ch 191 chapter 19 dc circuits © 2006, b.j. lieb some figures electronically reproduced by...
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Ch 19 1
Chapter 19
DC Circuits
© 2006, B.J. LiebSome figures electronically reproduced by permission of Pearson
Education, Inc., Upper Saddle River, New Jersey Giancoli, PHYSICS,6/E © 2004.
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Ch 19 2
EMF
• Devices that supply energy to an electric circuit are referred to as a source of electromotive force. Since this name is misleading, we just refer to them as source of emf (symbolized by and a slightly different symbol in the book.)
• Sources of emf such as batteries often have resistance which is referred to as internal resistance.
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Ch 19 3
Terminal Voltage
r a b
Vab
•We can treat a battery as a source of in series with an internal resistor r.•When there is no current then the terminal voltage is Vab= •But with current I we have:
r Vab
•The internal resistance is small but increases with age.
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Ch 19 4
Circuit Symbols
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Ch 19 5
Resistors in Series - Derivation
•We want to find the single resistance Req that has the same effect as the three resistors R1, R2, and R3.•Note that the current I is the same throughout the circuit since charge can’t accumulate anywhere.•V is the voltage across the battery and also V = V1 + V2 + V3
•Since V1 = I R1 etc., we can say
321321 IRIRIRVVVV )( 321 RRRIV
The equivalent equation is V=IReq and thus
321 RRRReq
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Ch 19 6
Summary - Resistors in Series
The current I is the same throughout the circuit since charge can’t accumulate anywhere.
321 RRRReq
321 VVVV
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Ch 19 7
Resistors in Parallel - Derivation
This is called a parallel circuit•Notice V1 = V2 = V3 = V•Since charge can’t disappear, we can say
321
1111
RRRReq
332211321 RVRVRVIIII
•We can combine these equations with V = IReq to give
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Ch 19 8
Summary - Resistors in Parallel
•The electric potential (voltage) is the same across each resistor
V1 = V2 = V3
•The current through the battery splits several ways I = I1 + I2 + I3
•Can be 2, 3 or more resistors in parallel.
321
1111
RRRReq
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Ch 19 9
321
1111
RRRREQ
2000
1
1000
1
500
1
286EQR
EQRIV
EQR
VI
286
00.3 VmA5.10
VIP )00.3()5.10( VmA
mWP 5.31
Example 19-1A. A 3.0 V battery is connected to three resistors as shown. Calculate the resistance of the equivalent circuit and the power dissipated in the equivalent circuit. R1 = 500 Ω, R2 = 1000 Ω and R3 = 2000 Ω.
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Ch 19 10
Example 19-1B Calculate the current and the power dissipated in each resistor and the total power dissipated in the circuit.
RIV
11 R
VI
500
00.3 V mA0.6
111 VIP )00.3()0.6( VmA mW0.18
22 R
VI
1000
00.3 VmA0.3
222 VIP )00.3()0.3( VmA mW0.9
33 R
VI
2000
00.3 V mA5.1
333 VIP )00.3()5.1( VmA mW5.4
321 IIII mA5.10
321 PPPP mW5.31
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Ch 19 11
seriesinareRandR 32
3223 RRR 10001000
200023R
parallelinareRandR 423
423324
111
RRR
2000
1
2000
1
1000234RseriesinareRandR 2341
2341 RRREQ 1000500 1500
RIV
EQR
VI
1500
0.3 VmA0.2
Example 19-2. A 3.0 V battery is connected to 4 resistors as shown. Calculate the resistance of the equivalent circuit and the current in the equivalent circuit. R1 =
500 Ω, R2 = 1000 Ω, R3 = 1000 Ω, and R4 = 2000 Ω.
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Ch 19 12
Ammeter
•To measure current ammeter must be in circuit.
•Must have small internal resistance or it will reduce current and give a faulty measurement.
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Ch 19 13
Voltmeters
•To measure voltage difference, it must be connected to two different parts of circuit.
• Must have high internal resistance or it will draw too much current which reduces voltage difference and gives a faulty measurement.
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Ch 19 14
Kirchhoff’s Junction Rule•Kirchhoff’s Rules are necessary for complicated circuits.
•Junction rule is based on conservation of charge.
•Junction Rule: at any junction, the sum of all currents entering the junction must equal the sum of all currents leaving the junction.
I3
I22
I11
a b
R1 R2
R3
Point a: I1 + I2 = I3
Point b: I3 = I1+ I2
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Ch 19 15
Kirchhoff’s Loop Rule•Loop rule is based on conservation of energy.
•Loop Rule: the sum of the changes in potential around any closed path of a circuit must be zero.
I3
I22
I11
a b
R1 R2
R3
All loops clockwise:
Upper Loop: +2 – I2 R3 – I3 R1 – I3 R2 = 0
Lower Loop: +1 + I2 R3 – 2 = 0
Large Loop: +1 – I3 R1 – I3 R2 = 0
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Ch 19 16
Using Kirchhoff’s Rules
• Current:• Current is the same between junctions.• Assign direction to current arbitrarily.• If result is a negative current, it means that the current actually flows in the
opposite direction. Don’t change direction, just give negative answer.• Branches with a capacitor have zero current.
• Signs for Loop Rule• Go around loop clockwise or counterclockwise.• IR drop across resistor is negative if you are moving in direction of the current.• Voltage drop across battery or other emf is positive if you move from minus to
plus.
• Simultaneous Equations• You will need one equation for each unknown.• It pays to generate “extra” equations because they may lead to a simpler solution.
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Ch 19 17
JUNCTIONLOWER132 III AII2 23
32 2 IAI CLOCKWISELOOPRIGHT
0)12()42(0.8 32 IIV
0)12()2()6(0.8 33 IIAV
0)126(120.8 3 IVV
AV
I 22.018
0.43
32 2 IAI 30.2 IA A78.1
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Ch 19 18
CLOCKWISELOOPLEFT
0)22.0()12()2)(4(1 AAE
VE 6.101
Continued
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Ch 19 19
Capacitors in Parallel
•V is the same for each capacitor•The total charge that leaves the battery is Q = Q1 + Q2 + Q3 = C1V + C2V + C3V •Combine this with Q = Ceq V to give:
321 CCCC
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Ch 19 20
Capacitors in Series
•The charge on each capacitor must be the same.•Thus Q = C1 V1 = C2 V2 = C3 V3
•Combine this with V = V1 + V2 + V3 to give:
321
1111
CCCC
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Ch 19 21
Charging a Capacitor (Qualitative)
•When switch is closed, current flows because capacitor is charging
•As capacitor becomes charged, the current slows because the voltage across the resistor is - Vc and Vc
gradually approaches .
•Once capacitor is charged the current is zero.
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Ch 19 22
RC Decay
•If a capacitor is charged and the switch is closed, then current flows and the voltage on the capacitor gradually decreases.• Since I VC we can say that:
CVQt
Q
•It is necessary to use calculus to find:
RCteVV /0
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Ch 19 23
Exponential Decay
RCteVV /0
•The value = RC is called the time constant of the decay. If R is in and C is in F, then has units of seconds.
•During each time constant, the voltage falls to 0.37 of its value at the start of the period.
•We can also define the half-life (1/2) by 1/2= 0.693 RC.
•During each half-life, the voltage falls to ½ of its value at the start of the period.
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Ch 19 24
Example 4
)1010()100.2( 66 Fs20
)20()693.0()693.0(2
1 sRC
s9.132
1
RCt
eVV
0
RCt
eV
V
0
RCt
V
V
0
ln
RC
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Ch 19 25
Example 4
0
lnV
VRCt
2
1ln)20(
21 st
st 9.132
1
4
1ln)20(
41 st
st 7.274
1
Continued
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Ch 19 26
Electric Hazards
•A current greater than 70 mA through the upper torso can be lethal.
•Wet skin: I = 120 V / 1000 = 120 mA
•Dry skin: I = 120 V / 10000 = 12 mA
•Your body can act as a capacitor in parallel with the resistance and this gives greater current for ac.
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Ch 19 27
Electric Hazards
•The key to safety is don’t let your body become part of the circuit.
•Standing in water can give path to ground which will complete circuit.
•Bathrooms can be dangerous
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Ch 19 28
Grounded Enclosures
•Metal cabinet grounded by 3-prong plug protects if there is loose wire inside because it causes short that trips circuit breaker.
•“Ground fault detector” should turn off current in time to protect you
•Circuit Breakers are to slow for personal safety