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=
CHAPTER
28 - DC CIRCUITS
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ε - emf ε is a voltage supplied by a voltage<=
/font>
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source
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![](3D"v3_slide0001_image00=) |
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| <=
/td>
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/td>
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/td>
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3
Cases –
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1
external (load) resistor, no internal resistance in the
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source
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![](3D"v3_slide0001_image002.g=) |
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VR -
voltage across resistor R
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RES=
UME EDITING
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<=
/td>
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NOTE:
YOUR TEXT MAY
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USE ΔV FOR EFFECTIVE
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OUT=
PUT VOLTAGE
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V -
effective output voltage
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=
16;V’s around the loop =3D
0 hence,
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V -
VR =3D 0
for only one load resistor, R
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| <=
/td>
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I =3D
VR / R
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(1) I
=3D V / R =3D ε
/ R
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![](3D"v3_slide0010_image00=) |
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![](3D"v3_slide0011_image008.gif") |
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| =
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EXAMPLE
- A 100 Ω resistor is con=
nected
to a DC source with an EMF
|
of 10 Volts an=
d no
internal resistance. Find I:
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| <=
/td>
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V =3D ε
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I =3D
VR / R =3D V / R
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I =3D
10 Volts / 100 Ω =
=3D 0.1 A
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(II=
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internal resistance
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/td>
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Consider
two internal resistors wit resistance, R1
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and =
R2
in series
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![](3D"v3_slide0012_image012.gif") |
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| <=
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also=
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+ V2 =3D V
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and=
: V1 =3D
R1 I; V<=
sub>2 =3D
R2 I
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Wha=
t is the
justification for this ?
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<=
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E=
XAMPLE –
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A 75 Ω
resistor and a 25 Ω resistor are
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connected
in series to a 10 Volt DC source
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with
no internal resistance ( V =3D ε)
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Fin=
d: (a)
I (b) V1 and V2
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(b) V1 =3D
R1 I =3D (75 Ω) (0.1 A) =3D
7.5 Volts
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V2 =3D R2=
I =3D
(25 Ω) (0.1 A) =3D=
2.5 Volts
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(III)
One external resistor of resistance, R and an internal resistance, r
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rce
which has EMF, ε
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/td>
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V is
the effective output voltage of the source, including any voltage dr=
op
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w=
ithin the
source
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cur=
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in the circuit including through the source
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+&#=
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/font>
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![](3D"v3_slide0015_image020.g=) |
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![](3D"v3_slide0015_image021.g=) |
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| =
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Then
effective output voltage, V is:
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V =3D ε
– r I
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and since there is only one
external resistor:
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VR =3D
V hence:
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(2)=
I
=3D VR / R =3D
V / R =3D
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| <=
/td>
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and
(2) can be solved for I
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(3)<=
/font>
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GO<=
span
style=3D'mso-spacerun:yes'> TO
NOTES FOR EXAMPLE OF “CHECKING V WITH
A METER”
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=
SECTION 28.2
- RESISTORS IN SERIES A=
ND
PARALLEL
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n series
resistors are connected to a DC source of Voltage, V. I
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is =
the current
in the circuit.
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![](3D"v3_slide000=) |
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| =
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FOR=
SERIES R’s:
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I is
the same for each resistor:
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I1
=3D I2 =3D
In =3D I
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| <=
/td>
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but V
may not be the same for
|
each=
resistor
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![](3D"v3_slide0002_image023.=) |
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| <=
/td>
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howe=
ver:
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V =3D
V1 + V2 …+… Vn
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![](3D"v3_slide002=) |
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Vol=
tage, Vi is
the same for each
resistor:
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| <=
/td>
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But I =
may
not be the same for each resistor:
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I1 ≠ I2 ≠
In
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| <=
/td>
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Howe=
ver:
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I1 + I2 + .=
..+ In =3D
I (total current in the group of resistors)
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I S=
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II
Parallel - Consider 2 resistors
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and=
for n
resistors in parallel:
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INSERT
SLIDE ON EXAMPLE OF SERIES AND PARLLEL R’s
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AND=
FIRST SLIDE
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![](3D"v3_slide001=) |
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![](3D"v3_slide0019_image037.g=) |
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R1 =3D
4.1 Ω; R2 =3D 2
Ω; R3 =3D 5.1 Ω; R4 =3D 3.5 Ω; R5
=3D 3.8 Ω and V =3D 10 Vo=
lts
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Fin=
d (a) Req (b) I1 I2 I34
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(a)=
R3
in series with R4
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R434
=3D R3 + R4 =3D 8.6 Ω
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R34
in parallel with R2:
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1 / R234 =3D
1/R2 + 1/R34 =3D
1/2 Ω-1
+ 1/8.6 Ω-1
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| <=
/td>
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R1
and R5 are in series with
|
R234
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![](3D"v3_slide0020_image038.gif") |
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| <=
/td>
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Req
=3D 4.1 Ω + 3.8 Ω + 1.62 Ω
|
Req =3D
9.52 Ω
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![](3D"v3_slide0020_image039.g=) |
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| <=
/td>
| <=
/td>
| <=
/td>
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(b)
HW: Use :
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I
=3D V / Req<=
span
style=3D'mso-spacerun:yes'> and
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I2 =3D V2 /=
R2, I34 =3D V34 / R34 and of course: V2 =3D V<=
sub>34
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=
SECTION
28.4 - RC
CIRCUITS
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| =
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<=
/td>
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A capacitor of capacitance, C is =
wired,
in series with a resistor of
|
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resistance,
R and a DC voltage source of of Voltage, V. The current in
|
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the
circuit is I and it causes the charge, q on the capacitor to accumulate=
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or
diminish with time. So q, I and the Voltages will be time
dependent.
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I C=
harging Capacitor – at t =3D 0, q =3D 0 and=
the
switch is closed
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(4) RI
+ q / C =3D
V and I
=3D dq / dt
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| <=
/td>
|
then=
(4) is:
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R dq
/ dt + q / C
=3D V
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dq =
/ dt =3D
(V / R - q / RC)
or:
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![](3D"v3_slide0008_image043=) |
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![](3D"v3_slide0008_image045.gif") |
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![](3D"v3_slide0009_image049=) |
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(6)=
gives q
on C as a function of time
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Now=
, define
τ =3D RC,
where τ is
the time constant
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![](3D"v3_slide0009_image05=) |
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| =
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By
definition the voltage on the capacitor is given by:
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| VC =3D
q / C then:<=
br>
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![](3D"v3_slide0009_image051=) |
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| =
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(7)=
and
(8) show that q and
VC rise exponentially with time, t
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We
are more interested in VC than q because VC can be
easily measured
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![](3D"v3_slide0017_image052.gif") |
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![](3D"v3_slide0017_image053=) |
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Con=
sider q
and VC at three different times: times
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![](3D"v3_slide0017_image054=) |
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![](3D"v3_slide0017_image055.=) |
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| =
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q =3D
CV =3D Qm, maximum charge
|
VC =3D V, the
source voltage
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![](3D"v3_slide0017_image056=) |
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III=
at
t =3D Τ
=3D RC, the
time constant
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| =
|
<=
/td>
|
so at
t =3D Τ, the capacitor
|
is
charged to 63% of its
|
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fin=
al maximum
value
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![](3D"v3_slide0017_image057.gif") |
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Use=
equation 8
to plot Vc vs t
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This
shows that τ is the ti=
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needed for the capacitor voltage to reach
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maximum value, V ( the source voltage)
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PowerPoint Presentation
SECTION
28.1 - EMF=
–
<=
/div>
=
ε - emf
ε is a voltage supplied by a voltag=
e
=
; &n=
bsp;
source
=
ε is the maximum effective output voltage of a source and
as such is an “ideal
case” quantity
3 Cases –
(I)1 external (load) resistor, no
internal resistance in the
source
![](3D"slide0001_image060.gif")
![](3D"slide0001_image061.gif")
![](3D"slide0001_image062.gif")
![](3D"slide0001_image063.gif")
![](3D"slide0001_image064.gif")
![](3D"slide0001_image065.gif")
![](3D"slide0001_image066.gif")
V
R
![](3D"slide0001_image067.gif")
I
ε
![](3D"slide0001_image068.gif")
VR
VR -
voltage across resistor R
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RESUME EDITING SLIDES HERE
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PowerPoint Presentation
![](3D"slide0010_image069.gif")
![](3D"slide0010_image070.gif")
![](3D"slide0010_image071.gif")
![](3D"slide0010_image072.gif")
![](3D"slide0010_image073.gif")
![](3D"slide0010_image074.gif")
![](3D"slide0010_image075.gif")
![](3D"slide0010_image076.gif")
I
ε
![](3D"slide0010_image077.gif")
VR
&nb=
sp;
V
V -<=
span
style=3D'mso-spacerun:yes'> effective output voltage
ΔV’s aro=
und
the loop =3D 0 hence,
V -
VR =3D
0 for only one load resistor, R
R
=
I =
=3D VR / R =
(1) =
;
I =3D V / R =3D
ε / R&#=
13;
(1)shows that with one external resistor and no internal resistance
Ohm’s
Law=
gives
the current to be “EMF
divided by R”
NOTE: YOUR TEXT MAY
USE ΔV FOR EFFECTIVE
OUTPUT VOLT=
AGE
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PowerPoint Presentation
![](3D"slide0011_image078.gif")
![](3D"slide0011_image079.gif")
![](3D"slide0011_image080.gif")
![](3D"slide0011_image081.gif")
![](3D"slide0011_image082.gif")
![](3D"slide0011_image083.gif")
![](3D"slide0011_image084.gif")
![](3D"slide0011_image085.gif")
I
ε
![](3D"slide0011_image086.gif")
VR
R
V
EXAMPLE - A 100 Ω resistor is connected to a DC source with an EMF
=
=
; &n=
bsp;
of 10 Volts and no internal resistance. Find I:
V =
=3D &#=
949;I =3D
VR / R
=3D V / R
I =
=3D 10 Volts / 100 Ω =3D 0.1 A
NOTE: IN GENERAL&=
nbsp;
I =3D VR / R<=
span
style=3D'mso-spacerun:yes'> WHERE VR IS T=
HE
VOLTAGE
ACROSS RESISTOR, R. IN THIS CASE VR =3D =
V, THE
SOURCE VOLTAGE.
THIS IS NOT, GENERALLY, THE CASE
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PowerPoint Presentation
![](3D"slide0012_image087.gif")
V1
R1<=
/span>
V2
R2<=
/span>
![](3D"slide0012_image088.gif")
![](3D"slide0012_image089.gif")
![](3D"slide0012_image090.gif")
![](3D"slide0012_image091.gif")
![](3D"slide0012_image092.gif")
ε
V
![](3D"slide0012_image093.gif")
![](3D"slide0012_image094.gif")
I
(II)
Multiple external resistors (series); no internal resistance=
Consider two internal resistors wit resistance, R<=
/span>1&=
#13;
and R2 in series
but now,
R =3D R1 + R<=
/span>2<=
/span>
![](3D"slide0012_im=)
![](3D"slide0012_image097.gif")
What is the justification for this ?
and: V1<=
/span> =3D=
R1 I; V<=
span
style=3D'font-size:55%;position:relative;top:.37em;mso-text-raise:-25%'>2<=
/span> =3D=
R2 I
![3D"Collapse](3D"v3_outline_collapse.gif")