Electricity and Magnetism
DC Circuits
EMF and Internal Resistance
Kirchhoff’s Laws
Lana Sheridan
De Anza College
Oct 27, 2015
Last time
• current
• current density
• drift speed
• resistance
• resistivity
• conductance
• Ohm’s Law
• power
Overview
• emf
• internal resistance of batteries
• potential drops
• Kirchhoff loop rule
• resistors in series and parallel
Power
P = I ∆V
The units for power are Watts, W.
1 W = 1 J/s.
Does this agree with the new equation?
Power
P = I ∆V
The units for power are Watts, W.
1 W = 1 J/s.
Does this agree with the new equation?
1 A V = (1 C/s) (1 J/C) = 1 J/s . Yes.
Power Dissipated
P = I ∆V
We can use this expression along with R =
dissipated as heat in a resistor.
∆V
I
to find the power
Power dissipated as heat in a resistor:
P = I 2R
or equivalently,
P=
(∆V )2
R
where I in the first equation is the current in the resistor and ∆V
in the second equation is the potential difference across the
resistor.
Example: Why High Voltage?
Example
A power station supplies current I = 5 A and potential difference
∆V = 1200 kV to a particular installation along the electric grid.
How much power is supplied to the installation?
Example
A power station supplies current I = 5 A and potential difference
∆V = 1200 kV to a particular installation along the electric grid.
How much power is supplied to the installation?
P = I ∆V = (5 A)(1.2 × 106 V) = 6 MW
Example
A power station supplies current I = 5 A and potential difference
∆V = 1200 kV to a particular installation along the electric grid.
How much power is supplied to the installation?
P = I ∆V = (5 A)(1.2 × 106 V) = 6 MW
Suppose the power station is 1000 km from the installation and
delivers the power over copper wires. Assume the resistivity of
copper is 1.69 × 10−8 Ω m and the radius of the high tension wire
is 2 cm. What is the resistance of the wire delivering the
electricity?
Example
A power station supplies current I = 5 A and potential difference
∆V = 1200 kV to a particular installation along the electric grid.
How much power is supplied to the installation?
P = I ∆V = (5 A)(1.2 × 106 V) = 6 MW
Suppose the power station is 1000 km from the installation and
delivers the power over copper wires. Assume the resistivity of
copper is 1.69 × 10−8 Ω m and the radius of the high tension wire
is 2 cm. What is the resistance of the wire delivering the
electricity?
R=
ρL
= 13.4 Ω
A
Example
How much power is dissipated as heat in the transmission lines to
the installation (current I = 5 A and potential difference
∆V = 1200 kV are supplied to the station)?
Example
How much power is dissipated as heat in the transmission lines to
the installation (current I = 5 A and potential difference
∆V = 1200 kV are supplied to the station)?
P = I 2 R = (5 A)2 (13.4 Ω) = 336 W
Example
How much power is dissipated as heat in the transmission lines to
the installation (current I = 5 A and potential difference
∆V = 1200 kV are supplied to the station)?
P = I 2 R = (5 A)2 (13.4 Ω) = 336 W
How much power would be dissipated as heat in the transmission
lines to the installation if instead the station supplied 6 MW of
power with current I = 500 A and potential difference
∆V = 12 kV?
Example
How much power is dissipated as heat in the transmission lines to
the installation (current I = 5 A and potential difference
∆V = 1200 kV are supplied to the station)?
P = I 2 R = (5 A)2 (13.4 Ω) = 336 W
How much power would be dissipated as heat in the transmission
lines to the installation if instead the station supplied 6 MW of
power with current I = 500 A and potential difference
∆V = 12 kV?
P = I 2 R = (500 A)2 (13.4 Ω) = 3.36 MW
Much more loss!
Example
This is why power stations transmit power at a very high voltage.
The voltage is “stepped down” before being delivered to your
house.
Mains electricity in the US is distributed throughout a house at
120 V. (The “line voltage”.)
Circuits (Ch 27)
Circuits consist of a collection of electrical components connected
by conducting wires through which charge is driven by an energy
source.
Right now we focus on direct-current (DC) circuits.
In a direct-current circuit current flows in one direction only.
This is the only type of situation we have been considering so far.
However, in the coming labs you may look at some situations with
alternating-current (AC), in which the current flows forward,
then backward, through the circuit.
te
an
Potential in a Circuit
re
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Chapter 28 Direct-Current Circuits
th
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Because
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e
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e with zero internal resistance
te
Ir
(called its terminal voltage)
equals its e
I
I
IR
voltage is not equal to the emf for a b
R
To understand why, consider the circu
Fr
tery as shown in the diagram; it is rep
e
f
is,
0
an ideal,
resistance-free emf in ser
a
a
b
resistance
R is connected across the t
be
The potential drops across each resistor in the circuit as each
through the battery from a to d and
Eq
Figure
28.1
(a)
Circuit
diagram
transforms
electrical
power
to
heat.
V e
locations.
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r
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f
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of aa battery),
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connectedby
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en
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(b) Graphical
terminal voltage of the battery DV 5 V
Ir
to
representation showing how the
IR
Batteries increase the potential energyelectric
of a charge
/
raise
the
re
DV
potential changes as the
potential.
circuit in (a) is traversed clockwise.
co
f
e
a
e
e
A closer look at batteries and power supplies
Batteries and power supplies fill a critical role in circuits.
They supply the energy to drive the charges around the circuit.
They do this by creating a charge imbalance and causing each
charge to experience a force.
Electomotive Force
We say that a battery or power supply contributes an
electromotive force (emf) and we can call batteries and power
supplies emf devices.
These devices act as “charge pumps” in a circuit.
emf device
A device that maintains a potential difference between two points
(terminals) in the circuit.
Electomotive “Force” (emf)
There is a force on each free charge in the system because there is
an electric field.
F = qE
The electric field exists because of the potential difference supplied
to the circuit by the battery.
But this is not what we mean by emf! The emf is not actually a
force.
Electomotive “Force”
We write an emf as E, and label the battery with it:
a
i
thermopile; o
Let us no
energy transf
tion of this c
device at its
must do an a
We define the
a'
+
R
i
–
i
Fig. 27-1
A simple electric circuit, in
Emf is actually
a energy
supplied
unitwork
charge!
which
a device
of emfper
! does
on the
(Measured in charge
volts.) carriers and maintains a steady
current i in a resistor of resistance R.
This makes calling it a “force” a bit misleading.
In words, the
does in movi
nal. The SI u
unit as the vo
An ideal
movement of
the terminals
Electomotive “Force”
To be clear: the electromotive “force” is not a force.
It is an energy supplied per unit charge! It has the units volts.
The name is an unfortunate choice that stuck.
EMF
We can define emf by the following relation:
E=
∆W
∆q
meaning, an emf device does a work ∆W on anamount of charge
∆q:
∆W = E ∆q
while moving the infinitesimal charge dq from the negative
terminal to the positive terminal. (Imagining dq to be positive.)
The amount of work that is done “lifting” this charge to the higher
potential terminal depends only on the potential difference, so E is
a potential difference measured in volts.
Power Supplied
This definition for emf gives the power supplied by an emf device.
E=
∆W
∆q
Power is the rate at which the work is done:
P=
∆W
∆q
=E
∆t
∆t
(assuming that the emf supplied by a source is constant.)
Then notice that I = ∆q
∆t , so
P = IE
This is the total power supplied by an emf device!
Compare to P = I (∆V ) as the power delivered to any component.
EMF
Why do we suddenly need to call potential difference ∆V of a
battery emf E?
EMF
Why do we suddenly need to call potential difference ∆V of a
battery emf E?
Usually, we introduce emf when we want to make the battery more
realistic: batteries have some internal resistance, so the
potential the supplied is not the same in all circumstances.
The emf gives the maximum potential a battery can supply.
EMF
Why do we suddenly need to call potential difference ∆V of a
battery emf E?
Usually, we introduce emf when we want to make the battery more
realistic: batteries have some internal resistance, so the
potential the supplied is not the same in all circumstances.
The emf gives the maximum potential a battery can supply.
There is one other important reason, however: we can now start to
encounter circumstances where we cannot define electric potential
- this will only be important when we come onto magnetic fields.
Internal resistance
How does internal resistance affect the supplied potential
difference?
(c
vo
R
To
How does internal resistance affect the supplied potential
te
e
difference?
f
an
a
It
is
another
resistance
that
is
in
series!
re
834
Chapter 28 Direct-Current Circuits
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Because
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f
r
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within the battery. This resistance is
tia
d
a b
c
battery
e with zero internal resistance
te
Ir
(called
its terminal voltage)
equals its e
I
I
IR
voltage is not equal to the emf for a b
R
To understand why, consider the circu
Fr
tery as shown in the diagram; it is rep
e
f
is,
0
an ideal,
resistance-free emf in ser
a
a
b
R is connected across the t
Let r be the internal resistance resistance
be
through the battery from a to d and
Eq
Figure
28.1
(a)
Circuit
diagram
V e
Passing
from
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negative t
Vr =locations.
Irof a source
r
R
of
emf
e
(in
this
po
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d
e
f
increases
byofan
amount
. As we move
a battery),
internal
resistance
Vr is the potential drop across thetial
internal
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by
anexternal
amount
Ir, where
r,decreases
connected
to an
resis-
Internal resistance
I
I
e
e
Internal resistance
834
Chapter 28
Direct-Current Circuits
Because a real battery i
within the battery. This
d
c
battery with zero intern
(called its terminal voltag
I
I
voltage is not equal to th
R
To understand why, con
tery as shown in the dia
e
f
an ideal, resistance-free
a
resistance R is connecte
Let V be the potential difference supplied by the battery
thebattery fro
throughtothe
V
rest of the circuit:
e
locations. Passing from
r
R
a b c
f
V = Ed − Ire
increases by an amount
tial decreases by an amou
V is the potential difference
between the terminalsterminal
of the battery
e
voltage of the b
Ir
e
! "
a b
r
at points a and d inIRthe diagram.
0
From this expression, n
is, the terminal voltage
Internal resistance
834
Chapter 28
Direct-Current Circuits
Because a real battery i
within the battery. This
d
c
battery with zero intern
(called its terminal voltag
I
I
voltage is not equal to th
R
To understand why, con
tery as shown in the dia
e
f
an ideal, resistance-free
a
resistance R is connecte
Let V be the potential difference supplied by the battery
thebattery fro
throughtothe
V
rest of the circuit:
e
locations. Passing from
r
R
a b c
f
V = Ed − Ire
increases by an amount
tial decreases by an amou
V is the potential difference
between the terminalsterminal
of the battery
e
voltage of the b
Ir
e
! "
a b
r
at points a and d inIRthe diagram.
V depends on the current that flows in the circuit!From this expression, n
0
is, the terminal voltage
Internal resistance
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Chapter 28
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a b
Direct-Current Circu
r
d
c
I
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R
f
e
a
Ideal battery
An ideal battery has
V no
e internalr resistance. R(r = 0)
a b c
d
e
f
Real batteries do have internal resistance.
e
Because a rea
within the ba
battery with
(called its term
voltage is not
To understan
tery as shown
an ideal, resis
resistance R i
through the
locations. Pas
increases by a
tial decreases b
Internal resistance and current
The current that flows in the circuit, I, will in turn depend on the
Chapter
Circuits
load resistance R, 834
ie. the resistance
in the28restDirect-Current
of the circuit.
e
r
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a b
d
c
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I
R
e
f
a
∆V = IR
V e
and so, IR = E − Ir and:
a
e
IR
b
c
r
d
E
I=
r +R
Ir
e
R
f
Because a real battery i
within the battery. This
battery with zero intern
(called its terminal voltag
voltage is not equal to th
To understand why, con
tery as shown in the dia
an ideal, resistance-free
resistance R is connecte
through the battery fro
locations. Passing from
increases by an amount
tial decreases by an amou
terminal voltage of the b
Internal resistance, potential difference, and power
E
r +R
The potential difference supplied to the circuit ∆V :
I=
∆V = IR =
ER
r +R
It depends on both the internal and external (“load”) resistances.
Power:
power supplied = total power delivered
IE = I 2 r + I 2 R
Question
Quick Quiz 28.1: To maximize the percentage of the power from
the emf of a battery that is delivered to a device external to the
battery, what should the internal resistance of the battery be?
(A) It should be as low as possible.
(B) It should be as high as possible.
(C) The percentage does not depend on the internal resistance.
Question
Quick Quiz 28.1: To maximize the percentage of the power from
the emf of a battery that is delivered to a device external to the
battery, what should the internal resistance of the battery be?
(A) It should be as low as possible.
←
(B) It should be as high as possible.
(C) The percentage does not depend on the internal resistance.
834
Chapter 28
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Internal resistance
r
! "
a b
e
! "
a b
I
c
r
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d
c
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f
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e
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f
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Direct-Current Circuits
Because a real battery is mad
within the battery. This resis
Becausewith
a real
battery
is mad
battery
zero
internal
re
within
the
battery.
This
resis
(called
its
terminal
voltage)
eq
I
battery iswith
re
voltage
not zero
equalinternal
to the em
(called
its
terminal
voltage)
eq
To
understand
why,
consider
I
voltage
is not equal
the em
tery
as shown
in thetodiagram
To understand
why, consider
an
ideal, resistance-free
emf
tery as shown
the diagram
resistance
R is in
connected
acr
an ideal, the
resistance-free
emf
through
battery from
a
resistance Passing
R is connected
locations.
from theacr
ne
through the
battery
from. As
a
increases
by an
amount
locations.
Passing
the n
tial
decreases
by anfrom
amount
Ir
increasesvoltage
by an amount
. As
terminal
of the batter
e
e
tial decreases by an amount Ir
terminal voltage of the batter
IR
From this expression, notice
is, the terminal voltage when
0
this for
expression,
aFrom
battery;
example,notice
the em
b difference supplied by the battery to the
Let ∆V be the potential
is, the terminal
voltage
when
between
a battery’s
terminals
0
rest of the circuit: Figure
a battery; 28.1.
for example,
the em
Equation
Figure
28.1b
i
28.1
(a) Circuit diagram
b
of a source of emf e (in this case,
between aasbattery’s
terminals
potential
the circuit
is trav
∆V
=
E
−
Ir
a battery),
of internal
resistance
Equation
28.1bthi
Figure
28.1
(a) Circuit
diagram
Figure 28.1.
28.1a Figure
shows that
r, connected
toemf
an e
external
resisof
a
source
of
(in
this
case,terminals
∆V is the potential difference
between
the
of
the
battery
potential
as
the
circuit
isresist
trav
ence
across
the
external
of resistance
R. (b) Graphical
ator
battery),
of internal
resistance
shows resistive
that th
torFigure
might 28.1a
be a simple
at points a and d in the
diagram.
representation
showing how the
IR
e
Ir
r, connected to an external resis-
R
e
f
To understand why, consider
tery as shown in the diagram
an ideal, resistance-free emf
resistance R is connected acr
through the battery from a
locations. Passing from the ne
increases by an amount e. As
tial decreases by an amount Ir
terminal voltage of the batter
Potential differencea between two points
V
a
e
IR
e
b
c
r
d
e
Ir
R
f
From this expression, notice
is, the terminal voltage when
0
a battery; for example, the em
b
between a battery’s terminals
Equation
28.1.points
Figure 28.1b i
Circuit diagram
For any circuit we canFigure
find 28.1
the (a)
potential
difference
between
of a source of emf e (in this case,
potential
as
the
in the circuit by finding
the potential
drop or jump across the circuit is trav
a battery),
of internal resistance
Figure 28.1a shows that th
r, connected
elements between those
points.to an external resisence
across the external resist
tor of resistance R. (b) Graphical
tor might be a simple resistive
representation showing how the
resistance of some electrical
electric potential changes as the
Two rules can help uscircuit
track
this.
in (a)
is traversed clockwise.
connected to the battery (or,
The resistor represents a load
to operate the device contain
load resistance is DV 5 IR. Com
e
f
a
tery as shown in the diagram; it is repre
an ideal, resistance-free emf in series
resistance R is connected across the term
through the battery from a to d and m
locations. Passing from the negative term
increases by an amount . As we move th
tial decreases by an amount Ir, where I is
terminal voltage of the battery DV 5 Vd 2
e
Potential difference between two points
V
a
e
IR
e
b
c
r
d
e
Ir
R
f
e
DV 5
e
From this expression, notice that is e
is, the terminal voltage when the curren
a battery; for example, the emf of a D c
b
between a battery’s terminals depends o
Equation 28.1. Figure 28.1b is a graphic
Figure 28.1 (a) Circuit diagram
“Voltage Drops”:
of a source of emf e (in this case,
potential as the circuit is traversed in th
a battery), of internal resistance
Figure 28.1a shows that the terminal
resistance rule
r, connected to an external resisence across the external resistance R, ofte
tor of resistance R. (b) Graphical
tor
be a simple resistive
representation
how the
Going through a resistance
R in showing
the direction
of might
the current,
the circuit ele
resistance of some electrical device (suc
electric potential changes as the
circuit in
is traversed
clockwise. direction
change in potential is −iR;
in(a)the
opposite
connecteditto is
the+iR.
battery (or, in the cas
The resistor represents a load on the batt
to operate the device containing the res
“Voltage jumps”:
load resistance is DV 5 IR. Combining th
0
e5
emf rule
Figure 28.1a shows a graphical represen
direction
of the emf
rent gives
Going through an ideal emf device in the
arrow, the change in potential is +E; in the opposite direction it is
−E.
I5
Equation 28.3 shows that the current i
The loop rule
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Chapter
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Chapter
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Because
a real a
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Because
real is
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the battery.
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re
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its terminal
its em
(called
its
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eq
voltage
is
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the
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for
a
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I
To understand
thethe
circuit
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notconsider
equal to
em
tery To
as shown
in the diagram;
it is repre
understand
why, consider
an ideal, resistance-free emf in series
tery as shown in the diagram
resistance R is connected across the term
an ideal,
resistance-free
emfm
through
the battery
from a to d and
locations.
PassingRfrom
the negative acr
term
resistance
is connected
increases
by an amount
. As we
moveath
through
the battery
from
tial decreases by an amount Ir, where I is
locations. Passing from the n
terminal voltage of the battery DV 5 Vd 2
R
e
e
f
R
f
e
increases by an amount . As
DV 5
tial decreases by an amount
Ir
Fromterminal
this expression,
notice
that
e
is e
voltage of the batter
is, the terminal voltage when the curren
a battery; for example, the emf of a D c
between a battery’s terminals depends o
From
this
expression,
notice
28.1b
is a graphic
Figure 28.1
Circuit
diagram
Notice in the lower diagram
that(a)we
we
come Equation
back
at28.1.
theFigure
right
end
of a source of emf e (in this case,
potential
as the
circuit isvoltage
traversedwhen
in th
is, the
terminal
0 a battery),
to the same potential that
weofstarted
at on the Figure
left
end.
internal resistance
28.1a shows
that the terminal
a battery;
for example,
the em
r, connected to an external resisb
ence across the external resistance R, ofte
tor of resistance R. (b) Graphical
between
a battery’s
terminals
tor
be ago
simple
resistive
ele
representation
showing how
the
In fact, it doesn’t matter
what
point
we
start
at:might
if we
around
acircuit
Equation
28.1.
Figure
28.1b
Figure
28.1
(a)changes
Circuitasdiagram
resistance
of some
electrical
device
(suci
electric
potential
the
in to
(a)
isthe
traversed
clockwise.
acircuit
source
of emf
e (in
this case,point,
closed loop, when we ofreturn
starting
we tomust
return
connected
theasbattery
(or, in the
cas
potential
the
circuit
is trav
a battery), of internal resistance
The resistor
represents
load onthat
the batt
Figure
28.1a ashows
th
to the starting potential
also.
r, connected to an external resis- to operate the device containing the res
IR 0
b
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Chapter 28
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a a bb
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a
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a
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a
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e
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b
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c
r
r
d
e
d Ir
Ir
Because
real is
battery
mad
Because
a real a
battery
made ofismatte
within
the battery.
This resistance
is ca
within
the battery.
This resis
battery
with zero
resistance, re
t
battery
withinternal
zero internal
(called its terminal voltage) equals its em
(called
its
terminal
voltage)
eq
I voltage is not equal to the emf for a batt
voltage iswhy,
notconsider
equal to
em
To understand
thethe
circuit
understand
why, consider
tery To
as shown
in the diagram;
it is repre
an ideal,
in series
tery resistance-free
as shown in emf
the diagram
resistance R is connected across the term
an ideal, resistance-free emf
through the battery from a to d and m
resistance
is connected
locations.
PassingRfrom
the negative acr
term
through
the battery
from
increases
by an amount
. As we
moveath
tial decreases
by an
amountfrom
Ir, where
locations.
Passing
theI nis
terminal
voltageby
of the
battery DV 5. VAs
d2
increases
an amount
e
e
f
Direct-Current Circuits
Direct-Current Circuits
R
e
e
f
R
f
e
DV Ir
5
tial decreases by an amount
voltage
of the
Fromterminal
this expression,
notice
thatbatter
e is e
is, the terminal voltage when the curren
a battery; for example, the emf of a D c
between a battery’s terminals depends o
From this expression, notice
Equation 28.1. Figure 28.1b is a graphic
28.1 (a) Circuit diagram
Kirchhoff’s loop rule: Figure
is, the
terminal
potential
as the
circuit isvoltage
traversedwhen
in th
0 of a source of emf e (in this case,
a battery), of internal resistance
Figure
28.1a shows
that the terminal
a battery;
for example,
the em
Loop Rule
r,
connected
to
an
external
resisb
encebetween
across the a
external
resistance
R, ofte
battery’s
terminals
tor of resistance R. (b) Graphical
tor might be a simple resistive circuit ele
representation showing how the
Equation
28.1. Figure 28.1b i
Figure
(a) Circuit
diagram
The sum of the changes
in 28.1
potential
encountered
in aofcomplete
resistance
some electrical device (suc
electric
potential changes
as the
of acircuit
source
of emf
e (in clockwise.
this case,
potential
the circuit
is trav
in (a)
is traversed
connected
to theasbattery
(or, in the
cas
traversal of any loop of
a
circuit
must
be
zero.
a battery), of internal resistance
Figure
28.1a ashows
th
The resistor
represents
load onthat
the batt
r, connected to an external resisIR 0
b
Multiloop Circuits
Single loop
Multiloop
R1
V
V
R2
R1
R2
Series and Parallel
Series
Parallel
When components are
connected one after the other
along a single path, they are
connected in series.
When components are
connected side-by-side on
different paths, they are
connected in parallel.
R1
V
V
R2
R1
R2
Resistors in Series
The current though resistors in series in a loop is the same.
Let the total potential difference across two resistors be ∆V , then
∆V = IR1 + IR2 = I(R1 + R2 )
Then the effective equivalent resistance of both together is just the
sum
Req = R1 + R2
For n resistors in series:
Req = R1 + R2 + ... + Rn =
n
X
i=1
Ri
Resistors in Parallel
The potential difference across two resistors in parallel is the same.
(Loop rule.)
Let i be the total current that flows through both resistors:
I = I1 + I2 . (Junction rule.)
I=
∆V
∆V
∆V
=
+
Req
R1
R2
Dividing the equation by V :
1
1
1
=
+
Req
R1 R2
For n of resistors in parallel:
n
X
1
1
1
1
1
=
+
+ ... +
=
Req
R1 R2
Rn
Ri
i=1
Resistors vs. Capacitors
Table of equivalent capacitances and resistances for series and
parallel.
resistors
series
parallel
Req =
1
Req
=
P
P
capacitors
Ri
1
Ceq
1
Ri
Ceq =
=
P
P
1
Ci
Ci
Example
Consider the circuit pictured with E = 12 V, and the following
resistor values: R1 = 20 Ω, R2 = 20 Ω, R3 = 30 Ω, and
R4 = 8.0 Ω.
i1
b
R1
R1
R3
+
–
R2
+
–
R4
a
R3
i2
R2
R4
c
a
c
i1
(a )
What is the current through the battery?
answer: I = 0.30 A
What is the current through resistor R2 ?
answer: I2 = 0.18 A
Applying the loop rule
yields the current.
i3
b
(b )
Kirchhoff’s Laws
The loop rule for potential difference and the junction rule for
current together are called Kirchhoff’s laws.
Loop Rule
The sum of the changes in potential encountered in a complete
traversal of any loop of a circuit must be zero.
Junction Rule
The sum of the currents entering any junction must be equal to
the sum of the currents leaving that junction.
Using both it is possible to discover many things about how a
circuit operates, for example how much power will be dissipated in
a particular component.
e
f
a
tery as shown in the diagram; it is repre
an ideal, resistance-free emf in series
resistance R is connected across the term
through the battery from a to d and m
locations. Passing from the negative term
increases by an amount . As we move th
tial decreases by an amount Ir, where I is
terminal voltage of the battery DV 5 Vd 2
e
Potential difference between two points
V
a
e
IR
e
b
c
r
d
e
Ir
R
f
e
DV 5
e
From this expression, notice that is e
is, the terminal voltage when the curren
a battery; for example, the emf of a D c
b
between a battery’s terminals depends o
Equation 28.1. Figure 28.1b is a graphic
Figure 28.1 (a) Circuit diagram
“Voltage Drops”:
of a source of emf e (in this case,
potential as the circuit is traversed in th
a battery), of internal resistance
Figure 28.1a shows that the terminal
resistance rule
r, connected to an external resisence across the external resistance R, ofte
tor of resistance R. (b) Graphical
tor
be a simple resistive
representation
how the
Going through a resistance
R in showing
the direction
of might
the current,
the circuit ele
resistance of some electrical device (suc
electric potential changes as the
circuit in
is traversed
clockwise. direction
change in potential is −iR;
in(a)the
opposite
connecteditto is
the+iR.
battery (or, in the cas
The resistor represents a load on the batt
to operate the device containing the res
“Voltage jumps”:
load resistance is DV 5 IR. Combining th
0
e5
emf rule
Figure 28.1a shows a graphical represen
direction
of the emf
rent gives
Going through an ideal emf device in the
arrow, the change in potential is +E; in the opposite direction it is
−E.
I5
Equation 28.3 shows that the current i
Example with Two Batteries
Find the current in the circuit.
gure 28.14.
circuit.
e1 # 6.0 V
a
! "
R 2 # 10 $
uess at the
he current
tion of the
ss is repre-
d
I
b
R 1 # 8.0 $
! "
e2 # 12 V
c
Figure
28.14
28.6)
Suppose the current
flows
in the (Example
direction shown.
it, but let’s
A series circuit containing two
Example with Two Batteries
own in Figure 28.14.
nt in the circuit.
and a guess at the
two, so the current
the direction of the
rrect guess is repre-
e1 # 6.0 V
a
! "
R 2 # 10 $
d
I
b
R 1 # 8.0 $
! "
e2 # 12 V
c
Figure 28.14 (Example 28.6)
XA series circuit containing two
ple circuit, but let’s
∆V = Eand
IR1resistors,
− E2 − IR2 = 0
1 −two
batteries
no junctions in this
where the polarities of the batnts.
teries are in opposition.
e 28.14. Traversing the circuit in the clockwise direcnce of 1e 1 , b S c represents a potential difference of
Example with Two Batteries
own in Figure 28.14.
nt in the circuit.
and a guess at the
two, so the current
the direction of the
rrect guess is repre-
e1 # 6.0 V
a
! "
R 2 # 10 $
d
I
b
R 1 # 8.0 $
! "
e2 # 12 V
c
Figure 28.14 (Example 28.6)
XA series circuit containing two
ple circuit, but let’s
∆V = Eand
IR1resistors,
− E2 − IR2 = 0
1 −two
batteries
no junctions in this
where the polarities of the batnts.
teries are inEopposition.
1 − E2
⇒ I=
= −0.33 A
R1 + R2
e 28.14. Traversing the circuit in the clockwise direcMinus sign means that the current flows opposite to the direction
nce of 1e 1 , b S c represents a potential difference of
shown in the diagram.
R2
2
pation rate
(b) resistor
(4.0and
") and
batteries
haveinemfs
!1 ! 121 V
!2 !(c) –
R1
2 (8.0
"), the
and
the energy
Using
Laws
examples
1
6.0resistor
V. Kirchhoff’s
What
are (a)
current,
the transfer
dissi- +
rate
in
(d)
battery
1
and
(e)
battery
2?
Is
R2
2
pation rate in (b) resistor 1 (4.0 ") and (c) –
– +
energy
being
supplied
or
absorbed
by
(f)
resistor 2 (8.0 "), and the energy transfer
1
battery
1battery
and#2
(g) 1battery
2? battery 2? Is
Fig. 27-25
rate
in (d)
and (e)
Page
726,
Problem
1.
– +
•2 In
Fig. supplied
27-26, the
batteries
energy
being
orideal
absorbed
by (f)
have 1emfs
!1 !
150 V2?
and !2 ! 50 V Q
battery
and (g)
battery
Fig. 27-25
and the resistances are R1 ! 3.0 "
R1
Problem
1.
•2 In Fig. 27-26, the ideal batteries
and R2 ! 2.0 ". If the potential at P is –
–
have emfs !1 ! 150 V and !2 ! 50 V Q
100 V, what is it at Q?
1
2
+
+
and the resistances are R1 ! 3.0 "
R1
battery
withataP12
and•3R2ILW
! 2.0A".car
If the
potential
is V –
R2
–
and an
internal
100emf
V, what
is it
at Q? resistance of 0.040 +
1
2
+ P
" is being charged with a current of 50
•3 ILW A car battery with a 12 V
R 2 Problem 2.
A. What are (a) the potential differ- Fig. 27-26
emf and an internal resistance of 0.040
P
ence V across the terminals, (b) the
" is being charged with a current of 50
rate P of energy dissipation inside the battery, and (c) the rate P
A. Whatr are (a) the potential differ- Fig. 27-26 Problem 2. emf
of energy conversion to chemical form? When the battery is used to
ence
V across the terminals, (b) the
supply 50 A to the starter motor, what are (d) V and (e) Pr?
rate Pr of energy dissipation inside the battery, and (c) the rate Pemf
of energy conversion to chemical form? When the battery is used to
supply 50 A to the starter motor, what are (d) V and (e) Pr?
tric
of t
R2
2
pation rate
(b) resistor
(4.0and
") and
batteries
haveinemfs
!1 ! 121 V
!2 !(c) –
R1
2 (8.0
"), the
and
the energy
Using
Laws
examples
1
6.0resistor
V. Kirchhoff’s
What
are (a)
current,
the transfer
dissi- +
rate
in
(d)
battery
1
and
(e)
battery
2?
Is
R2
2
pation rate in (b) resistor 1 (4.0 ") and (c) –
– +
energy
being
supplied
or
absorbed
by
(f)
resistor 2 (8.0 "), and the energy transfer
1
battery
1battery
and#2
(g) 1battery
2? battery 2? Is
Fig. 27-25
rate
in (d)
and (e)
Page
726,
Problem
1.
– +
•2 In
Fig. supplied
27-26, the
batteries
energy
being
orideal
absorbed
by (f)
have 1emfs
!1 !
150 V2?
and !2 ! 50 V Q
battery
and (g)
battery
Fig. 27-25
and the resistances are R1 ! 3.0 "
R1
Problem
1.
•2 In Fig. 27-26, the ideal batteries
and R2 ! 2.0 ". If the potential at P is –
–
have emfs !1 ! 150 V and !2 ! 50 V Q
100 V, what is it at Q?
1
2
+
+
and the resistances are R1 ! 3.0 "
R1
battery
withataP12
and•3R2ILW
! 2.0A".car
If the
potential
is V –
R2
–
and an
internal
100emf
V, what
is it
at Q? resistance of 0.040 +
1
2
+ P
" is being charged with a current of 50
•3 ILW A car battery with a 12 V
R 2 Problem 2.
A. What are (a) the potential differ- Fig. 27-26
emf and an internal resistance of 0.040
P
encerule:
V across
(b) =the
−E2 −the
IR2terminals,
E1 − IR
0, I = 20 A.
"Loop
is being
charged
with
a+current
of1 50
rate P of energy dissipation inside the battery, and (c) the rate P
A. Whatr are (a) the potential differ- Fig. 27-26 Problem 2. emf
of
energyatconversion
Potential
Q = −10 to
V.chemical form? When the battery is used to
ence
V across the terminals, (b) the
supply 50 A to the starter motor, what are (d) V and (e) Pr?
rate Pr of energy dissipation inside the battery, and (c) the rate Pemf
of energy conversion to chemical form? When the battery is used to
supply 50 A to the starter motor, what are (d) V and (e) Pr?
tric
of t
Example with a Multiloop Circuit
Find the currents I1 , I2 , and I3 in the circuit.
14.0 V
e
f
" #
4.0 !
b
(Example
ircuit containing
branches.
I1
# "
10.0 V
28.15
I2
6.0 !
a
2.0 !
c
I3
d
Suppose the currents flow in the direction shown.
in Figure
28.15.
Example with a Multiloop Circuit
Junction rule:
I1 + I2 = I3
(1)
10V − (6Ω)I1 + (2Ω)I3 = 0
(2)
−14V + (6Ω)I1 − 10V − (4Ω)I2 = 0
(3)
−14V − (2Ω)I3 − (4Ω)I2 = 0
(4)
Loops:
Example with a Multiloop Circuit
14.0 V
e
f
" #
4.0 !
b
A circuit containing
rent branches.
I1
# "
10.0 V
re 28.15 (Example
I2
6.0 !
a
2.0 !
c
I3
d
led in Figure 28.15.
1 I2 2 I3 5 0
I1 = +2.0 A I2 = −3.0 A I3 = −1.0 A
(a)
(b)
Using Kirchhoff’s Laws
examples
Fig. 27-20 Question 5.
(c)
6 Res-monster maze. In Fig. 27-21, all the resistors have a
resistance of 4.0 $ and all the (ideal) batteries have an emf of 4.0
V. What is the current through resistor R? (If you can find the
proper loop through this maze, you can answer the question with a
few seconds of mental calculation.)
(b) Are
sistances
to a batel. Rank
through
R
x
e
stion 4.
Fig. 27-21
Question 6.
rent i1 through R1 now more than, less than, or the same as previ-
Using ously?
Kirchhoff’s
Laws resistance
examples
(c) Is the equivalent
R12 of R1 and R2 more than,
less than, or equal to R1?
8 Cap-monster maze. In Fig. 27-22, all the capacitors have a
capacitance of 6.0 mF, and all the batteries have an emf of 10 V.
What is the charge on capacitor C? (If you can find the proper loop
through this maze, you can answer the question with a few seconds
of mental calculation.)
C
Fig. 27-22
Question 8.
R2 more
through
current t
10 Aft
closed on
through
gives tha
values of
and C0, (
2C0, (4) 2
with whic
11 Figu
tions of
nected in
via a swi
capacito
rium) ch
capacito
Summary
• Kirchhoff’s laws
• resistors in series and parallel
Midterm on Thursday May 14.
Homework Halliday, Resnick, Walker:
• Ch 26, onward from page 699. Problems: 41, 43, 45, 47, 55,
71
• Ch 27, onward from page 725. Questions: 1, 3; Problems: 1,
5, 7, 33