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General Physics (PHY 2140)
Lecture 6Lecture 6Electrodynamics
Direct current circuitsparallel and series connectionsKirchhoff’s rulesRC circuits
Chapter 18
http://www.physics.wayne.edu/~alan/2140Website/Main.htm
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Department of Physics and Astronomyannounces the Spring-Summer 2007 opening of
The Physics Resource CenterThe Physics Resource Centeron Monday, May 21 in
Room 172Room 172 of Physics Research Building.of Physics Research Building.
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Lightning ReviewLightning Review
Last lecture:
1.1.
Current and resistanceCurrent and resistance
Temperature dependence of resistanceTemperature dependence of resistance
Power in electric circuitsPower in electric circuits
2.2.
DC CircuitsDC Circuits
EMFEMF
Resistors in seriesResistors in series
QIt
Δ=Δ
( )1o oR R T Tα⎡ ⎤= + −⎣ ⎦( )2
2 VP I V I R
RΔ
= Δ = =
V IrΔ = −E
1 2 3 ...eqR R R R= + + +
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Introduction: elements of electrical circuits Introduction: elements of electrical circuits A branchA branch::
A branch is a single electrical element or device (resistor, etA branch is a single electrical element or device (resistor, etc.).c.).
A junctionA junction::
A junction (or node) is a connection point between two or moreA junction (or node) is a connection point between two or more
branches.branches.
If we start at any point in a circuit (node), proceed through coIf we start at any point in a circuit (node), proceed through connected nnected electric devices back to the point (node) from which we startedelectric devices back to the point (node) from which we started, , without without crossing a node more than one timecrossing a node more than one time, we form a closed, we form a closed--path (or path (or looploop).).
b b b b b
A circuit with 5 branches.
•
•
•
b
bb
A circuit with 3 nodes.
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Steady current (constant in magnitude and direction)• requires a complete circuit• path cannot be only resistance
cannot be only potential drops in direction of current flowElectromotive Force (EMF)• provides increase in potential E• converts some external form of energy into electrical energy
Single emf and a single resistor: emf can be thought of as a “charge pump”
I
V = IR
E
+ - V = IR = E
18.1 Sources of EMF18.1 Sources of EMF
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EMFEMF
Each real battery has some Each real battery has some internal resistanceinternal resistanceAB: potential increases by AB: potential increases by E
on the source of EMF, then on the source of EMF, then decreases by decreases by IrIr
(because of (because of the internal resistance)the internal resistance)Thus, terminal voltage on the Thus, terminal voltage on the battery battery ΔΔV isV is
Note: Note: E
is the same as the is the same as the terminal voltage when the terminal voltage when the current is zero (open circuit)current is zero (open circuit)
E
r
R
A
B C
DV IrΔ = −E
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EMF (continued)EMF (continued)
Now add a load resistance RNow add a load resistance RSince it is connected by a Since it is connected by a conducting wire to the battery conducting wire to the battery →→
terminal voltage is the same as terminal voltage is the same as the potential difference across the the potential difference across the load resistanceload resistance
Thus, the current in the circuit isThus, the current in the circuit is
E
r
R
A
B C
D
IR r
=+E
,V Ir IR orIr IR
Δ = − == +
EE
Power output:
2 2I I r I R= +E
Note: we’ll assume r negligible unless otherwise is stated
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Voltmeters measure Potential Difference (or voltage) across a device by being placed in parallel with the device.
V
Ammeters measure current through a device by being placed in series with the device.
A
Measurements in electrical circuitsMeasurements in electrical circuits
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ab eq
abeq
V IR
VRI
=
≡
Req
Ia
b
Direct Current CircuitsDirect Current Circuits
Two Basic Principles:Conservation of ChargeConservation of Energy
Resistance Networks
Ohm’s Law:
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1 21 2
eq
eq
V IR
IR IRVR R RI I
Δ =
+Δ≡ = = +
18.2 Resistors in series18.2 Resistors in series
1 2I I I= =
1. Because of the charge conservation, all charges going through the resistor R2
will also go through resistor R1
. Thus, currents
in R1
and R2
are the same,
1 2V IR IRΔ = +
2. Because of the energy conservation, total potential drop
(between A and C) equals to the sum of potential drops between A and B and B and C,
By definition,
Thus, Req
would be
1 2eqR R R= +
R2
R1
v2
v1+ +
+
_
_
_v i1
A B
C
I
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Resistors in series: notesResistors in series: notes
Analogous formula is true for any number of resistors,Analogous formula is true for any number of resistors,
It follows that the equivalent resistance of a series It follows that the equivalent resistance of a series combination of resistors is greater than any of the combination of resistors is greater than any of the individual resistorsindividual resistors
1 2 3 ...eqR R R R= + + + (series combination)
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Resistors in series: exampleResistors in series: example
R2
R1
v2
v1+ +
+
_
_
_v i1
A B
C
I
In the electrical circuit below, find voltage across the resistoIn the electrical circuit below, find voltage across the resistor Rr R11
in terms of in terms of the resistances Rthe resistances R11
, R, R22
and potential difference between the batteryand potential difference between the battery’’s s terminals V. terminals V.
1 2V V V= +
Energy conservation implies:
with 1 1 2 2andV IR V IR= =
Then, ( )1 21 2
, so VV I R R IR R
= + =+
Thus,1
11 2
RV VR R
=+
This circuit is known as voltage divider.
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1 2
1 2 1 2
eq
eq
VIR
V VV V VIR R R R R
=
= = + = +
18.3 Resistors in parallel18.3 Resistors in parallel
1 2V V V= =
1. Since both R1
and R2
are connected to the same battery, potential differences across R1
and R2
are the same,
1 2I I I= +
2. Because of the charge conservation, current, entering the junction A,
must equal the current leaving this junction,
By definition,
Thus, Req
would be
1 2
1 1 1
eqR R R= +
I
I2 I1
R2 R1V
+
_
I
ReqV
+
_
A
or 1 2
1 2eq
R RR
R R=
+
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Resistors in parallel: notesResistors in parallel: notes
Analogous formula is true for any number of resistors,Analogous formula is true for any number of resistors,
It follows that the equivalent resistance of a parallel It follows that the equivalent resistance of a parallel combination of resistors is always less than any of the combination of resistors is always less than any of the individual resistorsindividual resistors
(parallel combination)
1 2 3
1 1 1 1 ...eqR R R R= + + +
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Resistors in parallel: exampleResistors in parallel: example
In the electrical circuit below, find current through the resistIn the electrical circuit below, find current through the resistor Ror R11
in terms of in terms of the resistances Rthe resistances R11
, R, R22
and total current I induced by the battery. and total current I induced by the battery.
1 2I I I= +Charge conservation implies:
with 1 21 2
, andV VI IR R
= =
Then, 1 21
1 1 2
, witheqeq
IR R RI RR R R
= =+
Thus, 21
1 2
RI IR R
=+
This circuit is known as current divider.
I
I2 I1
R2 R1
+
_
V
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Direct current circuits: exampleDirect current circuits: exampleFind the currents IFind the currents I11
and Iand I22
and the voltage and the voltage VVxx
in the circuit shown below.in the circuit shown below.
I1I2
4 Ω 12 Ω20 V
7 Ω
+_
+
_
Vx
IStrategy:
1.
Find current I by finding the equivalent resistance of the circuit
2.
Use current divider rule to find the currents I1
and I2
3.
Knowing I2
, find Vx
.
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Direct current circuits: exampleDirect current circuits: exampleFind the currents IFind the currents I11
and Iand I22
and the voltage and the voltage VVxx
in the circuit shown below.in the circuit shown below.
I1I2
4 Ω 12 Ω20 V
7 Ω
+_
+
_
Vx
IFirst find the equivalent resistance seen by the 20 V source:
4 (12 )7 1012 4eqR Ω Ω
= Ω+ = ΩΩ+ Ω
Then find current I by,20 20 2
10eq
V VI AR
= = =Ω
We now find I1 and I2 directly from the current division rule:
1 2 12 (4 ) 0.5 , and 1.5
12 4AI A I I I AΩ
= = = − =Ω+ Ω
Finally, voltage Vx is ( ) ( )2 4 1.5 4 6xV I A V= Ω = Ω =
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R1 =4Ω
R2 =3Ω R3 =6Ω
E=18V
Example: Determine the equivalent resistance of the circuit as shown.Determine the voltage across and current through each resistor.Determine the power dissipated in each resistorDetermine the power delivered by the battery
+
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Example: Determine the equivalent resistance of the circuit as shown.Determine the voltage across and current through each resistor.Determine the power dissipated in each resistorDetermine the power delivered by the battery
R1 =4Ω
R2 =3Ω R3 =6ΩE=18V
+R1 =4Ω
R23 =2Ω
+
Req =6Ω+
E=18V So, Ieq = E/Req =3A
P=Ieq E
= 108W
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18.4 Kirchhoff18.4 Kirchhoff’’s rules and DC currentss rules and DC currents
The procedure for analyzing complex circuits is based on The procedure for analyzing complex circuits is based on the principles of conservation of charge and energythe principles of conservation of charge and energyThey are formulated in terms of two KirchhoffThey are formulated in terms of two Kirchhoff’’s rules:s rules:
1.1.
The sum of currents entering any junction must equal the The sum of currents entering any junction must equal the sum of the currents leaving that junction (current or sum of the currents leaving that junction (current or junction rule) .junction rule) .
2.2.
The sum of the potential differences across all the The sum of the potential differences across all the elements around any closedelements around any closed--circuit loop must be zero circuit loop must be zero (voltage or loop rule).(voltage or loop rule).
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1. Junction rule1. Junction rule
• The sum of the currents entering a node (junction point) The sum of the currents entering a node (junction point) equal to the sum of the currents leaving. equal to the sum of the currents leaving.
Ia
Ib
Ic
Id
Ia, Ib, Ic, and Id can each be either a positiveor negative number.
Ia + Ib = Ic + Id
As a consequence of the Law of the conservation of charge, we haAs a consequence of the Law of the conservation of charge, we have: ve:
Similar to the water flow in a pipe.
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2. Loop rule2. Loop rule
1.1.
Assign symbolsAssign symbols
and and directionsdirections
of currents in the loopof currents in the loopIf the direction is chosen wrong, the current will come out withIf the direction is chosen wrong, the current will come out with the the correct magnitude, but a negative sign (itcorrect magnitude, but a negative sign (it’’s ok).s ok).
2.2.
ChooseChoose
a direction (a direction (cwcw
or or ccwccw) ) for going aroundfor going around
the loop. the loop. Record drops and rises of voltage according to this:Record drops and rises of voltage according to this:
If a If a resistorresistor is traversed is traversed in the directionin the direction of the current: of the current: V = V = --IRIRIf a If a resistorresistor is traversed is traversed in the direction oppositein the direction opposite to the current:to the current: V=+IRV=+IRIf If EMFEMF is traversed is traversed ““from from –– to +to + ””: : + + EIf If EMFEMF is traversed is traversed ““from from + to + to –– ””: : -- E
•The sum of the potential differences across all the elements around any closed loop must be zero.
As a consequence of the Law of the conservation of energy, we haAs a consequence of the Law of the conservation of energy, we have: ve:
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Simplest Loop Rule Example:Simplest Loop Rule Example:
Single loop, start at point ASingle loop, start at point ABattery traversed from Battery traversed from ––
to +to +So use + So use + E (a voltage gain)(a voltage gain)
Resistor traversed from + to Resistor traversed from + to ––So use So use –– IR IR (a voltage drop)(a voltage drop)
For the loop we have:For the loop we have:
E R
A
B C
D
+
-
+
-
I
IR−+= E0
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Loops can be chosen arbitrarily. For example, the circuit below contains a number of closed paths. Three have been selected for discussion.
+
+
+
+ +
+
+
+
+
+
+
-- -
-
-
-
--
-
-
-v1
v2
v4
v3
v12
v11 v9
v8
v6
v5
v7
v10
+
-
Path 1
Path 2
Path 3
Loop rule: more complex illustrationLoop rule: more complex illustration
Suppose that for each element, respective current flows from +
to -
signs.
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+
+
+
+ +
+
+
+
+
+
+
-- -
-
-
-
--
-
-
-v1
v2
v4
v3
v12
v11 v9
v8
v6
v5
v7
v10
+
-
“a”•
Blue path, starting at “a”
+ v7 - v10 + v9 - v8 = 0
•“b”
Red path, starting at “b”
-v2 + v5 + v6 + v8 - v9 + v11+ v12 - v1 = 0
Yellow path, starting at “b”
- v2 + v5 + v6 + v7 - v10 + v11+ v12 - v1 = 0
Using sum of the drops = 0
Loop rule: illustrationLoop rule: illustration
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Example: For the circuit below find I, V1 , V2 , V3 , V4 and the power supplied by the 10 volt source.
1. For convenience, we start at point “a” and sum voltage drops =0 in the direction of the current I.
-10 - V1 + 30 - V3 - V4 + 20 - V2 = 0 (1)
KirchhoffKirchhoff’’s Ruless Rules: Single: Single--loop circuitsloop circuits
2. We note that: V1 = 20I, V2 = 40I, V3 = 15I, V4 = 5I (2)
3. We substitute the above into Eq. 1 to obtain Eq. 3 below.
-10 - 20I + 30 - 15I - 5I + 20 - 40I = 0 (3)
Solving this equation gives, I = 0.5 A
+ +
+
_ _
_
V1
V4
V3 V2
30 V 10 V
15 Ω 40 Ω
5 Ω
20 Ω
20 V
I
"a"•
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Using this value of I in Eq. 2 gives:
V1 = 10 V
V2 = 20 V
V3 = 7.5 V
V4 = 2.5 V
P10(supplied) = -10I = - 5 W
(We use the minus sign in –10I because the current is entering the + terminal)In this case, power is being absorbed by the 10 volt supply.
KirchhoffKirchhoff’’s Ruless Rules: Single: Single--loop circuits (cont.)loop circuits (cont.)
+ +
+
-
+
+
-_ _
+
_
_
_
+V1
V4
V3 V2
30 V 10 V
15 Ω 40 Ω
5 Ω
20 Ω
20 V
I
"a"•
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18.5 RC circuits18.5 RC circuitsWhen switch is closed, current flows because When switch is closed, current flows because capacitor is capacitor is chargingcharging
As capacitor becomes charged, the current As capacitor becomes charged, the current slows because the voltage across the resistor is slows because the voltage across the resistor is E
--
VVcc
and and VVcc
gradually approaches gradually approaches E..Once capacitor is charged the current is zeroOnce capacitor is charged the current is zero
( )1 t RCq Q e−= −
RC is called the time constant
Cha
rge
acro
ss c
apac
itor Q
0.63 Q
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Discharging the capacitor in RC circuitDischarging the capacitor in RC circuit
If a capacitor is charged and the switch is closed, then If a capacitor is charged and the switch is closed, then current flows and the voltage on the capacitor gradually current flows and the voltage on the capacitor gradually decreases.decreases.
This leads to This leads to decreasing chargedecreasing charge t RCq Qe−=
Cha
rge
acro
ss c
apac
itor
Q
0.37Q
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Example : charging an unknown capacitorExample : charging an unknown capacitor
A series combination of a 12 kA series combination of a 12 kΩΩ
resistor and an unknown capacitor is resistor and an unknown capacitor is connected to a 12 V battery. One second after the circuit is connected to a 12 V battery. One second after the circuit is completed, the voltage across the capacitor is 10 V. Determine tcompleted, the voltage across the capacitor is 10 V. Determine the he capacitance of the capacitor.capacitance of the capacitor.
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A series combination of a 12 kA series combination of a 12 kΩΩ
resistor and an unknown capacitor is resistor and an unknown capacitor is connected to a 12 V battery. One second after the circuit is comconnected to a 12 V battery. One second after the circuit is completed, the pleted, the voltage across the capacitor is 10 V. Determine the capacitance voltage across the capacitor is 10 V. Determine the capacitance of the of the capacitor.capacitor.
Given:
R =12 kΩΕ
=
12 V V =10 Vt = 1 sec
Find:
C=?
Recall that the charge is building up according to
Thus the voltage across the capacitor changes as
This is also true for voltage at t = 1s after the switch is closed,
( )1 t RCq Q e
Q CV
−= −
=
( ) ( )1 1t RC t RCq QV e eC C
− −= = − = −E
1 t RCV e−= − ⇒E
( )
1 46.510log 1 12,000 log 112
t sC FV VR
V
μ= − = − =⎛ ⎞ ⎛ ⎞− Ω −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠E
log 1t VRC− ⎛ ⎞= −⎜ ⎟
⎝ ⎠E1t RC Ve− = − ⇒
E
I
RC
E
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Chapter 19: MagnetismChapter 19: Magnetism
Magnetic field pattern for a horse-shoe magnet
