Download Lecture08: Multi-Loop and RC Circuits

Physics 121 - Electricity and Magnetism
Lecture 08 - Multi-Loop and RC Circuits
Y&F Chapter 26 Sect. 2 - 5
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Definitions of Circuit Terms
Kirchhoff Rules
Problem solving using Kirchhoff’s Rules
Multi-Loop Circuit Examples
RC Circuits
– Charging a Capacitor
– Discharging a Capacitor
Discharging Solution of the RC Circuit Differential
Equation
The Time Constant
Examples
Charging Solution of the RC Circuit Differential Equation
Features of the Solution
Examples
Copyright R. Janow Fall 2015
Definitions of Circuit Terms
Basic circuit elements have two terminals.
Generic symbol:
ECE
2
1
Node: A point where 2 or more circuit elements are joined.
i1
Essential Node (Junction): A node where at least 3 circuit elements are joined.
i2
Path: A route (or trace) through adjacent basic elements with no element included
more than once. May pass through essential and/or non-essential nodes.
Branch: A path that connects two nodes (essential or not). Includes 1 or more elements
Essential Branch:
• A path that connects two essential nodes without passing through another essential node.
• A circuit section between two essential nodes with 1 or more elements in series.
• Non-essential (trivial) nodes connect circuit elements in “series”.
• There is exactly one current per essential node.
Loop: A closed path whose last node is the same as the starting node.
Mesh: A loop that does not enclose any other loops.
Planar Circuit: A circuit whose diagram can be drawn on a plane with no crossing branches.
Copyright R. Janow Fall 2015
i
Example: Circuit Terms
b
a
+
v1
+
f
i1
R1
-
c
v2
Source: Nilsson & Reidel
Electric Circuits
i2
d
R2
R4
i3
i5
R3
i4
R5
i7
e
R7
i6
R6
g
Ideal, Independent Current Sources: i6
Ideal, Independent Voltage Sources (ideal EMFs) : v1, v2
Nodes: a, b, c, d, e, f, g
Essential Nodes (Junctions): b, c, e, g
Branches: v1, v2, R1, R2, R3, R4, R5, R6, R7, i6
Paths: Many, traversing 1, 2, …..n branches
Essential Branches (7): v1-R1, R2-R3, v2-R4, R5, R6, R7, i6
Currents: 1 per essential branch, total of 7: i1 . . . i7
Meshes (subset of loops): v1-R1-R5-R3-R2,
v2-R2-R3-R6-R4, R5-R7-R6, R7-i6
Loops: Meshes + V1-R1-R5-R6-R4-V2, V1-R1-R7-R4-V2, v1-R1-i
R6Janow
-R5-i6 Fall 2015
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Copyright
Solving Circuit Problems using Kirchhoff’s Rules
i
CIRCUITS CONSIST OF:
JUNCTIONS (ESSENTIAL NODES)
... and …
ESSENTIAL BRANCHES (elements
connected in series, one current/branch)
1
i
i2
i
The current through all series elements in an essential branch is the same;
i.e., the number of currents = the number of essential branches
Often, all element capacities are specified but the currents are unknowns.
Analysis strategy:
1) Use Kirchhoff Rules to generate N independent equations in N unknowns
• Junction Rule or Current rule or Node Law (charge
iin 
iout
conservation): At any junction the algebraic sum of the
currents at any (essential) node equals zero
• Loop Rule (energy conservation): The algebraic sum
V  0
of all potential changes is zero for every closed path
around a circuit. Corollary: voltage difference is
the same for all paths connecting a pair of nodes.
2) Solve the resulting set of simultaneous equations (you need a strategy).



• Linear, algebraic for resistances and EMFs only Copyright R. Janow
Fall 2015
Applying Kirchhoff’s Rules:
We use “active” sign convention: voltage drops considered negative
“passive” sign convention: voltage drops considered positive
For brevity: “branch” means “essential branch
“junction” means “essential node”
Procedure for Generating Circuit Equations
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Name the currents or other unknowns
In each branch, arbitrarily assign direction to current
- a negative result  opposite flow
Apply Junction Rule.
Apply Loop Rule to write down terms in equations:
- choose direction for traversing a closed loop. possibly traverse
branches with or against assumed current directions.
When crossing resistances:
- voltage drop (V = - iR) negative when following assumed current
- positive voltage change V = +iR for crossing opposite to assumed current
- when crossing EMFs from – to +, V = +E. Otherwise V= -E
Keep generating equations until you have N independent ones
After solving, calculate power or other quantities as needed
Dot product i.E determines whether EMFs supply or dissipate power
For Later: when crossing C with current write –VC= -Q/C.
Copyright R. Janow Fall 2015
when crossing inductance write VL=-Ldi/dt
Example: Equivalent resistance for resistors in series
Junction Rule: The current through all of the resistances in series (a single branch)
is identical. No information from Junction/Current/Node Rule
i  i1  i2  i3
Loop Rule: The sum of the potential differences around a
closed loop equals zero. Only one loop path exists:
  iR 1  iR 2  iR 3  0
i

R1  R 2  R 3
The equivalent circuit replaces the series resistors with a
single equivalent resistance: same E, same i as above
  iR eq  0
i

R eq
CONCLUSION: The equivalent resistance for a series
combination is the sum of the individual resistances and is
always greater than any one of them.
R eq  R 1  R 2  R 3
R eq 
n
 Ri
i 1
Copyright R. Janow Fall 2015
inverse of series capacitance rule
Example: Equivalent resistance for resistors in parallel
Loop Rule: The potential differences across each of the 4 parallel branches
are the same. Four unknown currents. Apply loop rule to 3 paths.
E  i1R1  0
i1 
E  i2R 2  0
E
E
E
, i2 
, i3 
R1
R2
R3
E  i3R 3  0
i not in
these
equations
Junction Rule: The sum of the currents flowing in equals the
sum of the currents flowing out. Combine equations for all the
upper junctions at “a” (same at “b”).
 1
1
1 

i  i1  i2  i3  E 


R
R
R
2
3
 1
The equivalent circuit replaces the series resistors with a
single equivalent resistance: same E, same i as above.
i
  iR eq  0

R eq
CONCLUSION: The reciprocal of the equivalent resistance for a
parallel combination is the sum of the individual reciprocal
resistances and is always smaller than any one of them.
1
1
1
1



R eq
R1 R 2 R 3
1

R eq
n
1
R
i 1 i
inverse of parallel capacitance rule
R eq 
R 1R 2
R1  R 2
Copyright R. Janow Fall 2015
EXAMPLE: MULTIPLE BATTERIES
SINGLE LOOP
i
+
-
R1= 10 W
E1 = 8 V
+
-
i
E2 = 3 V
+
R2= 15 W
A battery (EMF) absorbs power (charges up) when I is opposite to E
Pemf
E is opposite to Vdrop
 
  Ei  E  i
-3.0x0.2
Copyright R. Janow Fall 2015
Example: Multi-loop circuit with 2 EMFs
+
R1
i1
E2
A
-
E1
-
D
R3
i3
+
Given all resistances and EMFs in circuit:
• Find currents (i1, i2, i3), then potential
drops and power dissipated by resistors
• 3 unknowns (currents)
imply 3 independent equations needed
E
R2
i2
Apply Procedure:
C
B
• Identify essential branches (3) & junctions (2).
• Name all currents (3) and other variables.
• Same current flows through all elements in any series branch.
• Assume arbitrary current directions; negative result means opposite direction.
• At junctions, write Current Rule (Junction Rule) equations.
iin

iout
• Same equation at junctions A and B (not independent).
• Junction Rule yields only 1 of 3 equations needed
• Are points C, D, E, F junctions? (not essential nodes)
i2  i1  i3
(1)
Copyright R. Janow Fall 2015
F
Procedure, continued:
 V
E1
- +
equations that include all the unknowns (3).
• Traversal direction is arbitrary.
• When following the assumed current direction
voltage change = - iR. When going against assumed
current voltage steps up by +IR
• EMF’s count positive when traversed from – to + side
• EMF’s count negative when traversed from + to - sides
D
R1
i1
A
E2
R3
i3
C
+ -
• Apply Loop Rule as often as needed to find
B
 0
Loop equations for the example circuit:
ADCBA - CCW
ADCBFEA - CCW
ABFEA - CCW
E1 i1 R1  i3R 3  0
E1 i1 R1  i2R2 E2  0
i3 R 3  i2R 2 E2  0
Solution: (after a lot of algebra)
Define:
   R1R 2  R 2R 3  R1R 3
• Only 2 of these
three are independent
• Now have 3 equations
in 3 unknowns
i1 
E1R 2  E1R 3  E2R 3
i2 

E1R 3  E2R 3  E2R1

 E2R1  E1R 2
i3Copyright

R. Janow Fall 2015

E
R2
i2
F
Example: find currents, voltages, power
6 BRANCHES  6 CURRENTS.
•JUNCTION RULE:
Junctions C & E are the
same point, as are D & F
-> 4 currents left.
Remaining 2 junction
equations are dependent
-> 1 junction equation
i  i1  i2  i3
LOOP RULE:
ABCDA - CW
CEFDC - CW
E  i1 R1  0  E  i1 R1  i1  E/R1  12 / 3  4.0 A.
 i2R2  i1R1  0  i2  i1R1/R2  4x3 / 8  1.5 A.
 i3R 3  i2R 2  0  i3  i2R 2/R 3  1.5x8 / 6  2.0 A.
R eq  1.6 W
CHECK: i  i1  i2  i3  4.0  1.5  2.0  7.5 A.
E should  VR1  i1R1  4.0x3.0  12.0 Volts
EGHFE - CW
POWER:
PR1  i12R1  48.0 Watts
PR2  i22R 2  18.0 Watts
PR3  i23R 3  24.0 Watts

PE  E  i  90.0 Watts
 PR1  PR2  PR3
Copyright R. Janow Fall 2015
Multiple EMF Example: find currents, voltages, power
R2 = 4 W
E2 = 6 V
R1 = 2 W
E1 = 3 V
MULTIPLE
EMF
CIRCUIT
JUNCTION RULE at A & B:
USE THE
SAME RULES
i3  i1  i2
LOOP ACDBA:
 i1R1  E1  i1R1  E2  i2R 2  0
i2  i1  3/4
LOOP BFEAB:
 i3R1  E2  i3R1  i2R 2  E2  0
USE JUNCTION EQUATION:
i3   i2  i1  i2
EVALUATE
i  1/2, i2  - 1/4,
NUMERICALLY: 1
i3   i2
i1   2 i2
i3   1/4
For power use:
V  iiRi
Pi  ii2Ri
Copyright R. Janow Fall 2015
RC Circuits: Time dependance
27.8
i a
Can constant current flow through
R
+
+
E
-
b
C
-
i
a capacitor indefinitely?
Vc • Given Capacitance + Resistance + EMF
• Use Loop Rule + Junction Rule
• New term Vc= Q/C in Loop Rule
• Find Q, i, Vc, U for capacitor
as functions of time
First charge up C (switch to “a”) then discharge (switch to “b”)
Charging: “Step Response”
Switch to “a” then watch.
Loop equation:
E  iR  Vc  0
Discharging: Switch to “b”.
no EMF, Loop equation:
 iR  Vc  0
• Assume clockwise current i through R
• Vcap E as t  infinity, Q  Qinf = CE
• Expect zero current as t  infinity
• Expect largest current at t = 0,
• Energy is stored in C, some is dissipated in R
• Energy stored in C is now dissipated in R
• Arbitrarily assume current is still CW
• Vcap= E at t =0, but it must die away
• Q0= full charge = CE
• Result: i through R is actually CCW
Copyright R. Janow Fall 2015
RC Circuit: solution for discharging
Loop Equation is :
Substitute :
iR  Vc  0
dQ
i(t ) 
dt
Circuit Equation:
dQ
Q(t )

dt
RC
Q( t )
V c (t ) 
C
First order differential equation, form is Q’ = -kQ  Exponential solution
Charge decays
exponentially:
• t/RC is
dimensionless
Q(t )  Q 0e  t / RC
RC = t = the TIME CONSTANT
Q falls to 1/e of original value
Voltage across C
also decays
exponentially:
Current also
decays
exponentially:
E  Q0 / C
Vc (t)  Q(t) / C
dQ
i(t ) 
 i0e  t / RC
dt
Q0
e 1  0.37
Q
t
2t
3t
t
Vc (t)  E e  t / RC
E
Q0
i0 

R RC
Copyright R. Janow Fall 2015
Solving for discharging phase by direct integration
dQ
Q( t )

dt
RC
dQ
dt

Q
RC
RC is
constant
Initial conditions (“boundary conditions”)
At t  0 : Q(t)  Q0  CE
dQ'
1 t


dt'
Q0 Q'

0
RC
exponentiate both sides of above right
ln (
e
Q
)
Q0
Q
t
ln (
) 
Q0
RC
Q
Q

e
Q0

t
RC
eln (x)  x
Q( t )  Q 0 e
 t / RC
exponential
decay
RC = time constant = time for Q to fall to 1/e of its initial value
RC  t
e 1
1
1
 
 .37
e 2.71828
Time
t
2t 3t
Value e-1 e-2 e-3
% left 36.8 13.5 5.0
4t
5t
e-4
e-5
1.8 0.67
After 3-5 time constants the action is over
Copyright R. Janow Fall 2015
Units for RC
8-1: We defined t = RC, which of the choices best captures the
physical units for the decay constant t ?
[t] = [RC] =[(V/i)(Q/V)]=[Q/Q/t]=[t]
A.
B.
C.
D.
E.
WF
C/A
WC/V
VF/A
s
(ohmfarad)
(coulomb per ampere)
(ohmcoulomb per volt)
(voltfarad per ampere)
(second)
Copyright R. Janow Fall 2015
Examples: discharging capacitor C through resistor R
a) When has the charge fallen to half of it’s initial value Q0?
1
 e t / t
2
1
2
set: Q(t )  Q 0  Q 0e  t / t
take log: ln( 12 )   t / t
ln(1)  0
 ln( 2)  - t/t
(solve for t - depends only on
t)
ln(a/b)  ln(a) - ln(b)
ln(2)  0.69
 t  0.69 t
b) When has the stored energy fallen to half of its original value?
Q2
recall: U(t ) 
2C
and
Q(t )  Q 0e  t / RC
at any time t: U(t)  U0 e 2t / RC
evaluate for: U(t) 
U0
 U0 e 2 t / t
2
take log: ln( 1 )   2t / t
2
where
Q20
U0  U(t  0) 
2C
 t  0.69 t /2  0.35 t
c) How does the power delivered by C vary with time?
power: P  dU  U0 d [ e  2t / t ]  U0 [  2 ]e  2t / t   2 Q 0 Q 0 e  2 t / t
dt
dt
t
2 C RC
Q0
C supplies rather than absorbs power
recall: Q 0  E
 i0
Drop minus sign
C
RC
power supplied
by C:
P  i0e  t / t  Ee  t / t  i(t)  V(t)
 P0e
2 t / t
P(t )  i(t) V(t)
Copyright R. Janow Fall 2015
RC Circuit: solution for charging
Loop Equation is :
Substitute :
E - iR  Vc  0
dQ
i(t ) 
dt
Circuit Equation:
dQ
Q( t ) E


dt
RC R
Q( t )
V c (t ) 
C
• First order differential equation again: form is Q’ = - kQ + constant
• Same equation as for discharge, but with i0 = E / R added on right side
• At t = 0: Q = 0 & i = i0. Large current flows (C acts like a plain wire)
• As t  infinity: Current  0 (C acts like a broken wire)
Q  Qinf = CE = limiting charge
Solution: Charge starts from zero, grows as a saturating exponential.

Q(t)  Qinf 1  e t / RC

Qinf
Q
• RC = t = TIME CONSTANT
describes time dependance again
• Q(t)  0 as t  0
• Q(t)  Qinf as t  infinity
1  e 1  0.63
t
2t
t
Copyright R. Janow Fall 2015
3t
RC Circuit: solution for charging, continued
Voltage across C while charging:
Vc (t)  Q(t) / C
and
Vc (t )  E (1 - e  t / RC )
E  Qinf / C
Voltage across C also starts from zero and saturates exponentially
Current in the charging circuit:

dQ(t )
d
i(t ) 
 Qinf
1  e  t / RC
dt
dt
1  t / RC
 Qinf
e
RC
i(t )  i0 e  t / RC

i0 
E
R

Qinf
RC
Current decays exponentially just as in discharging case
Growing potential Vc on C blocks current completely at t = infinity
At t=0 C acts like a wire. At t=infinity C acts like a broken wire
Voltage drop VR across the resistor:
VR (t)  E e t / RC
VR (t )  i(t )R  i0 Re t / RC
Voltage across R decays exponentially, reaches 0 as t infinity
Form factor: 1 – exp( - t / t )
Factor
.63
.865
.95
.982
.993
Time
t
2t
3t
4t
5t
After 3-5 time
constants the
Copyright
R.
Janow
Fall 2015
6t
action
is over
.998
RC circuit – multiple resistors at t = 0
8-2: Consider the circuit shown, The battery has no internal resistance.
The capacitor has zero charge.
Just after the switch is closed, what is the current through the battery?
A. 0.
C
B. /2R.
C. 2/R.
D. /R.
E. impossible to determine

R
R
Copyright R. Janow Fall 2015
RC circuit – multiple resistors at t = infinity
8-3: Consider the circuit shown. The battery has no internal resistance.
After the switch has been closed for a very long time, what is the current
through the battery?
A. 0.
B. /2R.
C
C. 2/R.
D. /R.

R
R
E. impossible to determine
Copyright R. Janow Fall 2015
Discharging Example: A 2 mF capacitor is charged and
then connected in series with a resistance R. The
original potential across it drops to ¼ of it’s starting
value in 2 seconds. What is the value of the
resistance?
Use:
Vc (t )  V0 e  t / RC
Vc (t ) 1
  e  t / RC
V0
4
Set:
Take natural log of both sides, t = 2 s:
2
ln(1)  ln(4)  ln[ e  2 / RC ] 
RC
ln(4)  1.39
1.39 RC  2
ln(1)  0
 R 
ln[ ex ]  x
2
1
1.39 2x10 6
R  0.72 MW
Define: 1 MW = 106 W
Copyright R. Janow Fall 2015
Example: Discharging
C = 500 mF
R = 10 KW
E = 12 V
Capacitor C is charged for a long
time to E, then discharged.
E
a) Find current at t = 0
i(t ) 
dQ
 i0e  t / RC
dt
i0 
E
R

Q0
RC
i(t  0) 
E 0
12
e 
 1.2 mA
4
R
10
b) When does VCap (voltage on C) reach 1 Volt?
Vcap (t )  E e  t / RC
Vcap
1

 et / 5
E
12
RC  104 x 5 x 102 x 106  5 sec
 ln(12)   t / 5
V0  E  12
Volts
t  5 ln(12)  12.4 sec
c) Find the current in the resistor at that time
dQ
i(t ) 
 i0e  t / RC
dt
i(t  12.4 sec)  1.2 mA x e12.4 / 5
i(t  12.4 sec)  0.1 mA
Copyright R. Janow Fall 2015
Charging Example: How many time constants does it take
for an initially uncharged capacitor in an RC circuit to
become 99% charged?
Use:
Require:

Q(t)  Q 1  e t / t
Q( t )
 0.99  1  e  t / t
Q
Take natural log of both sides:
ln (0.01)  - 4.61  - t/t

t  RC  time constant
0.01  e  t / t
 t/t  4.61 
# of time constants
Did not need specific values of RC
Copyright R. Janow Fall 2015
Example: Charging a 100 mF
R
capacitor in series with a 10,000 W
resistor, using EMF E = 5 V.
E
C
a) How long after voltage is applied does Vcap(t) reach 4 volts?
Vc (t )  E (1 - e  t / RC )
Vc (t )
RC  104 x 100 x 10-6  1.0 sec
4
 0.8  1 - e  t / RC
E
5
Take natural log of both sides:

ln( 0.2 )   1.61  ln [e - t/RC ] 
 e  t / RC  0.2
t
t

RC
1.0
t  1.61 sec
b) What’s the current through R at t = 2 sec?
i(t )  i0 e  t / RC
i0 
i(t  2)  i0e  2.0 / 1.0 
E
R
E  2.0 / 1.0
5
e
 4 (0.37)2  6.77 x 10- 5
R
10
i(t  2)  6.8 mA.
Copyright R. Janow Fall 2015
Example: Multiple loops and EMFs
• Open switch S for a long time.
• Capacitor C charges to potential of battery 2
• Then close S for a long time
What is the CHANGE in charge on C?
First: E2 charges C to have:
Vc  E2  3 volts with current i1  0
Q 0  final charge for first phase  CE2  3.0 x 10-5
Q0  inital charge for
second phase
 30 mC
Second: Close switch for a long time
At equilibrium, current i3 though capacitor  zero
Find outer loop current i = i1 = 12 using loop rule
E2  iR 2  iR 1  E1  0
3  i(0.4  0.2)  1  0
i  2.0/0.6  3.33 A.
Now find Voltage across C, same as voltage across right hand branch
Vb  Va  E2  iR 2  3 - 3.33x0.4  1.67 V
Final charge on C:
Q final  C (Vb  Va )  10x10-6 x 1.67
Q final  16.7 mC
Copyright
Q final Q0  - 13.3
mC R. Janow
Fall 2015