PHY 2049 Lecture Notes
Chapter 28: Page 1 of 9
Electromotive Force E
I
I
+
Emf
∆V
-
V
Emf + ∆V = 0
Electromotive Force:
The electromotive force EMF of a source of electric potential energy is defined as the
amount of electric energy per Coulomb of positive charge as the charge passes
through the source from low potential to high potential.
Emf = E = dU/dq
(The units for EMF is Volts)
I
Single-loop circuit with a resistor:
ε
+
R
-
• Power Delivered by Emf (P = E I):
dW
dq
dW = ε dq P =
=ε
= εI
dt
dt
• Power Dissipated in Resistor (P = I2R)
• Combining the two:
E - IR = 0 and I = E /R
(Kirchhoff's Rule)
A. Korytov
PHY 2049 Lecture Notes
Chapter 28: Page 2 of 9
Delivering power to a load: high voltage lines
I
+
Emf
R
Useful Load
(motor, light bulb, etc)
• Power Delivered by Emf:
P0 = E I
• Power Dissipated in wire resistance: P = I2R= (P0/E
E )2R
• Losses:
P/P0 = P0R/E
E2
• Delivering power at high voltage results in lower losses
• Example:
Emf=120 V
Power= 100 kW (this building)
Line ~ 2⋅⋅10-4 Ohm/m (1 cm × 1 cm copper)
Losses <10%
Maximum length: L<0.1⋅⋅1202 /105/0.0002 = 50 m
A. Korytov
PHY 2049 Lecture Notes
Chapter 28: Page 3 of 9
Circuit Rules
Loop Rule:
The algebraic sum of the changes in potential
encountered in a complete traversal of any loop of a
circuit must be zero.
Switch
+
C
Q
− IR = 0 .
C
R
-
Junction
Rule:
The sum of the currents entering any junction must be
currents leaving that junction.
∑ I =∑ I
i
in
i
out
equal the sum of the
I
∆V=-IR
Resistor:
If you move across a resistor in the direction of the
potential change is ∆ VR = - IR.
Capacitor:
I
Q
current flow then the
+
-
∆V=Q/C
If you move across a capacitor from minus to plus then the
potential change is
∆VC = Q/C,
and the current leaving the capacitor is I = -dQ/dt.
Inductor (Chapter 31):
If you move across an inductor in the direction of the
current flow then the potential change is
∆VL = - L dI/dt.
I
L
∆VL=-LdI/dt
A. Korytov
PHY 2049 Lecture Notes
Chapter 28: Page 4 of 9
Examples of multi-loop circuits
I2
I1
R1
+
E1
R2
I1+I2
-
+
-
E2
R3
To find currents through all resistors:
• Simplify: replace elements connected in series and parallel with
corresponding equivalent elements
• Denote all “independent” loops
• Assign currents and their directions
(directions can be picked arbitraryà
à solution will be negative if you picked
wrong direction, that all)
• Apply the rule of current conservation at all junctions
• Go along each of the loops and add up all ∆V’s encountered;
Sum of ∆V’s for a complete loop should be zero
First Loop :
E 1 − I 1 R1 − ( I 1 + I 2 ) R 3 = 0
Second Loop :
E 2 − I 2 R2 − ( I1 + I 2 ) R3 = 0
• Solve the system of equations
A. Korytov
PHY 2049 Lecture Notes
Chapter 28: Page 5 of 9
Voltmeter
Voltmeter:
- device to measure difference of potentials between two points
- connects in parallel (spy device)
- should have large internal resistance
I
+
Emf
R1
∆V=?
R2
RV
Error in the measurement:
•
Before connection: V0=ER2/(R1+R2)
•
After connection:
RV R 2
R2
R2
R 2 → R 2′
=
=
≈ R 2 (1 − δ )
RV + R 2 1 + R 2 / RV
1+δ
V meas = E

R 2′
R 2 (1 − δ )
R1
R2 

≈ E
= ... = V 0  1 −
′
R1 + R 2
R1 + R 2 (1 − δ )
R1 + R 2 RV 

A. Korytov
PHY 2049 Lecture Notes
Chapter 28: Page 6 of 9
Ammeter
Ammeter:
- device to measure current
- a circuit is to be cut and the ammeter is inserted
- should have small internal resistance
I=?
R1
+
Emf
R2
RA
Error in the measurement:
•
Before connection: I0=E/(R1+R2)
•
After connection:
I =
E
=
R1 + R 2 + R A
E

RA
( R 1 + R 2 )  1 +
R1 + R 2




=

I0
RA
≈ I 0 (1 − δ ) = I 0  1 −
1+ δ
R1 + R 2




A. Korytov
PHY 2049 Lecture Notes
Chapter 28: Page 7 of 9
Using Voltmeter and Ammeter to measure R
One has a battery, voltmeter, ammeter, and a resistor of
unknown value. How should one connect the battery, resistor,
voltmeter, and ammeter to evaluate R2 and to minimize error of
the measurement?
Expected values of Emf (~9 V), R2 (~10 kOhm), RV (~100 MOhm),
and RA (~10 Ohm) are known with about 10% uncertainty.
Homework problem…
A. Korytov
PHY 2049 Lecture Notes
Chapter 28: Page 8 of 9
RC-circuit: charging up a capacitor
Switch
+
C
R
-
After the switch is closed the current is entering the
capacitor so that I = dQ/dt, where Q is the charge on the
capacitor and summing all the potential changes in
going around the loop gives
Q
− IR = 0 ,
C
where I(t) and Q(t) are a function of time.
If the switch is closed at t=0 then Q(0)=0 and
Q
dQ
+ R
= 0 , which can be written in the form
C
dt
dQ
1
= − Q , where I have define τ=RC.
dt
τ
Dividing by (Q-εC) and multiplying by dt and integrating gives
∫
Q
Q0
t
 Q
t
dQ
1
 = − .
= − ∫ dt , which implies ln 
τ
 Q0 
Q
τ
0
Solving for Q(t) gives
Charging a Capacitor
1.50
Q ( t ) = Q0e − t / τ .
1.25
1.00
Q(t) 0.75
0.50
0.25
The curent is given by
I(t)=dQ/dt which yields
Q
I ( t ) = 0 e − t / τ . The quantity
RC
0.00
0
1
2
3
4
Time
τ=RC is call the time constant and has dimensions of time.
A. Korytov
PHY 2049 Lecture Notes
Chapter 28: Page 9 of 9
RC-circuit: discharging a capacitor
Switch
+
C
R
-
After the switch is closed the current is leaving the capacitor
so that I = -dQ/dt, where Q is the charge on the capacitor and
summing all the potential changes in going around the loop
gives
Q
− IR = 0 ,
C
where I(t) and Q(t) are a function of time.
If the switch is closed at t=0 then Q(0)=Q0 and
Q
dQ
+ R
= 0 , which can be written in the form
C
dt
dQ
1
= − Q , where I have defined τ=RC.
dt
τ
Dividing by Q and multiplying by dt and integrating gives
∫
Q
Q0
t
 Q
t
dQ
1
= − ∫ dt , which implies ln 
 = − .
τ
Q
τ
 Q0 
0
Solving for Q(t) gives
Discharging a Capacitor
1.00
Q ( t ) = Q0e − t / τ .
0.75
Q(t)
0.50
0.25
The current is given by
I(t)=-dQ/dt which yields
I (t ) =
Q0 − t / τ
e
. The
RC
0.00
0
1
2
3
4
Time
quantity τ=RC is call the "time constant" and has dimensions of time.
A. Korytov