CH25 Summary: Resistance and Current
Current and Current Density:
dQ
I=
= n | q | vd A
dt


J = nqv d
E
J
ρ(T) = ρ 0 [1+ α (T − T0 )]
ρ=
Resistivity:
€
Resistor:
Very small
L
A
V = IR
I = V /R,R = V /I
R=ρ
€
R becomes smaller when
A is? Or L is?
Vab = ε − Ir
Circuits and emf (eletromotive
€
Internal
resistance reduces
force):
the potential difference of the two terminals
Energy and power in Circuits:
€
Power into a resistor
P = Vab I
Power to a general element
Vab2
P=I R=
R
2
Chapter 26: Direct-Current Circuits
•  Resistors in series and parallel
•  Compare that with Capacitors
•  Kirchoff’s Rules for complicated circuits
•  electronic measuring instruments
•  Reistor-Capacitor (RC) circuits
•  pplications of circuits in household wiring
Introduction
•  Circuits in CH25 are way too
simple!
•  Modern PCB are designed are
so complicate that you have to
design it with software aids.
–  What are the principles
behind it?
–  We will start with resistors in
series and parallel
Resistors in series and parallel I
•  Resistors in Series:
I1 = I2 = I3 ≡ I
Vab = Vax + Vxy + Vyb = IR1 + IR2 + IR3
⇒ Vab = I(R1 + R2 + R3 ) ≡ IReq
⇒ Req = R1 + R2 + R3
Req (Series) is larger than all R’s; Why?
•  €Resistors in Parallel
V1 = V2 = V3 ≡ Vab
V
V
V
I = I1 + I2 + I3 = ab + ab + ab
R1 R2 R3
I
1 1
1
1
⇒
= +
+
≡
Vab R1 R2 R3 Req
i.e.
1
1 1
1
= +
+
Req R1 R2 R3
Compare with the relation
between resistor and
resistivity (Does it
contradict with our new
knowledge?)
L
Req (Parallel) is smaller than all R’s; Why?
€
R≡ρ
A
Resistors in series and parallel II: Examples
What is a better design for Xmas lights:
Lights in series, or in parallel?
What happens if R2 is broken?
What happens if R2 is broken?
Q26.1
Which of the two
arrangements shown has the
smaller equivalent resistance
between points a and b?
A. the series arrangement
B. the parallel arrangement
C. The equivalent resistance
is the same for both
arrangements.
D. The answer depends on
the values of the individual
resistances R1, R2, and R3.
Q26.2
Three identical resistors,
each of resistance R, are
connected as shown. What is
the equivalent resistance of
this arrangement of three
resistors?
A. 3R
B. 2R
C. 3R/2
D. 2R/3
E. R/3
Q26.3
A 120-V, 60-W light bulb, a 120-V,
120-W light bulb, and a 120-V,
240-W light bulb are connected in
parallel as shown.
120 V
60 W
The voltage between points a and b
is 120 V. Through which bulb is
there the greatest voltage drop?
120 V
120 W
A. the 120-V, 60-W light bulb
B. the 120-V, 120-W light bulb
a
120 V
240 W
C. the 120-V, 240-W light bulb
D. All three light bulbs have the same voltage drop.
b
Q26.5
A 120-V, 60-W light bulb, a 120-V, 120-W
light bulb, and a 120-V, 240-W light bulb are
connected in series as shown.
The voltage between points a and b is 120 V.
Through which bulb is there the greatest
voltage drop?
a
120 V
60 W
120 V
120 W
A. the 120-V, 60-W light bulb
B. the 120-V, 120-W light bulb
C. the 120-V, 240-W light bulb
D. All three light bulbs have the same voltage drop.
120 V
240 W
b
Q26.6
A 120-V, 60-W light bulb, a 120-V, 120-W
light bulb, and a 120-V, 240-W light bulb are
connected in series as shown.
The voltage between points a and b is 120 V.
Which bulb glows the brightest?
a
120 V
60 W
120 V
120 W
A. the 120-V, 60-W light bulb
B. the 120-V, 120-W light bulb
C. the 120-V, 240-W light bulb
D. All three light bulbs glow with equal brightness.
120 V
240 W
b
Resistors in series and parallel III—combinations
•  A battery without internal resistance (r=0) is
connected to the following resistors. What is the
current through each resistor, and the potential
difference across each resistor?
R2
I1=?V1=Vac=?
I2=?V2=Vcb=?
R1
R3
I3=?V3=Vcb=?
What we have learned so far about resistance
Resistors in series
Req = ∑ Ri
If i=1,2 Req = R1 + R2
i
(Req > Ri )
(Think of a longer traffic jam)
Resistors in parallel
If i=1,2
1
1
=∑
Req
i R
1
1 1
= +
Req R1 R1
(Req < Ri )
(Think of easier traffic
with more lanes)
€
€
R1R2
Req =
R1 + R2
Challenging example: Please do this offline
•  What is the resistance of the following resistor network?
•  What are the currents I1, I2, I3?
R0
R0
R0
R0
R0
R0
R0
R0
R0
R0
symmetry
R0
R0
R0
R0
R0
R0
R0
Req = R0 +
Req R0
+ R0
Req + R0
⇒ Req2 − 2R0 Req − 2R02 = 0
Req
⇒ Req = (1 ± 3)R0
Req
€
Which one?
Let’s do some experiments together
RA=0, I1
A
r=?
RA=0, I2
A
R0=?
R0=?
r=?
ε=6V
ε=6V
R0=?
The left network has only one bulb connected to the battery of 6V.
The right network has two identical bulbs.
Should the readings of the ammmeter be such that
Because
€
I1 = 2I2
ε
ε
I1 =
=2
= 2I2 ?
R0
2R0
If our experiment doesn’t show this, why?
A simple method of measuring the internal resistance
RA=0, I1
A
r=?
R0=?
R0=?
ε=6V
I1 = 2I2
ε
ε
I1 =
=2
= 2I2
R0
2R0
€
RA=0, I2
A
Wrong if internal resistance
r is not 0!
Instead, we have:
r=?
ε=6V
R0=?
ε
ε
⇒ r + R0 = .........a)
r + R0
I1
ε
ε
I2 =
⇒ r + 2R0 = .........b)
r + 2R0
I2
b) − a) ⇒
ε ε
R0 = −
I2 I1
ε
ε
ε ε
a) ⇒ r = − R0 = − ( − )
I1
I1
I2 I1
ε ε
r =2 −
I1 I2
I1 =
Q26.7
Three identical light bulbs are
connected to a source of emf as
shown. Which bulb is brightest?
A. light bulb A
B. light bulb B
C. light bulb C
D. both light bulbs B and C
(Both are equally bright and are
brighter than light bulb A.)
E. All bulbs are equally bright.
Kirchoff’s Rules I—junctions rule and loop rule
•  Resistor networks are often complicate and not just series-parallel
combinations.
•  What happens if it is neither?
Such as the bridge network below?
Junction rule:
∑I = 0
No charge buildup.
Loop rule:
∑V = 0 If we come back to the
starting point, the potential doesn’t change
Kirchoff’s Rules II—loops
•  The algebraic sum of the potential differences in any loop,
including those associated with emfs and those of resistive
elements, must equal zero.
Current flows from high potential (+)
To low potential (-)
You loop can travel both directions
Kirchoff’s Rules III: Multiple EMFs
A 12V power supply with internal resistance r is connected to a
run-down rechargable battery with unknown emf ε. The battery
carries current of 1A, and the light indicator (3Ω) carries current
of 2A.
What is the current through the power supply? How many equations
do we need here?
What is value of r?
What is the value of E of the emf (remaining voltage)
Junction a: I+1A+2A=0I=-3A.
what does “-” mean here?
Loop 1: +12V-(3A)r-(2A) (3Ω) =0
r=2Ω
Loop 2: -ε+(1A)(1Ω)-(2A) (3Ω) =0
ε=-5V
What does this “-” mean?
Kirchoff’s Rules III: Multiple EMFs
A 12V power supply with internal resistance r is connected to a
run-down rechargable battery with unknown emf ε. The battery
carries current of 1A, and the light indicator (3Ω) carries current
of 2A.
r, ε, I: Three unknowns. So we need three equations!
I=-3A
“-” means current is flowing
out of junction a
ε=-5V
“-” means the polarity is opposite
to our assumption. Therefore,
top terminal is positive instead!
We could have used either
Loop 2+3, or Loop 1+3.
They will give the same results as Loop 1+2.
Kirchoff’s Rules IV– Bridge network
What is the currents through each
Resistor, and the emf?
Using junction rules, we can easily
deduce that we only need to know
three current independently:I1,2,3
All the rest can be obtained from
these three (using what rules?)
What happens if this resistor
is 1Ω too? Can you guess?
Loop 1: 13V-I1(1Ω)-(I1-I3)(1Ω)=0
Loop 2: 13V-I2(1Ω)-(I1+I3)(2Ω)=0 
I1=6A, I2=5A, I3=-1A
Loop 3: -I1(1Ω)-I3(1Ω)+I2(1Ω)=0
(meaning)
What is the equivalent resistance of the network?
Req=E/Itot=13V/(I1+I2)=13V/11A=1.2Ω
Another bridge? example
c
R2=3Ω
R1=2Ω
a
ε=6V
R1=2Ω
A
b
R2=3Ω
R1=2Ω
d
In the resistor network above, what is the reading of the current?
Another bridge? example
c
R2=3Ω
R1=2Ω
A
a
ε=6V
b
R2=3Ω
R1=2Ω
d
Vcd = ε = 6V
Vcd
Vca =
= 3V
2
V
Vcb = cd = 3V
2
∴Vab = (Vca − Vcb ) = 0V
V
Iab = ab = 0A
R1
Imagine the middle resistor and ammeter did not exist:
€
Due to the symmetry on both sides of the
bridge, there is
no potential difference between point a and point b
No current flowing from a to b!
€
R-C (Resistor-Capacitor) circuits
In a circuit with both resistor and capacitor (flashing traffic light,
Car turn signal, pacemaker, etc.), will the
charging and discharging instantaneously?
Vab = iR
Q0 = Cε
ε
I0 =
R
V0 = ε
Vbc = q' /C
q'
ε − iR − = 0
Loop rule
C
dq' ε
q'
q' −Cε
i≡
= −
=−
⇒
dt' R CR
RC
What does it mean?
q
t
dq'
t'
dq'
dt'
=−
⇒ ∫
=−∫
q' −Cε
RC
q'
−C
ε
At t=0, q=0, Vbc=0,
0
0 RC
q − Cε
t
=> ln(
)=−
0 − Cε
RC
q = Cε (1 − e −t / RC )
ε
i = dq /dt = e −t / RC
R
Vbc = Q /C = ε (1 − e −t / RC )
€
The capacitor acted like
a wire, so i=ε/R
Time constant: τ=RC
If your turn light starts to flash crazily, what’s your
guess and plan to fix it?
(hint, this means τ is abnormally small)
Discharging and RC timing graphs
q(t) = Q0e −t / RC
Discharging
i(t) = dq /dt = (−
Q0 −t / RC
)e
≡ I0e −t / RC
RC
Vbc (t) = ?
Vbc (t) =
€
q(t) q(t) −t / RC
=
e
= V0e −t / RC
C
C
I0 = (−
V0 =
€
Q0
)
RC
Q0
C
Negative current
Means it is
flowing out of
the positive plate
Charging power:
εi = i 2 R + iVbc = i 2 R + i
Dissipated by resisstor
€
Q
C
Q26.9
A battery, a capacitor, and a resistor are connected in series.
Which of the following affect(s) the maximum charge stored
on the capacitor?
A. the emf ε of the battery
B. the capacitance C of the capacitor
C. the resistance R of the resistor
D. both ε and C
E. all three of ε, C, and R
Q26.10
A battery, a capacitor, and a resistor are connected in series.
Which of the following affect(s) the rate at which the
capacitor charges?
A. the emf ε of the battery
B. the capacitance C of the capacitor
C. the resistance R of the resistor
D. both C and R
E. all three of ε, C, and R
€
Power distribution in RC circuit
Instantaenous charging power:
εi = i 2 R + iVbc = i 2 R + i
Q
C
Stored by capacitor
Dissipated by resisstor
€
ε −t / RC 2
ε2
E R = ∫ i Rdt = ( e
) Rdt =
R
R
ε 2 RC ε 2C
ER =
=
R 2
2
2
∫
∝ −2t / RC
0
e
dt
ε −t / RC Cε (1 − e −t / RC )
E C = ∫ iQ /Cdt = ( e
)
dt
R
C
ε 2 ∝ −t / RC
−2t / RC
EC =
(e
−
e
)dt
∫
R 0
ε 2 RC
1 ε 2C
EC =
(1 − ) =
R
2
2
If we integrate
from t=0 to t=infinity:
What do we find?
They are the same! E R = EC
Independent of the R, C, ε, values!
€
€
Power distribution systems—what’s your money
saving example after learning physics?
Fuses, circuit breakers, and GFI
•  Fuse: Will melt if there is excess
current, therefore break the circuit
•  2prong plug VS 3prong plug
•  GFI (Ground-Fault Interrupter)
circuit breaker:
•
stops further current flow when a
sudden drop in resistance
indicates that someone has offered
a new path to ground.
No ground you could become
The path to the ground when
Your drill is disfunctioning.
CH26 Summary: Resistance and Current
Resistors in series and parallel:
Req = R1 + R2 + R3 + ... Series
1
1 1
1
= +
+
+ ...
Parallel
Req R1 R2 R3
Kirchhoff’s rules:
∑I = 0
∑V = 0
€
€
q = Cε (1 − e
RC circuits:
Time constant τ=RC
−t / RC
)
ε
i = dq /dt = e −t / RC
R
€
Vbc = ?
Charging:
€ Starting: q=0, i=ε/R
Ending:q=Cε, i=0€
Junction rule
Loop rule
q(t) = Q0e −t / RC
ε
i(t) = dq /dt = − e −t / RC
R
Q
= − 0 e −t / RC
RC
DisCharging:
Starting: q=Q0, i=Q0/RC
Ending:q=0, i=0
(Take the limit of t=0 or t=∞)