Slide 1

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= CHAPTER 18�� -� DC CIRCUITS

=
�� ε� -�� emf�� ε� is a voltage supplied by a voltage<= /font>

�� source

=
ε� is the maximum effective out= put voltage of a source and

as = such is an �ideal case� quantity

V is the act= ual output voltage of the source

<= /td>		<= /td>	<= /td>
3 Cases �

(I)	1 external (load) resistor, no internal resistance in the

�� source

I

VR� -� voltage across resistor R

VR

R

ε= ;

V

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I

=	<= /td>
NOTE: YOUR TEXT MAY
USE ΔV� FOR EFFECTIVE
OUT= PUT VOLTAGE

R

VR

ε= ;

V� -� effective output voltage

Here= , V� =3D� ε

=
= 16;V�s� around the loop� =3D� 0 hence,
V� -� V_R�=3D� 0�� for� only� one load resistor, R

V� =3D� V_R

<= /td>
I� =3D� V_R / R
(1)�� I� =3D� V / R� =3D� ε / R

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I

VR

R

ε= ;

V

=
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:

<= /td>
V� =3D� ε
I� =3D� V_R / R� =3D� V / R

I� =3D� 10 Volts / 100 Ω� = =3D� 0.1 A

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(II= )� Multiple external resistors (series= ); no internal resistance

<= /td>
Consider two external resistors wit resistance, R₁
and = R₂ in series

I

Agai= n, V =3D ε

V1

R1

but= now,� R� =3D� R₁ + R2

V

and:=

ε= ;

V2

R2

<= /td>
also= :
V₁ + V₂� =3D� V

and= :� V₁� =3D� R₁ I;�� V<= sub>2� =3D� R₂ I

Wha= t is the justification for this ?

OHM�= S LAW

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I

<= /td>	=	<= /td>
E= XAMPLE �
A 75 Ω resistor and a 25 Ω resistor are
connected in series to a 10 Volt DC source
with no internal resistance ( V =3D ε)

V1

R1

Fin= d: (a) I�� (b) V₁ and V2

V

ε= ;

V2

R2

(a)<= /font>

I� =3D� 0.1 A

<= /td>	=
(b) V₁� =3D� R₁ I� =3D� (75 Ω) (0.1 A)� =3D� 7.5 Volts

�� 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
�� inside the voltage sou= rce which has EMF, ε

3

I

3

I

r

VR

R

V

2

ε= ;

4

1

<= /td>
V is the effective output voltage of the source, including any voltage dr= op
w= ithin the source

cur= rent I flows in the circuit including through the source

&= #931; ΔV�s:

&= #916;V

+&#= 949;�� voltage rise<= /font>

1 - = 2

� 2� - 3�� -r I�� voltage drop

3 -= � 4�� -V_{R=
�� voltage
drop}

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(III= )

3

I

3

I

r

VR

R

V

2

ε= ;

4

1

=
Then effective output voltage, V is:
�� V� =3D� ε � r I

=
�and since there is only one external resistor:
�� V_R��=3D� V�� hence:

(2)= �� I� =3D� V_R / R� =3D�� V / R�� =3D

<= /td>
and (2) can be solved for I

(3)<= /font>

See Eq. 18.2 p.= 595

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=	=
INS= ERT� SLIDES� ON� �OPEN� SWITCH� AND
�CL= OSED SWITCH� OUTPUT VOLTAGES AS CHECKED
BY A VOLT METER=

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= SECTIONS� 18.2 & 18.3� -� RESISTORS IN SERIES AND PARALLEL

=
n series resistors are connected to a DC source of Voltage, V. I
is = the current in the circuit.

I

<= /td>	<= /td>
for = series R�s:
I is the same for each resistor:
�� I_1��=3D� I₂� =3D� I_n��=3D� I

R1

V1

R2

V2

<= /td>
but V may not be the same for
each= resistor

V

Vn

Rn

<= /td>
howe= ver:
V� =3D� V₁ + V₂ + V_n

=
NOT= E: Starting in Sec.18.2 The text uses ΔV� instead of V
for the source = output voltage.

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=
n resistors are connected in parallel to a DC source of voltage, V.� I
is = the total current in the circuit.

I1

I2

In

I

R1

R2

Rn

V

V1

V2

Vn

=
for parallel R�s:� V across each re= sistor is the same:

�� V_{1<= /sub>� =3D� V₂� =3D� V_n}

but= I may not be the same for each resistor:

how= ever:� I� =3D� I₁ + I2 +��.+ I_n

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= Equivalent Resistance,� R_eq_�-

Equ= ivalent Circuit

I S= eries -� Consider 2 resistors

I

Com= bination Circuit

Req

I

V

V1

R1

V

V2

R2

Use = I� =3D� V / R_eq

Req �=3D V /I

or R= _eq� =3D� R₁ + R₂

and = V =3D V₁ + V₂

=
for= n resistors in series:
R_eq� =3D� R₁ + R₂ + �..+ R_n

<= /td>
then from Ohm�s Law:
V� =3D� R_1�I� +� R₂ I

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II Parallel� -� Consider 2 resistors

Com= bination Circuit

Equ= ivalent Circuit

I

I2

I1

I

Req

V1

V2

R1

V

R2

V

I� =3D� V / R_eq

and= for n resistors in parallel:

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file:///C:/C6F2A658/PHYSICS1112K-WEB-CH18_files/v3_slide0023.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="windows-1252" Slide 1

=
INS= ERT SLIDES ON EXAMPLE OF SERIES AND PARALLEL
�� RESISTORS

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= SECTION 18.5� -� RC� CIRCUITS

=		<= /td>
�� A capacitor of capacitance, C is = wired, in series with a resistor of
resistance, R and a DC voltage source of� Voltage, V. The current in
the circuit is I and it causes the charge, q on the capacitor to accumulate=
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

R

VC

VR

C

V

S

�V � RI � q / C� =3D� 0

�� (4) RI� +� q / C� =3D� V�� and� I� =3D� dq / dt

<= /td>
then= (4) is:
R dq / dt� +� q / C�� =3D� V

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It can be shown= that:

NOTE : the text= uses Q� for� Q_m;� see eq. = 18.7 p. 607

It can also be = shown that:

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(6)= gives� q� on� C� as a function of time

Now= ,� define� τ� =3D� RC,�� where� τ� is� the� time� constant

then= � (6)� is :

=
By definition the voltage on the capacitor is given by:
�� V_C� =3D� q / C�� then:<= br>

=
(7)= and (8)� show that q�� and� V_C� rise� exponentially� with time, t
We are more interested in V_C than q because V_C can be easily measured

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Con= sider� q� and� V_C� at three different times:

I� at� t =3D 0 :

q� =3D 0� and V_C�=3D 0=

II� at� t� =3D

=
q� =3D� CV� =3D�� Q_m,� maximum charge
�V_C =3D V,� the� source voltage

III= � at� t� =3D� Τ� =3D� RC,�� the� time� constant

=	<= /td>
so at t =3D Τ,� the� capacitor
is charged to 63% of its
fin= al maximum value

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so q� and� V_C� increa= se exponentially� with time

or:

<= /td>	=
Now= � Ohm�s Law says :� V_R� =3D� R I
whe= re V_R is the voltage across the resistor.
So, from (9) :

=
So (7) and (= 8) show:� q and V_C rise exponentially with time
= �� (9) and (10) show: I and V<= b>_R drop exponentially with time

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Plot of� V_C vs� time

V_C

V

0.63V

t

from (8)� V_C at� t =3D

is: V_C�=3D� V(1 � e^-1)� = =3D� 0.63 V

so

<= /td>
is = the time needed� for the capacitor to ch= arge to 63% of its
final value, V

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Plot of� V_R vs� time

V_R

V

0.37 V

<= /td>
is = the time it takes for the resistor voltage to drop to 37% of its
initial value (V= )

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Slide 1

CHAPTER 18 - DC CIRCUITS

SECTION 18.1 - EMF –

= ε - emf &= #949; 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

V

R

I

ε

VR

VR - voltage across resistor R

V is the actual output voltage of the source

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I

ε

VR

V

V - effective output voltage

Here, V =3D &= #949;

ΔV’s around the loop =3D 0 hence,

V - VR =3D 0   for only one load resistor, R

V =3D VR

R

= I =3D VR / R

(1)        = ; I =3D V / R =3D &= #949; / R

(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 VOLTAGE

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I

ε

VR

R

V

EXAMPLE - A 100 Ω resistor is connec= ted to a DC source with an EMF&#= 13;

        = ;            &n= bsp; of 10 Volts and no internal resistance. Find I:

V =3D &= #949; <= /span>

I =3D VR / R =3D V / R

I =3D 10 Volts / 100 Ω =3D 0.1 A

NOTE: IN GENERAL I =3D VR / R WHERE VR IS THE VOLTAGE

ACROSS RESISTOR, R. IN THIS CASE VR =3D V, THE SOURCE VOLTAGE.

THIS IS NOT, GENERA= LLY, THE CASE

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V1

R1

V2

R2

ε

V

I

(II) Multiple external resistors (seri= es); no internal resistance

Consider two extern= al resistors wit resistance, R1

and R2 in series

Again, V =3D ε

but now, R =3D R1 + R2

and:

also:
V1 + V2 =3D V

What is the justifi= cation for this ?

OHM’S LAW

and: V1 =3D R1 I;      V<= span style=3D'font-family:"Times New Roman";font-size:74%;position:relative; top:.37em;mso-text-raise:-25%'>2 =3D R2 I