Today in Physics 122: Kirchhoff’s rules
and RC circuits
 For completeness: how the
pros solve for currents in
complex circuits
 The RC time constant
 Using the Kirchhoff rules
in RC circuits
 Dielectric relaxation: an
example of circuit analysis
reveals an important
microphysical property of
materials
18 October 2012
Physics 122, Fall 2012
I1
-
+
R1
+
- V0
dQ dt
+
-
I2
R2
+
C
-
t=0
1
Reminder: applying Kirchhoff’s rules
 Identify the unknown quantities – N, say – in the circuit,
and count them.
 Then write the node rule and/or the loop rule to generate
as many relations between the voltages and currents as
there are unknowns (N).
• Use both the node rule and the loop rule, at least once
each.
 This gives a system of N equations in N unknowns, which
is an algebra problem you first learned how to solve in
middle school. (No calculus required! Yet.)
 Important point: it doesn’t matter whether you correctly
guess the direction of each current. If you guess wrong,
your answer will just be a negative number.
18 October 2012
Physics 122, Fall 2012
2
When it gets ugly
If a complicated DC circuit has no evident symmetry, there’s
no way out but to hack through the algebra jungle.
Fortunately there are powerful math tools that can help.
Here’s how it’s done in practice:
I
I2
 Example: 5-R circuit from last
R1
lecture, this time with all
R2
I1
different resistances.
I3
V
 Independent currents in each
R3
R4
R5
resistor, so five more
unknowns, to a total of six
I4
I5
unknown currents.
 Need six independent equations.
18 October 2012
Physics 122, Fall 2012
3
When it gets ugly (continued)
 Let’s choose three node equations, from the top three
nodes:
=
I I +I
1
2
I=
1 I3 + I4
I2 + I3 =
I5
I
 … and three loop equations:
left, all around, and top right.
V − I 1 R1 − I 4 R4 =
0
V − I 2 R2 − I 5 R5 =
=
0
=0
I 1 R1 − I 2 R2 + I 3 R3 =
18 October 2012
Physics 122, Fall 2012
+
V-
R1
I1
R4
I2
+
+
R2
+
I3
+ R3 +R5
-
I4
I5
4
When it gets ugly (continued)
 Rearranged just slightly, it looks as if this system of linear
equations is a single matrix equation:
I − I1 − I 2 =
0
I1 − I 3 − I 4 =
0
I2 + I3 − I5 =
0
R1 I 1 + R4 I 4 =
V
R2 I 2 + R5 I 5 =
V
R1 I 1 − R2 I 2 + R3 I 3 =
0
I
V
I2
R1
I1
R4
R2
R3
I3
R5
I4
I5
 Don’t worry, we’re not going
to use matrices in PHY 122, but
for those have taken or will take MTH 165 and/or 235…
18 October 2012
Physics 122, Fall 2012
5
When it gets ugly (continued)
 … in matrix form, our six equations are
 1 −1 −1

0
0 1
0 0
1

0
 0 R1
0 0
R2

 0 R −R
1
2

0
−1
1
0
0
R3
0
−1
0
R4
0
0
0  I   0 
   
0   I1   0 
−1   I 2   0 
  =  
0   I3  V 
R5   I 4   V 
   
0   I 5   0 
 In those math classes you will learn how to solve such
equations, without the hideous prospect of the procedure
we can use in three-current circuits. Here it would involve
finding the inverse of the 6×6 matrix, and multiplying the
equation through from the left by this inverse.
18 October 2012
Physics 122, Fall 2012
6
When it gets ugly (continued)
 The answer, using those fancy linear-algebra methods:
 I 
 R1 R3 + R1 R4 + R2 R4 + R3 R4 + R2 R3 + R1 R5 + R2 R5 + R3 R5 
 


I
R
R
R
R
R
R
R
R
+
+
+
2 3
2 5
3 5
2 4
 1


 I2  V 

R1 R3 + R1 R4 + R3 R4 + R1 R5
 = 

I
R
R
−
R
R
D
2 4
1 5
 3


I 


R2 R3 + R1 R5 + R2 R5 + R3 R5
4
 


I 


R1 R3 + R1 R4 + R2 R4 + R3 R4
 5


where
D = R1 R2 R3 + R1 R2 R4 + R1 R2 R5 + R1 R3 R5
+ R2 R3 R4 + R1 R4 R5 + R2 R4 R5 + R3 R4 R5
 Note that I3 = 0 if R2 R4 = R1 R5 , so we got it right in our
first “symmetry” example (all Rs the same).
18 October 2012
Physics 122, Fall 2012
7
RC circuits and the RC time constant
In a steady state, ideal capacitors
draw no current. But if a circuit
is assembled and switched on,
one will observe that it take time
for the capacitor to receive its
full charge.
 Key: since capacitors
accumulate a charge Q, the
resistors in series with them
carry a current I = dQ/dt.
 The current in the series R
should decrease to zero as C
reaches its full charge.
18 October 2012
t=0
V0
Physics 122, Fall 2012
I = dQ dt
+
-
+
-
R
+
C
-
8
RC circuits and the RC time constant (continued)
Let’s apply the loop rule to
this simple circuit, in which
the switch is closed at t = 0.
Q
V0 − IR − =
0
C
dQ
RC = CV0 − Q
dt
Q
t
0
0
t=0
V0
dQ′
1
∫ CV0 − Q′ = RC ∫ dt′
Change variables:
18 October 2012
I = dQ dt
+
-
+
-
R
+
C
-
−dQ′
q=
CV0 − Q′ dq =
As Q′ =
0 → Q, q =
CV0 → CV0 − Q
Physics 122, Fall 2012
9
RC circuits and the RC time constant (continued)
Thus the integral is recognizable:
CV0 −Q
−
∫
CV0
CV0 −Q
ln q CV
0
CV0 − Q
t
= ln
= −
CV0
RC
CV0 − Q =
CV0 e −t
(
V0
RC
Q ( t ) CV0 1 − e −t
=
RC
dQ CV0 −t
I=
e
(t ) =
dt
RC
18 October 2012
t=0
dq
t
=
q
RC
I = dQ dt
+
-
+
-
R
+
C
-
)
RC
Physics 122, Fall 2012
10
RC circuits and the RC time constant (continued)
R⋅C
18 October 2012
Physics 122, Fall 2012


Q(t)
Qmax ⋅  1−
0
R⋅C
2
4
6
1

e
8
10
t (RC)
I(t)
The results are plotted at right.
 The current in the resistor
eventually drops to zero; it
reaches 1/e of its initial value in
t = RC.
 The charge on the capacitor
eventually reaches CV0 ; it
comes within 1/e of this value
at t = RC.
The quantity τ = RC is called the
circuit’s time constant, and is a
handy measure of how long it takes
to charge the capacitor.
1
⋅I
e max
0
2
4
6
8
10
t (RC)
11
C with series and parallel R
In more complex circuits with capacitors one can still use
Kirchhoff’s rules confidently in much the same way as with
purely resistive circuits.
+  Example (26-50 in the book):
R1
Determine the time constant,
+
+
+
and the maximum charge on
R2
C
- V0
the capacitor, in this circuit.
t=0
[Intuitively, we can tell that
the maximum charge should
be that for which all the current flows through the
resistors; in this case the voltage across C would be
charge Q CV0 R1 ( R1 + R2 ) .]
V0 R1 ( R1 + R2 ) , and the=
18 October 2012
Physics 122, Fall 2012
12
C with series and parallel R (continued)
 Like the other three-branch
circuits we’ve dealt with, we
use the node equation once
and the loop equation twice:
+
R1
dQ dt
dt =
I 1 − I 2 − dQ
0
=0
V0 − I 1 R1 − I 2 R2 =
=0
I 2 R2 − Q C =
 As before, the algebra
strategy is to target one
variable, and substitute out
the other two in its favor.
This time we target Q.
18 October 2012
I1
-
+
- V0
dQ dt
+
-
I2
R2
+
C
-
t=0
Three unknown currents.
Physics 122, Fall 2012
13
C with series and parallel R (continued)
 Second loop equation:
I 2 = Q R2C
I1
-
+
R1
 Node equation:
dQ
Q
dQ
I 1 =I 2 +
=
+
dt
R2C dt
 Results into first loop
equation, now in Q alone.
+
- V0
dQ dt
+
-
I2
R2
+
C
-
t=0
 Q
dQ  Q
+
V0 − R1 
0
− =
 R2C dt  C
18 October 2012
Physics 122, Fall 2012
14
C with series and parallel R (continued)
 Solve for dQ/dt:
dQ V0  Q
Q 
= −
+

dt
R1  R2C R1C 
+
 R1 + R2 
V0
=
−Q

R1
R
R
C
 1 2 
R1 + R2 
1 
=
Q
 CV0 −
R1C 
R2

 Separate and integrate:
Q
I1
R1
+
- V0
dQ dt
+
-
I2
R2
+
C
-
t=0
t
dQ′
1
=
dt ′
R1 + R2
R1C
Q′
0 CV0 −
0
R2
∫
18 October 2012
∫
Physics 122, Fall 2012
15
C with series and parallel R (continued)
 Same integral as before. Change variables:
R1 + R2
q CV0 −
Q′
=
I1
R2
+ R1 + R2
R1
dq = −
dQ′
R2
+
+
V0
′
As Q = 0 → Q ,
-
R1 + R2
q = CV0 → CV0 −
Q:
R2
R2
−
R1 + R2
18 October 2012
CV0 −
R1 + R2
Q
R2
∫
CV0
dQ dt
I2
R2
+
C
-
t=0
dq
t
=
q
R1C
Physics 122, Fall 2012
16
C with series and parallel R (continued)
CV0 −
ln q CV
0
R1 + R2
Q
R2
R1 + R2
t
t≡−
=−
R1 R2C
RC
R1 + R2
CV0 −
Q
R2
t
= −
ln
CV0
RC
R1 + R2
CV0 −
Q
R2
−t RC
=e
CV0
R2
−t RC
Q (t )
CV0 1 − e
R1 + R2
(
18 October 2012
I1
-
+
R1
+
- V0
dQ dt
+
-
I2
R2
+
C
-
t=0
)
Physics 122, Fall 2012
17
C with series and parallel R (continued)
Comments on the solution:
 Indeed the maximum charge
I1
on the capacitor comes out to
+ R2
R1
Qmax lim
Q (t )
CV0 ,
=
=
R1 + R2
t →∞
+
+
- V0
as we expected.
t=0
 The time constant is
R1 + R2
t
−
=
τ=
t,
RC
R1 R2C
dQ dt
I2
R2
+
C
-
as if determined by the parallel
combination of the Rs. (!?!)
18 October 2012
Physics 122, Fall 2012
18
Dielectric relaxation
An application of RC circuits to the physics of imperfect
conductors/insulators (problem 26-47 in the book):
Materials that have very large – but not infinite – resistivity
can exhibit significant polarization, characterized by a
dielectric constant > 1.
 Even the “bad” conductors out of which resistors are
made, like graphite, have resistivity small enough that
polarization effects are small: K= 1.
Consider a parallel-plate capacitor (area A, separation d)
filled with such material (dielectric constant K, resistivity ρ).
At t = 0 it has charge Q. What is the time constant with
which the charge leaks away?
18 October 2012
Physics 122, Fall 2012
19
Dielectric relaxation (continued)
Solution:
 We worked out the
capacitance and resistance
of this arrangement
before, on 2 October and 4
October respectively:
Kε 0 A
ρd
C =
R
d
A
and we noted that the R
and C could be considered
to be in parallel in the
circuit-element sense.
18 October 2012
Physics 122, Fall 2012
dQ dt
R
+
C
+Q
-Q
t=0
Note dQ/dt is presumed to
have the direction that
increases +Q.
20
Dielectric relaxation (continued)
 Only one K.R. equation here, and it’s a loop:
dQ dt
dQ Q
R
0
+ =
dt C
Q( t )
∫
Q0
t
dQ′
1
dt ′
= −
Q′
RC
∫
R
0
 Q (t ) 
t
ln 
= −
 Q 
RC
 0 
+
C
+Q
-Q
t=0
Q ( t ) = Q0 e −t RC
so τ = RC , as we might have expected.
18 October 2012
Physics 122, Fall 2012
21
Dielectric relaxation (continued)
 What’s interesting about this result – at least to physicists,
materials scientists, and electrical engineers – is that
everything having to do with the shape of the capacitor
vanishes from the expression for the time constant:
ρ d Kε 0 A
τ RC
=
=
= ρ Kε 0 .
A d
so we’d get the same result no matter what the electrode
shape.
 That is, this time constant is a property of the material, not
of the shape: it is the time constant associated with how
quickly charges can be moved around in high-resistivity
material. That there is a limit to how quickly, is a
phenomenon called dielectric relaxation.
18 October 2012
Physics 122, Fall 2012
22