?
Practice 5