Two Basic Principles: Conservation of Charge Conservation...
Transcript of Two Basic Principles: Conservation of Charge Conservation...
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p212c26: 1
Ch26: Direct Current Circuits
Two Basic Principles:Conservation of ChargeConservation of Energy
Resistance Networks
V IR
R VI
ab eq
eqab
=
≡Req
Ia
b
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p212c26: 2
Resistors in seriesConservation of Charge
I = I1 = I2 = I3
Conservation of Energy Vab = V1 + V2 + V3
RVI
V V VI
VI
VI
VI
VI
VI
VI
R R R R
eqab
eq
≡ =+ +
= + + = + +
= + +
1 2 3
1 2 3 1
1
2
2
3
3
1 2 3
a
b
I
R1
V1=I1R1
R2
V2=I2R2
R3
V3=I3R3
Voltage Divider: VV
RRab eq
1 1=
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p212c26: 3
Resistors in parallelConservation of Charge
I = I1 + I2 + I3
Conservation of Energy Vab = V1 = V2 = V3
1
1 1 1 1
1 2 3
1 2 3 1
1
2
2
3
3
1 2 3
RI
VI I I
VI
VI
VI
VIV
IV
IV
R R R R
eq ab ab
ab ab ab
eq
≡ =+ +
= + + = + +
= + +
aI
R1
V1=I1R1
R2
V2=I2R2
R3
V3=I3R3
b
Current Divider: II
RR
eq1
1
=
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p212c26: 4
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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p212c26: 5
Kirchoff’s Rules
Some circuits cannot be represented in terms of series and parallel combinations
E1
E2
R1
R2
R3
R1
R3
R2
R4
R5E
Kirchoff’s rules are based upon conservation of energy and conservation of charge.
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I1 I2
I3
Conservation of Charge: Junction ruleThe algebraic sum of the currents into any
junction is zero.
Σ I = 0 = + I1 + I2 − I3
Σ I = 0 = − I1 − I2 + I3
R1
R2
R3
Alternatively: The sum of the currents into a junction is equal to the sum of the currents out of that junction.
I1 + I2 = I3
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p212c26: 7
Conservation of Energy: Loop ruleThe algebraic sum of the potential
differences around any closed loop is zero.
I1
I2
I3
E1
E2
R1
R2
R3
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p212c26: 8
Potential Differences in the direction of travel
E
+ −
direction of travel
V = − E
E
+−
direction of travel
V = + E
I
+ −
direction of travel
V =− IR
I
+−
direction of travel V = + IR
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p212c26: 9
I1
I2
I3
E1
E2 R2
R1
R3
+ I1 + I2 − I3 =0−E1 + I1 R1 − I2 R2 + E2 =0− I3 R3 − I2 R2 + E2 =0− I3 R3 − I1 R1 + E1 =0
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p212c26: 10
I1
I2
I3
E1 =12V
E2 =5V R2 =1Ω
R1 =2Ω
R3 =3Ω
+ I1 + I2 − I3 =0− 12 + I1 2 − I2 1 + 5 =0− I3 3 −I2 1 + 5 =0
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p212c26: 11
Standard linear form:+ 1 I1 + 1 I2 − 1 I3 = 0+ 2 I1 −1 I2 + 0 I3 = 7+ 0 I1 + 1 I2 + 3 I3 = 5
+ I1 + I2 − I3 =0− 12 + I1 2 − I2 1 + 5 =0− I3 3 −I2 1 + 5 =0
TI-89: [2nd] [MATH] [4](selects matrix) [5] (selects simult)
Entry line: simult([1,1,-1;2,-1,0;0,1,3],[0;7;5])
I1= 3
I2= −1
I3= 2
or (using [F2])
solve(i1+i2-i3=0 and -12+2*i1-i2+5=0 and -3*i3-i2+5=0,i1,i2,i3)
yields i1=3 and i2=-1 and i3 =2
−
−
−
21
3
570
,310012111
simult
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p212c26: 12
R1 R2 R3
I 3A (-)1A 2AVP
P1
P2
I1
I2
I3
E1 =12V
E2 =5V R2 =1Ω
R1 =2Ω
R3 =3Ω
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Electrical InstrumentsGalvanometer:
Torque proportional to current IRestoring torque proportional to angle=> (Equilibrium) angle proportional to angle
θ
Maximum Deflection =“Full Scale Deflection”Ifs = Current to produce Full Scale DeflectionRc = Resistance (of coil)Vfs = Ifs Rc
example: Ifs = 1.0 mA, Rc = 20.0 Ω Vfs = 20 mV = .020 V
G
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p212c26: 14
Wheatstone Bridge
Circuit Balanced: Vab = 0
R2
bR1
R3 R4
E a G
4
2
3
1
44
22
33
11
4
2
3
1
4321
4231
0
RR
RR
RIRI
RIRI
VV
VV
VVVVIIII
IG
=
⇒=⇒=
====
=
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p212c26: 15
AmmeterMeasures current through device
Measure currents larger than Ifs by bypassing the galvanometer with another resistor (shunt Resistor Rs).
Ia = ammeter full scale
When I = Ia , we need Ic = Ifs VG = Vsh => Ifs Rc = (Ia -Ifs)Rsh
G
Rsh
I
Ic
I-Ic
afs II=
θθ
Design a 50 mA ammeter from example galvanometerIfs = 1.0 mA, Rc = 20.0 Ω, Vfs = 20 mV = .020 V
I
θ
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p212c26: 16
VoltmeterMeasures potential difference across device
Measure V larger than Vfs by adding a resistor (Rs) in series to the galvanometer.
Vm = Voltmeter full scaleWhen V = Vm , we need Ic = Ifs
Ic = Is , Vm = Vs + VG => Vm = Ifs Rs + Ifs Rc
Rs = (Vm / Ifs )- Rc
GRs
I
V
θθ fs fs m
II
VV
= =
V
Design a 10 V voltmeter from example galvanometerIfs = 1.0 mA, Rc = 20.0 Ω, Vfs = 20 mV = .020 V
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p212c26: 17
Ohmmeter
Measures resistance across terminalsMeter supplies an EMF, resistance is determined
by response.
When R = 0 , we need Ic = Ifs E = (Ifs Rs + Ifs Rc)=> Vm = Ifs Rs + Ifs Rc
Rs = (E / Ifs )- Rc
GRs
I
R
θθ fs fs
s c
s c
II
R RR R R
= = ++ + R
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p212c26: 18
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p212c26: 19
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p212c26: 20
Resistance Capacitance Circuits
Charging Capacitor
i E
RC
switch
+q -q
vc = q/C vR = iR
Capacitor is initially uncharged.Switch is closed at t=0 Apply Loop rule:
E − q/C− iR=0
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p212c26: 21
tRCC
Ctq
uduu
CqudtRCCq
dq
dtRCCq
dq
CqRCdt
dqdtdq
RCq
Ri
iRCq
ttq
1))((ln
ln1,1)(
1)(
)(1
=-
0=--
0
)(
0
−=−
−
=−=−=−
−=−
−−=
=
∫∫∫
EE
EE
E
E
E
E
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p212c26: 22
i E
RC
switch
+q -q
vc = q/C vR = iR
τ
τ
τ
//
/
/
/
)1(
"ntTimeConsta")1(
)1()(
1))((ln
to
RCt
RCt
tfinal
RCt
eIeR
eRC
Cdtdqi
RCeQ
eCtq
tRCC
Ctq
−−
−
−
−
==
==
==
−=−=
−=−
−
E
E
EEE
0 0.2 0.4 0.6 0.8 1
t
q
0 0.2 0.4 0.6 0.8 1
t
i
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p212c26: 23
Resistance Capacitance Circuits
Discharging Capacitor
i
RC
switch
+q -q
vc = q/C vR = iR
Capacitor has an initial charge Qo.Switch is closed at t=0 Apply Loop rule:
+ q/C-iR=0
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p212c26: 24
tRCQ
tq
dtRCq
dq
dtRCq
dq
qRCdt
dqdtdq
RCqi
iRCq
o
ttq
Qo
1)(ln
1
1
1
=
0=
0
)(
−=
−=
−=
−=
−=
−+
∫∫
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p212c26: 25
τ−
−
τ−
−
=
=−=
==τ=
=
−=
/
/
/
/
Constant" Time" same
)(
1)(ln
to
RCto
to
RCto
o
eI
eRCQ
dtdqi
RCeQ
eQtq
tRCQ
tq
0 0.2 0.4 0.6 0.8 1
t
q
i
RC
switch
+q -q
vc = q/C vR = iR
0 0.2 0.4 0.6 0.8 1
t
i