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p212c27: 1
Direct Current Circuits
Two Basic Principles:Conservation of ChargeConservation of Energy
Resistance Networks
V IR
RV
I
ab eq
eqab
Req
Ia
b
p212c27: 2
Resistors in seriesConservation of Charge
I = I1 = I2 = I3
Conservation of Energy Vab = V1 + V2 + V3
RV
I
V V V
IV
I
V
I
V
I
V
I
V
I
V
I
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:V
V
R
Rab eq
1 1
p212c27: 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
R
I
V
I I I
V
I
V
I
V
I
V
I
V
I
V
I
V
R R R R
eq ab ab
ab ab ab
eq
aI
R1
V1=I1R1
R2
V2=I2R2
R3
V3=I3R3
b
Current Divider:I
I
R
Req1
1
p212c27: 4
R1=4
R2=3 R3=6
=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
p212c27: 5
Kirchoff’s Rules
Some circuits cannot be represented in terms of series and parallel combinations
1
2
R1
R2
R3
R1
R3
R2
R4
R5
Kirchoff’s rules are based upon conservation of energy and conservation of charge.
p212c27: 6
I1 I2
I3
Conservation of Charge: Junction ruleThe algebraic sum of the currents into any
junction is zero.
I = 0 = + I1 I2 I3
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
p212c27: 7
Conservation of Energy: Loop ruleThe algebraic sum of the potential
differences around any closed loop is zero.
I1
I2
I3
1
2
R1
R2
R3
p212c27: 8
Potential Differences in the direction of travel
+
direction of travel
V =
+
direction of travel
V = +
I
+
direction of travel
V =IR
I
+
direction of travel
V = + IR
p212c27: 9
I1
I2
I3
1
2 R2
R1
R3
I1 I2 I3 =01 + I1 R1 I2 R2 + 2 =0 I3 R3 I2 R2 + 2 =0 I3 R3 I1 R11 =0
p212c27: 10
I1
I2
I3
1 =12V
2 =5V R2 =1
R1 =2
R3 =3
I1 I2 I3 =012 I1 2I2 1 + 5 =0 I3 3 I2 1 + 5 =0
p212c27: 11
Standard linear form:
I1 1 I2 1 I3 = 0 2 I1 1 I2 0 I3 = 7 0 I1 1 I2 3 I3 = 5
I1 I2 I3 =012 I1 2I2 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
2
1
3
5
7
0
,
310
012
111
simult
p212c27: 12
R1 R2 R3
I 3A (-)1A 2AVP
P1
P2
I1
I2
I3
1 =12V
2 =5V R2 =1
R1 =2
R3 =3
p212c27: 13
Electrical InstrumentsGalvanometer:
Torque proportional to current IRestoring torque proportional to angle=> (Equilibrium) angle proportional to angle
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 Vfs = 20 mV = .020 V
G
p212c27: 14
Wheatstone Bridge
Circuit Balanced: Vab = 0
R2
b
R1
R3 R4
a G
4
2
3
1
44
22
33
11
4
2
3
1
4321
4231
0
R
R
R
R
RI
RI
RI
RI
V
V
V
V
VVVV
IIII
IG
p212c27: 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 I
I
Design a 50 mA ammeter from example galvanometerIfs = 1.0 mA, Rc = 20.0 Vfs = 20 mV = .020 V
I
p212c27: 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
fs fs m
I
I
V
V
V
Design a 10 V voltmeter from example galvanometerIfs = 1.0 mA, Rc = 20.0 Vfs = 20 mV = .020 V
p212c27: 17
Ohmmeter
Measures resistance across terminalsMeter supplies an EMF, resistance is determined
by response.
When R = 0 , we need Ic = Ifs = (Ifs Rs + Ifs Rc)=> Vm = Ifs Rs + Ifs Rc
Rs = (E / Ifs )- Rc
GRs
I
R
fs fs
s c
s c
II
R RR R R
R
p212c27: 18
p212c27: 19
p212c27: 20
Resistance Capacitance Circuits
Charging Capacitor
i
RC
switch
+q -q
vc = q/C vR = iR
Capacitor is initially uncharged.Switch is closed at t=0 Apply Loop rule:
-q/C-iR=0
p212c27: 21
tRCC
Ctq
uduu
CqudtRCCq
dq
dtRCCq
dq
CqRCdt
dqdtdq
RCq
Ri
iRCq
ttq
1))((ln
ln1
,1
)(
1)(
)(1
=-
0=--
0
)(
0
p212c27: 22
i
RC
switch
+q -q
vc = q/C vR = iR
//
/
/
/
)1
(
"ntTimeConsta"
)1(
)1()(
1))((ln
to
RCt
RCt
tfinal
RCt
eIeR
eRC
Cdtdq
i
RC
eQ
eCtq
tRCC
Ctq
0 0.2 0.4 0.6 0.8 1
t
q
0 0.2 0.4 0.6 0.8 1
t
i
p212c27: 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
p212c27: 24
tRCQ
tq
dtRCq
dq
dtRCq
dq
qRCdt
dqdt
dq
RC
qi
iRC
q
o
ttq
Qo
1)(ln
1
1
1
=
0=
0
)(
p212c27: 25
/
/
/
/
Constant" Time" same
)(
1)(ln
to
RCto
to
RCto
o
eI
eRC
Q
dt
dqi
RC
eQ
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