Transcript Physics 261 - Purdue University

Lecture 12-1
Resistors in Parallel



1
1

Req
i Ri
1
1
1
1


or
 

R3 Req
Req R1 R2 R3
i
R1
R2
ε
i
Req  ( R1  R2 )
Req  R1  R2  R3  ...
Lecture 12-2
DOCCAM 2
Lecture 12-3
Another example (with parallel R combos)
I3
I2
• Sketch the diagram
• Simplify using equivalent resistors
I1
• Label currents with directions
• Use Junction Rule in labeling
• Choose independent loops
• Use Loop Rule
Replace by equivalent
R=2Ω first.
• Solve simultaneous linear equations
I3 = I 1 + I2
18  12( I1  I 2 )  6I1  0
3I1  2 I 2  3
3I 2  21  2 I 2  6I1  0
6I1  5I 2  21
 I 2  3( A), I1  1( A)
Lecture 12-4
Loop current example (with parallel R combos)
I1
I2
I1 -I2
• Sketch the diagram
• Simplify using equivalent resistors
• Label loop currents with directions
• Use Loop currents I1 and I2
• Choose interior clockwise loops
. Set up cononical equations in
Replace by equivalent
R=2Ω first.
. I1 I2
Ε format
I1 (12 +6) +I2 (-6) = +18 (Emf)
Left loop
I1 (-6) + I2 (6+3+2) =+21 (Emf)
Right loop
Note Symmetry of Equations
Lecture 12-5
Finding Potential and Power in a Circuit
Va  0  12  I 1 V 
But what is I? Must
solve for I first!
12  4
I
 0.5 ( A)  0
1 5  5 1 4
Va  12  0.5 1  11.5(V )
Vb  Va  I  5  9(V )
P12V
The rest?
Just means
0 V here
supplied by
 12  0.5  6(W ) 12V battery
PR  0.52 16  4(W ) dissipated by
into 4V battery
(charging)
resistors
P4V  4  0.5  2(W )
Lecture 12-6
Ammeter and Voltmeter
Ammeter: an instrument used to
measure currents
• It must be connected in series.
• The internal resistance of an
ammeter must be kept as small as
possible.
Voltmeter: an instrument used to
measure potential differences
• It must be connected in parallel.
• The internal resistance of a
voltmeter must be made as large
as possible.
Lecture 12-7
Galvanometer Inside Ammeter and Voltmeter
Galvanometer: a device that detects small currents
and indicates its magnitude. Its own resistance Rg
is small for not disturbing what is being measured.
galvanometer
Ammeter: an instrument
used to measure currents
shunt resistor
Voltmeter: an instrument
used to measure potential
differences
galvanometer
Lecture 12-8
Capacitor in RC Circuits
I
I
charging
discharging
ε
 Switch closed at t=0. C initially
uncharged, thus V0 = 0 across C
and I0 = ε/R initially.
 Switch closed at t=0. C
initially charged, thus V0 = Q0/C
across C and I0 = V0/R initially.
 After a long time, C is fully
charged and VC = ε across C
and I0 = 0.
 After a long time, C is fully
discharged and VC = 0 across C
and I∞ = 0.
time constant
Q  Q0 e
t / RC
Lecture 12-9
Discharging a Capacitor in RC Circuits
1.
Switch closed at t=0. Initially C
is fully charged with Q0
2.
Loop Rule:
3.
Convert to a differential equation
dQ
I 
dt
4.
Q
 IR  0
C
Q
dQ
R
0
C
dt
Solve it!
Q  Q0 e  t / RC
dQ Q0  t / RC 
I 

e
dt RC
I
Lecture 12-10
Q(t) during Discharging
Q(t )  Q0e
 dQ 
Q0 /  
 
 dt t 0 
t /
Q(t   )  Q(t )  e1  0.37Q(t )
time constant
0.37Q0
    RC   
C C
 s
V A
Lecture 12-11
Current I(t) during Discharging
Q(t )  Q0e
t /
dQ Q0  t / 
I 

e
 I 0e  t / 
dt

I (t   )  I (t )  e1  0.37  I (t )
I 0  Q0 /   (Q0 / C) / R  V0 / R
Lecture 12-12
Warm-up quiz 1
What’s the time constant in the following circuits.
All resistors have the same resistance R, all
capacitors have the same capacitance C.
A.
B.
C.
D.
E.
3/5RC
2RC
RC
4/9RC
6/4RC
Lecture 12-13
Charging a Capacitor in RC Circuits
1.
Switch closed at t=0
C initially uncharged, thus zero
voltage across C.
I0   / R
2.
Loop Rule:
  IR 
Q
0
C
3. Convert to a differential equation
dQ
I 
dt
4.
dQ Q
 R
 0
dt C
Solve it!
Q  C 1  e  t / RC  , I 
dQ   t / RC
 e
dt R
(τ=RC is the time constant again)
Lecture 12-14
Charge Q(t) during Charging
time constant
Q(t )  C (1  et / )  Q f (1  et / )
Q( )  Q f (1  e )  0.63  Q f
1
 dQ
Qf / 
 dt

 
t 0 
Lecture 12-15
Current I(t) during Charging
dQ Q f  t /
I

e
 I 0e  t / 
dt

I (t   )  I (t )  e1  0.37  I (t )
I0  Q f /   (Q f / C) / R   / R
Lecture 12-16
Broken circuit = RC
110V
q
0 A
q
q
q
q0 A q
C   C very
 small. i.e. RC very
C 

Vd Ed
d
V Ed  q 
q  discharged
small.Instantly
  Ad
  Ad
 0 
 0 
Lecture 12-17
Behavior of Capacitors
• Charging
– Initially, the capacitor behaves like a wire.
– After a long time, the capacitor behaves like an open switch.
• Discharging
– Initially, the capacitor behaves like a battery.
– After a long time, the capacitor behaves like an open switch
Lecture 12-18
DOCCAM 2
Lecture 12-19
Energy Conservation in Discharging a Capacitor
Discharging:
Energy
lost by C
U
Q0V0 1
 CV0 2
2
2
Power dissipated by R
PR  I (t )2  R  I 02e2t /  R


WR   PR (t ) dt  I 0 R  e 2 t /  dt
2
0
0

U  WR
2
2
V
R
C 1


 I 02 R    0 
 CV0 2
2 R 2
2
Lecture 12-20
Energy Conservation in Charging a Capacitor
Charging:
Work done
by battery
Energy
stored in C
The rest?
W  Q f   C 2
U
Qf
1
 C 2
2
2
PR  I (t )2  R  I 02e2t /  R
Power dissipated by R


WR   PR (t ) dt  I 0 R  e 2 t /  dt
2
0
0

 Independent of R
What if R=∞ ?
What if R=0 ?
2

R
C 1 2


 I 02 R    
 C
2 R 2
2
2
Lecture 12-21
PHYS 241 - 10:30 Quiz 2
All the capacitors below are identical and so are all
the resistors. Which circuit have the shortest time
constant?
B
A
C
E
D
Lecture 12-22
PHYS241 – 11:30 Quiz 2
All the capacitors below are identical and so are all
the resistors. Which circuit have the longest time
constant?
A
C
B
D
E
Lecture 12-23
PHYS241 - Quiz 12c
All the capacitors below are identical and so are all
the resistors. Which circuit have the shortest time
constant?
A
C
B
D
E
Lecture 12-24
Reading quiz 1
Consider the charging of a capacitor C by battery with emf E through
an external resistor R. The final charge of the capacitor is QF. Which of the
following statements is incorrect?
A| The energy delivered by the battery is E QF .
B| The energy stored in the capacitor 1/2 QF 2 /C.
C| The energy dissipated in the resistor is EQF /2.
D| The fraction of the energy delivered by the battery depends
the value of the resistor R.