Direct-Current Circuits
Physics 231
Lecture 6-1
Fall 2008
Resistors in Series and Parallel
As with capacitors, resistors are often in series and
parallel configurations in circuits
Series
Parallel
The question then is what is the equivalent resistance
Physics 231
Lecture 6-2
Fall 2008
Resistors in Series
Since these resistors are in series, we have the same current in
all three resistors
I1 = I 2 = I 3 = I
We also have that the sum of the potential differences across
the three resistors must be the same as the potential difference
between points a and b
Vab = Vax + V xy + V yb
Physics 231
Lecture 6-3
Fall 2008
Resistors in Series
Then using Vax = I R1; V xy = I R2 ; V yb = I R3
We have that
Vab = I ( R1 + R2 + R3 )
Now the equivalent resistor, R, will also have the same
potential difference across it as Vab, and it will also have the
same current I Vab = I R
Equating these last two results, we then have that
R = R1 + R2 + R3 = ∑ Ri
i
The equivalent resistance for a sequence of resistors in series
is just the sum of the individual resistances
Physics 231
Lecture 6-4
Fall 2008
Resistors in Parallel
Here we have that the voltage across each resistor has to be
the same (work done in going from a to b is independent of
the path, independent of which resistor you go through)
V1 = V2 = V3 = Vab
Physics 231
Lecture 6-5
Fall 2008
Resistors in Parallel
We now deal with currents through the resistors
At point a the current splits up into three distinct currents
We have that the sum of theses three currents must add to the
value coming into this point
I = I1 + I 2 + I 3
We also have that
Vab
Vab
Vab
I1 =
; I2 =
; I3 =
R1
R2
R3
The equivalent resistor, R, will have also have the
current I going through it
Physics 231
Lecture 6-6
Fall 2008
Resistors in Parallel
Using
Vab
I=
R
and combining with the previous equations, we then have
or
Vab Vab Vab Vab
=
+
+
R
R1 R2 R3
1 1
1
1
1
=
+
+
=∑
R R1 R2 R3
i Ri
The inverse of the effective resistance is given by the sum of the
inverses of the individual resistances
Physics 231
Lecture 6-7
Fall 2008
Solving Resistor Networks
Make a drawing of the resistor network
Determine whether the resistors are in series or
parallel or some combination
Determine what is being asked
Equivalent resistance
Potential difference across a particular
resistance
Current through a particular resistor
Physics 231
Lecture 6-8
Fall 2008
Solving Resistor Networks
Solve simplest parts of the network first
Then redraw network using the just calculated
effective resistance
Repeat calculating effective resistances until only
one effective resistance is left
Physics 231
Lecture 6-9
Fall 2008
Solving Resistor Networks
Given the following circuit
What is the equivalent resistance and what is the current
through each resistor
We see that we have two resistors in parallel with each other
and the effective resistance of these two is in series with the
remaining resistor
Physics 231
Lecture 6-10
Fall 2008
Solving Resistor Networks
Step 1: Combine the two
resistors that are in parallel
1
1
1
1
=
+
=
; Reff = 2 Ω
Reff 6 Ω 3 Ω 2 Ω
yielding
Step 2: Combine the two resistors that are in series
Reff = 4 Ω + 2 Ω = 6 Ω
Physics 231
yielding
Lecture 6-11
Fall 2008
Solving Resistor Networks
Current through this effective resistor is given by
V
18
I=
= = 3 Amps
R eff
6
The current through the resistors in the intermediate
circuit of Step 1 is also 3 Amps with the voltage drop
across the individual resistors being given by
V4 Ω = 3 ⋅ 4 = 12Volts;
V2 Ω = 3 ⋅ 2 = 6Volts
Physics 231
Lecture 6-12
Fall 2008
Solving Resistor Networks
To find the current through the resistors of the parallel section
of the initial circuit, we use the fact that both resistors have the
same voltage drop – 6 Volts
I6 Ω
6Volts
=
= 1 Amp;
6Ω
I3Ω
6Volts
=
= 2 Amps
3Ω
Physics 231
Lecture 6-13
Fall 2008
Consistency Check
There is a check that can be made to see if the answers for
the currents make sense:
The power supplied by the battery should equal the total
power being dissipated by the resistors
The power being supplied by the battery is given by P = I V
where I is the total current
P = I V = 3 ⋅ 18 = 54 Watts
The power being dissipated by each of the resistors is
given by P = I 2 R
P4Ω = 32 ⋅ 4 = 36 Watts; P3Ω = 22 ⋅ 3 = 12 Watts;
P6Ω = 12 ⋅ 6 = 6 Watts; PTotal = 54 Watts
Physics 231
Lecture 6-14
Fall 2008
Example 1
Two identical light bulbs are
represented by the resistors
R2 and R3 (R2 = R3 ). The
switch S is initially open.
If switch S is closed, what happens to the brightness of the
bulb R2?
a) It increases b) It decreases c) It doesn’t change
The power dissipated in R2 is given by
2
V
P=
R
When the switch is closed neither V nor R changes
So the brightness does not change
Physics 231
Lecture 6-15
Fall 2008
Example 2
Two identical light bulbs are
represented by the resistors
R2 and R3 (R2 = R3 ). The
switch S is initially open.
What happens to the current I, after the switch is closed ?
a) Iafter = 1/2 Ibefore b) Iafter = Ibefore c) Iafter = 2 Ibefore
Initially the current is given by
I before = ε R2
After the switch is closed the net resistance is given by
1
1
1
2
=
+
=
since R2 = R3
Rnet R2 R3 R2
Rnet = R2
2
⎞ = 2I
The new current is then I after = ε
= 2⎛⎜ ε
⎟
before
Rnet
⎝ R2 ⎠
Physics 231
Lecture 6-16
Fall 2008
Kirchoff’s Rules
Not all circuits are reducible
There is no way to reduce the four resistors to one
effective resistance or to combine the three voltage
sources to one voltage source
Physics 231
Lecture 6-17
Fall 2008
Kirchoff’s Rules
First some terminology
A junction, also called a
node or branch point, is
is a point where three
or more conductors
meet
A loop is any closed
conducting path
Physics 231
Lecture 6-18
Fall 2008
Kirchoff’s Rules
Kirchoff’s Rules are basically two statements
1. The algebraic sum of the
currents into any junction is zero
∑I =0
A sign convention:
A current heading towards a junction, is considered to be
positive,
A current heading away from a junction, is considered to
be negative
I1 + I 2 − I 3 = 0
Be aware that all the junction equations
for a circuit may not be independent of
each other
Physics 231
Lecture 6-19
Fall 2008
Kirchoff’s Rules
2. The algebraic sum of the potential differences in any loop
including those associated with emfs and those of resistive
elements must equal zero
∑V = 0
Procedures to apply this rule:
Pick a direction for the current in each branch
If you picked the wrong direction, the current will come
out negative
Physics 231
Lecture 6-20
Fall 2008
Kirchoff’s Rules
Pick a direction for traversing a loop – this direction must
be the same for all loops
Note that there is a third loop along the outside branches
As with the junction equations not all the loop
equations will be independent of each other.
Physics 231
Lecture 6-21
Fall 2008
Kirchoff’s Rules
Starting at any point on the loop add the emfs and IR terms
An IR term is negative if we traverse it in the same sense as
the current that is going through it, otherwise it is positive
An emf is considered to be positive if we go in the
direction - to +, otherwise it is negative
Need to have as many independent
equations as there are unknowns
Physics 231
Lecture 6-22
Fall 2008
Kirchoff’s Rules
For loop I we have − I1R1 − I1R2 − I 3 R4 + ε1 − ε 3 = 0
For loop II we have
− I 2 R3 + I 3 R4 − ε 2 + ε 3 = 0
Junction equation at a gives us
I1 − I 2 − I 3 = 0
We now have three equations for the three unknown currents
Physics 231
Lecture 6-23
Fall 2008
Kirchoff’s Rules
Assume that the batteries are: ε1 = 19 V; ε2 = 6 V; ε3 = 2 V
and the resistors are: R1 = 6Ω; R2 = 4Ω; R3 = 4Ω; R4 = 1Ω
you should end up with: I1 = 1.5 A; I2 = -0.5 A; I3 = 2.0 A
The minus sign on I2 indicates that the current is in fact in
the opposite direction to that shown on the diagram
Complete details can be found here
Physics 231
Lecture 6-24
Fall 2008
RC Circuits
Up until now we have assumed that the emfs and resistances
are constant in time so that all potentials, currents, and powers
are constant in time
However, whenever we have a capacitor that is being charged
or discharged this is not the case
Now consider a circuit that
consists of a source of emf, a
resistor and a capacitor but with
an open switch
With the switch open the current in the circuit is zero and
zero charge accumulates on the capacitor
Physics 231
Lecture 6-25
Fall 2008
RC Circuits
Now close the switch Initially the full potential will be
across the resistor as the potential
across the capacitor is zero since q is
zero
Initially the full potential is across the
resistor
The initial current in the circuit is then
given by I 0 = ε / R
As the current flows a charge will accumulate on the
capacitor
At some time t, the current in the circuit will be I and the
charge on the capacitor will be q
Physics 231
Lecture 6-26
Fall 2008
RC Circuits
According to Kirchoff’s 2nd rule we have
Using a counterclockwise loop
ε − Vresistor − Vcapacitor = 0
ε − IR −
Solving for the current
q
=0
C
ε
q
I= −
R RC
As time increases, the charge on the capacitor increases,
therefore the current in the circuit decreases
Current will flow until the capacitor has a charge on it given
by Q = C ε
Physics 231
Lecture 6-27
Fall 2008
RC Circuits
We remember that
dq
I=
dt
dq ε
q
1
So we then have
(q − Cε )
= −
=−
dt R RC
RC
dq
dt
=−
Rearranging we have
q − Cε
RC
q
t
dq
dt
Setting up the integration we have ∫
= −∫
q − Cε
RC
0
0
The resultant integration yields
Physics 231
Lecture 6-28
1
⎛ q − Cε ⎞
ln⎜
⎟=−
RC
⎝ − Cε ⎠
Fall 2008
RC Circuits
We exponentiate both sides of this last equation and rearrange
to obtain
(
)
(
q = Cε 1 − e −t / RC = Q f 1 − e −t / RC
)
where Qf is the final charge on the capacitor given by Cε
The constant RC is known as the time constant of the circuit
We see that the charge on the
capacitor increases
exponentially
Physics 231
Lecture 6-29
Fall 2008
Example 3
At t = 0 the switch is closed in
the circuit shown. The initially
uncharged capacitor then
begins to charge.
I1
I2
I3
ε
C
R2
R1
What will be the voltage across the capacitor a long time after
the switch is closed?
(a) VC = 0
(b) VC = ε R2/(R1+ R2)
(c) VC = ε
After a long time the capacitor is completely charged, so no
current flows through it
The circuit is then equivalent to a battery with two resistors in
series
The voltage across the capacitor equals the voltage across R2
(since C and R2 are in parallel)
Physics 231
Lecture 6-30
Fall 2008
RC Circuits
The current in the circuit is given by
d q ε −t / RC
I=
= e
= I 0 e −t / RC
dt R
and looks like
Note that is also how the voltage
across the resistor behaves
Vresistor = IR
Physics 231
Lecture 6-31
Fall 2008
RC Circuits – Charging Summary
For the simple RC circuit we have
the following for the voltage drops
across the capacitor and the
resistor
Physics 231
Lecture 6-32
Fall 2008
RC Circuits
We now start from a situation
where we have a charged capacitor
in series with a resistor and an open
switch
The capacitor will now act as a source of emf, but one whose
value is not constant with time
Physics 231
Lecture 6-33
Fall 2008
RC Circuits
We now close the switch and a
current will flow
Kirchoff’s 2nd rule gives us
q
− IR − = 0
C
Rearranging we have
dq
q
=−
dt
RC
To find q as a function of time we integrate the above equation
Physics 231
Lecture 6-34
Fall 2008
RC Circuits
q
∫
Q0
t
⎛ q ⎞
1
d q'
t
=−
d t ' ⇒ ln⎜⎜ ⎟⎟ = −
∫
q'
RC 0
RC
⎝ Q0 ⎠
Exponentiation of both sides of the equation on the right yields
q = Q0e −t / RC
We see that the charge on
the capacitor decreases
exponentially
Physics 231
Lecture 6-35
Fall 2008
RC Circuits
The current in the circuit is obtained by taking the derivative
of the charge equation
I=
dq
Q
=− 0 e
dt
RC
−
t
RC
The quantity Q0 / C is just the initial voltage, Vo , across the
capacitor
But then V0 / R is the initial current I0
So we then have that I = I 0 e
−
t
RC
The voltage across the resistor is given by V = V0 e
Physics 231
Lecture 6-36
−
t
RC
Fall 2008
Example 4
The two circuits shown below contain identical fully charged
capacitors at t = 0. Circuit 2 has twice as much resistance as
circuit 1.
Compare the charge on
the two capacitors a
short time after t = 0
a) Q1 > Q2
b) Q1 = Q2
c) Q1 < Q2
Initially, the charges on the two capacitors are the same. But the two
circuits have different time constants: τ1 = RC and τ2 = 2RC
Since τ2 > τ1 it takes circuit 2 longer to discharge its capacitor
Therefore, at any given time, the charge on capacitor 2 is larger than that on
capacitor 1
Physics 231
Lecture 6-37
Fall 2008
Example 5
a
The capacitor in the circuit shown is initially
charged to Q = Q0. At t = 0 the switch is
connected to position a.
At t = t0 the switch is immediately flipped
from position a to position b.
b
R
3R
C
Q
t0
time
Q0
c)
Q
b)
Q
a)
Q0
Q0
a) Which of the following graphs best represents the time dependence of the
charge on C?
t0
time
t0
time
b) Which of the following correctly relates the value of t0 to the time constant
τa while the switch is at a?
(a) t0 < τa
Physics 231
(b) t0 = τa
Lecture 6-38
(c) t0 > τa
Fall 2008
Example 5
a
The capacitor in the circuit shown is initially
charged to Q = Q0. At t = 0 the switch is
connected to position a.
At t = t0 the switch is immediately flipped
from position a to position b.
b
R
3R
C
Q
t0
time
Q0
c)
Q
b)
Q
a)
Q0
Q0
a) Which of the following graphs best represents the time dependence of the
charge on C?
t0
time
t0
time
For 0 < t < t0, the capacitor is discharging with time constant t = RC
For t > t0, the capacitor is discharging with time constant τ = 3RC,
i.e., much more slowly
Therefore, the answer is a)
Physics 231
Lecture 6-39
Fall 2008
Example 5
a
time
3R
Q0
C
c)
Q
Q
t0
R
b)
Q
a)
Q0
Q0
The capacitor in the circuit shown is initially
charged to Q = Q0. At t = 0 the switch is
connected to position a.
At t = t0 the switch is immediately flipped
from position a to position b.
b
t0
time
t0
time
b) Which of the following correctly relates the value of t0 to the time constant
τa while the switch is at a?
(a) t0 < τa
(b) t0 = τa
(c) t0 > τa
We know that for t = τa, the value of the charge is e-1 = 0.37 of the value at t = 0
Since the curve shows Q(t0) ~ 0.6 Q0, t0 must be less than τa
Physics 231
Lecture 6-40
Fall 2008
Capacitors Circuits, Qualitative
Basic principle:
Capacitor resists rapid change in Q Æ resists rapid changes in V
Charging
It takes time to put the final charge on
Initially, the capacitor behaves like a wire (∆V = 0, since Q = 0).
As current starts to flow, charge builds up on the capacitor
Æ it then becomes more difficult to add more charge
Æ the current decreases
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 a wire.
Physics 231
Lecture 6-41
Fall 2008