Transcript V 1 = V 2 = V 3

Chapter 19
DC Circuits
© 2006, B.J. Lieb
Some figures electronically reproduced by permission of Pearson
Education, Inc., Upper Saddle River, New Jersey Giancoli, PHYSICS,6/E ©
2004.
Ch 19
1
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.
Ch 19
2
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:
V   r
ab
•The internal resistance is small but increases with age.
Ch 19
3
Circuit Symbols
Ch 19
4
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 + V 2 + V 3
•Since V1 = I R1 etc., we can say
V  V1  V2  V3  IR1  IR2  IR3
V  I ( R1  R2  R3 )
The equivalent equation is V=IReq and thus
Ch 19
Req  R1  R2  R3
5
Summary - Resistors in Series
The current I is the same throughout the circuit since
charge can’t accumulate anywhere.
V  V1  V2  V3
Req  R1  R2  R3
Ch 19
6
Resistors in Parallel - Derivation
This is called a parallel circuit
•Notice V1 = V2 = V3 = V
•Since charge can’t disappear, we can say
I  I1  I 2  I 3  V1 R1  V2 R2  V3 R3
•We can combine these equations with
V = IReq to give
1
1
1
1



Req
R1
R2
R3
Ch 19
7
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 = I 1 + I2 + I3
•Can be 2, 3 or more resistors in parallel.
Ch 19
1
1
1
1



Req
R1
R2
R3
8
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.
1
1
1
1



REQ R1 R2 R3
1
1
1



500  1000 2000
REQ  286 
V  I REQ
V
3.00 V
I

 10.5 m A
REQ
286 
P  I V  (10.5 mA) (3.00 V )
Ch 19
P  31.5 m W
9
Example 19-1B Calculate the current and the power dissipated in
each resistor and the total power dissipated in the circuit.
V IR
I1 
3.00 V
V

500 
R1
 6.0 m A
P1  I1 V1  (6.0 mA) (3.00 V )  18.0 m W
V
3.00 V
I2 

 3.0 m A
R2
1000
P2  I 2 V2  (3.0 mA) (3.00 V )  9.0 m W
3.00 V
V

 1.5 m A
I3 
2000 
R3
P3  I 3 V3  (1.5 m A) (3.00 V )  4.5 m W
I  I1  I 2  I3  10.5 m A
P  P1  P2  P3  31.5 m W
Ch 19
10
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.
R2 and R3 are in series
R23  R2  R3  1000  1000
R23  2000
R23 and R4 are in parallel
1
1
1
1
1




2000 2000
R324 R23 R4
R234 1000
R1 and R234 are in series
REQ  R1  R234  500   1000 1500
V IR
I
Ch 19
V
3.0 V

 2.0 m A
REQ 1500
11
Ammeter
•To measure current ammeter must be in circuit.
•Must have small internal resistance or it will
reduce current and give a faulty measurement.
Ch 19
12
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.
Ch 19
13
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
R1
R2
I2
2
R3
b
a
I1
1
Ch 19
Point a:
I1 + I2 = I3
Point b:
I 3 = I 1+ I2
14
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.
R1
a
R3
I3
R2
I2
2
1
b
I1
All loops clockwise:
Upper Loop: +2 – I2 R3 – I3 R1 – I3 R2 = 0
Lower Loop: +1 + I2 R3 – 2 = 0
Ch 19
Large Loop: +1 – I3 R1 – I3 R2 = 0
15
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
Ch 19
16
solution.
LOWER JUNCTION
I 2  I 3  I1
I 2  I3  2 A
I 2  2 A  I3
RIGHT LOOP  CLOCKWISE
 8.0 V  (2   4 ) I 2  (12 ) I 3  0
 8.0 V  (6 ) (2 A  I 3 )  (12 ) I 3  0
 8.0 V  12V  (6   12 ) I 3  0
I3 
4.0 V
 0.22 A
18 
I 2  2 A  I3
Ch 19
 2.0 A  I 3
 1.78 A
17
Continued
LEFT LOOP  CLOCKWISE
 E1  (4 )(2 A)  (12 ) (0.22 A)  0
E1  10.6 V
Ch 19
18
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:
Ch 19
C  C1  C2  C3
19
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:
Ch 19
1
1
1
1
 

C
C1 C2 C3
20
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.
Ch 19
21
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:
Q
 Q  VC
t
•It is necessary to use calculus to find:
Ch 19
V  V0 e
t / RC
22
Exponential Decay
V  V0 et / RC
•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 23
the start of the period.
Ch 19
Example 4
  RC  (2.0106 ) (10106 F )
  20 s
 1  (0.693) RC  (0.693) (20 s)
2
 1  13.9 s
2
V  V0 e
t
RC
t
V
 e RC
V0
Ch 19
V
ln
 V0
 t
 
RC

24
Example 4
Continued
V
t   RC ln
 V0
t1
2



1
 (20 s) ln 
2
t 1  13.9 s
2
t1
Ch 19
4
1
 (20 s) ln 
4
t 1  27.7 s
4
25
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
Ch 19
•Your body can act as a capacitor in parallel with the
resistance and this gives greater current for ac.
26
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
Ch 19
27
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
Ch 19
28