This week
Electrical Circuits
Series or parallel that’s the question.
Current, Power and Energy
Why does my laptop battery die?
Transmission of power to your home
Why do we have big transmission towers?
Household Appliances
Why do fuses blow?
4/29/2016
Physics 214 Summer 2016
1
Electrical power and energy
In order to separate charge we have to do
work and energy is stored. The simplest
example is a battery. The quantity we call
voltage is related to the energy stored
∆V = ∆PE/q (joules/coulomb = volt)
When positive charge moves to the
negatively charged plate or vice versa then the
stored energy is released. In order for this to
happen the two sides of the battery must be
joined by a conductor so that the charge can
move easily. This is a simple circuit.
4/29/2016
Physics 214 Summer 2016
2
Electrical circuits
All circuits are basically the same.
There is an external source of energy
which produces a voltage.
In a charged battery there is
separation of charge. When the
circuit is made positive charge flows
from high to low voltage or negative
charge flows from low to high
voltage releasing the stored energy.
Normally it is electrons which flow.
As they move they “collide” with the
atoms of the wire and lose some of
their energy in the form of heat.
There is resistance to the flow.
4/29/2016
Physics 214 Summer 2016
3
Current and resistance
The rate of flow of charge determines how much
energy is released per unit time
I = q/t amperes (coulombs/sec)
The direction of I is the flow of positive charge or
opposite to the flow of negative charge.
OHM’s Law
R = ∆V/I
ohms
Every part of a circuit has resistance including an
internal resistance in the battery. The higher the
resistance the lower is the current for a given
voltage difference
4/29/2016
Physics 214 Summer 2016
4
Electromotive force
The electromotive force of a battery , ε
, is the voltage difference between the
two terminals when no current is
being drawn.
When it is connected to a simple
circuit
I = ε/(Rcircuit + Rbattery)
A voltage difference is the energy
stored or the energy released per unit
charge. So if charge +q goes from
high to low voltage then the energy
released is qε or q∆V
4/29/2016
Physics 214 Summer 2016
5
Series circuit
If we add more light bulbs in the circuit in series the total resistance
increases and the current will be reduced. The current is the same
in all parts of the circuit and the voltage difference across each
light is the same
ε = I(Rbattery + R + R + R)
The voltage difference across any light = IR
4/29/2016
Physics 214 Summer 2016
6
Parallel circuit
The current q at A
divides into three and
then recombines at B
In the circuit shown the voltage difference across each light is the same
and the total current is the sum of the three currents
I = I1 + I2 + I3 and since a current = ∆V /R
∆VAB /Rcircuit = ∆VAB /R1 + ∆VAB /R2 + ∆VAB /R3
and
1/Rcircuit = 1/R1 + 1/R2 + 1/R3
That is the total resistance of the circuit is smaller than any of the single
resistances. It is also true that if one bulb fails the other two will stay lit
at the same brightness.
4/29/2016
Physics 214 Summer 2016
7
What is voltage?
We have seen that the definition of voltage is
∆V = ∆PE/q when a charge q is moved in an electrical force field.
So energy is stored as potential energy as a positive charge is
moved in the opposite direction to E or a negative charge is
moved in the same direction as E.
If we move a positive charge toward a positive charge potential
energy and ∆V increase or if we move a negative charge away
from a positive charge.
Just as in the gravitational field there are only differences in PE.
So normally we use the term ∆V. But very often for circuits we
choose one point, usually the negative terminal, to be zero and
then instead of ∆V we just use V.
When charge is free to move, that is positive charge moving to a
lower voltage or negative charge moving to a higher voltage the
PE will transform into KE just like dropping something. In a simple
circuit with resistance this KE is turned into heat and light so
there is a voltage drop across every element in the circuit.
4/29/2016
Physics 214 Summer 2016
8
Voltage drop
If we have a circuit with many different
resistors then there is a voltage drop
across each resistor and there is also a
voltage drop for the whole circuit.
Current only flows if there is a voltage
difference and in a time t charge q
passes through the resistor.
I = q/t and ∆V = IR
Case1
I = 6/60 = 1/10
∆V15 = 15/10 ∆V20 = 20/10 ∆V25 = 25/10
Case 2
I = 12/8 I24 = 12/24
Case 3
I = VAB/5.5 ∆V31 = 3I/2 ∆V3 = 3I ∆V33 = 3I/3
4/29/2016
Physics 214 Summer 2016
9
Series plus parallel
A
I
B V
B
A
V
R1
R2
R3
I
R4
R4
B
VAB = I1R1
VAB = I2R2
VAB = I3R3
I = I1 + I2 + I3
4/29/2016
Rparallel
Physics 214 Summer 2016
V4 = IR4
V = VAB + V4
V = I(R4 + Rparallel)
10
Measuring current and Voltage
It is often very important to know
the current in a circuit or the
voltage difference between two
points.
A hand held meter is very useful to
test batteries or a circuit.
An ammeter is a device inserted
into a circuit. The resistance of an
ammeter is very small so as to
minimize the effect on the circuit.
A voltmeter is attached in parallel
and V is found by measuring the
current and V = ImeterRmeter. The
resistance has to be much larger
than the circuit resistance so that
the current is very small and does
not disturb the main circuit.
4/29/2016
Both meters actually
measure current
Physics 214 Summer 2016
11
Power
Electromotive force or voltage difference between two points is
the difference in potential energy/unit charge.
So the energy delivered if charge q is transferred is
energy = ∆Vq
power = ∆Vq/t = ∆VI watts
For any voltage difference ∆V and current I.
power = ∆VI
For circuits that obey OHM’s law ∆V
=IR
P = ∆VI = I2R = ∆V2/R watts
The power used appears as heat or light
Practical Unit
1 kilowatt
Energy Unit
1 kilowatt hour = 1000 x 3600 joules
4/29/2016
Physics 214 Summer 2016
12
Summary Chapter 13
I = q/t amperes (coulombs/sec)
OHM’s Law
R = ∆V/I
ohms
I = ε/(Rcircuit + Rbattery)
P = ∆VI = I2R = (∆V)2/R
4/29/2016
watts
Physics 214 Summer 2016
13
Mini Review
In order for a current to flow there must
be a voltage difference. The voltage
difference is the amount of potential
energy/unit charge. As the charge,
usually electrons move along a
conductor they lose energy
continuously so the voltage along the
conductor decreases continuously. For
a simple circuit we ignore losses in the
conductor but there is a voltage drop
across every resistor.
V
∆V
I = q/t
Energy liberated is qV. So the power is qV/t = IV
For a resistor V = IR and power = I2R = VI = V2/R
4/29/2016
Physics 214 Summer 2016
14
Simple circuits
The current is the same everywhere
R = R1 +R2 + R3
V = V1 + V2 +V3
V
V is the same across all resistors
V = IRcircuit
1/Rcircuit = 1/R1 + 1/R2 + 1/R3
I1 = V/R1 I2 = V/R2
I3 = V/R3
R = R1 + R2 + R3
Vab = IR
R2
4/29/2016
R1
R3
Physics 214 Summer 2016
15
Our World
4/29/2016
Physics 214 Summer 2016
16
Transmission
In the distribution of electric power the
goal is to deliver to the user as large a
fraction as possible of the generated
power.
Practical cables have a specific
resistance so the power losses
will be I2Rcable and we need I to be as
small as possible.
But we also need the delivered power
P = VsourceIsource
to be as high as possible, therefore,
the electrical power is distributed at
very high voltage and low current.
The voltage is reduced from
250,000volts to 220volts for
households by using a transformer.
The current increases by the same
factor since for an ideal transformer no
power is lost. Transformers are the
dominant reason electrical
transmission is alternating current
4/29/2016
i
Rcable
Vuser
Vsource
Ruser
i
Vuser = Vsource – IRcable
Puser = iVuser = iVsource – i2Rcable
Physics 214 Summer 2016
17
Household appliances
Household circuits are wired in parallel so that when more than
one appliance is plugged in each sees the same voltage and can
get the required current.
As we plug in more and more appliances the current in the circuit
increases and the I2R losses could cause a fire. This is why we
have fuses and why major appliances use 220 volts and many
parts of the world use 220 volts for all household use.
As many people turn on appliances (air conditioners) the grid has
to supply more power by increasing the current.
Puser = iVuser = iVsource – i2Rcable
This results in a higher fraction of the power being lost in the
cable
In cases of very heavy load the power station reduces the
transmission voltage resulting in a “brown out” and in extreme
cases there are rolling blackouts.
4/29/2016
Physics 214 Summer 2016
18
Worked Questions and Problems
4/29/2016
Physics 214 Summer2016
19
Questions Chapter 13
Q3 In a simple battery-and-bulb circuit, is the electric current that
enters the bulb on the side nearer to the positive terminal of the
battery larger than the current that leaves the bulb on the opposite
side?
The current is the same
Q4 Are electric current and electric charge the same thing?
No. electric current is the flow of charge I = q/t
4/29/2016
Physics 214 Summer 2016
20
Q6 Consider the circuit shown. Could we increase the brightness of
the bulb by connecting a wire between points A and B? Explain.
A
•
Є
B
•
No. The voltage drop from A to B is stiil the same so the
current through the bulb does not change.
4/29/2016
Physics 214 Summer 2016
21
Q7 Two circuit diagrams are shown. Which one, if either, will
cause the light bulb to light? Explain you analysis of each case.
Open
Switch
1.5 V
1.5 V
(b)
In a) the battery is not connected. In b) the bulb is lit
(a)
4/29/2016
Physics 214 Summer 2016
22
Q11 A dead battery will still indicate a voltage when a good
voltmeter is connected across the terminals. Can the battery still be
used to light a bulb?
A battery has internal resistance so although one can measure an
open circuit voltage when connected to a circuit the voltage will
drop and the current flow will be very low.
Q12 When a battery is being used in a circuit, will the voltage
across its terminals be less than that measured when there is no
current being drawn from the battery? Explain.
The voltage will be less because of the voltage drop due to the
internal resistance
4/29/2016
Physics 214 Summer 2016
23
Q13 Two resistors are connected in a series with a battery as
shown in the diagram. R1 is less than R2.
A. Which of the two resistors, if either, has the greater current
flowing through it? Explain.
B. Which of the two resistors, if either, has the greatest voltage
difference across it? Explain.
R1
Є
R2
4/29/2016
The current is the same in
both. Since V = IR the greatest
voltage drop will be across R2
Physics 214 Summer 2016
24
Q14 In the circuit shown below, R1, R2, and R3 are three resistors of
different values. R3 is greater than R2, and R2 is greater than R1. Є
is the electromotive force of the battery whose internal resistance
is negligible. Which of the three resistors has the greatest current
flowing through it?
R1
Є
R2
R3
I3 = I1 + I2 so I3 is the largest
4/29/2016
Physics 214 Summer 2016
25
Q15 In the circuit shown in question 14, which of the three
resistors, if any, has the largest voltage difference across it?
V = IR and both I and R are the largest for R3
Q16 If we disconnect R2 from the rest of the circuit shown in the
diagram for question 14, will the current through R3 increase,
decrease, or remain the same?
The resistance of the
circuit will increase
so the current
through R3 will
decrease.
R1
Є
R2
R3
4/29/2016
Physics 214 Summer 2016
26
Q18 In the circuit shown, the circle with a V in it represents a
voltmeter. Which of the following statements is correct? Explain.
A. The voltmeter is in the correct position for measuring the voltage
difference across R.
B. No current will flow through the meter, so it will have no effect.
C. The meter will draw a large current.
+
Є
+
R
V
The voltmeter is in the correct position. Current will flow
through the meter but the current will be very small because the
resistance is very high.
4/29/2016
Physics 214 Summer 2016
27
Q19 In the circuit shown, the circle with an A in it represents an
ammeter. Which of these statements is correct? Comment on each.
A. The meter is in the correct position for measuring the current
through R.
B. No current will flow through the meter, so it will have no effect.
C. The meter will draw a significant current from the battery.
+
Є
+
R
A
The meter is in the wrong position. A large current will flow
because the meter resistance is very low.
4/29/2016
Physics 214 Summer 2016
28
Q21 Is electric energy the same as electric power?
Power is the rate at which energy is used. Your electrical bill
is for the total energy you use.
Q22 If the current through a certain resistance is doubled, does the
power dissipated in that resistor also double?
P = I2R so the power increases by a factor of 4
Q23 Does the power being delivered by a battery depend on the
resistance of the circuit connected to the battery?
Yes because increasing the resistance lowers the current
4/29/2016
Physics 214 Summer 2016
29
Q29 Suppose that the appliances connected to a household circuit
were connected in series rather than in parallel. What disadvantages
would there be to this arrangement?
First if one failed then all the appliances will not function. Secondly the
current will be determined by the whole string of appliances by the
voltage drop from end to end and will depend on the number of
appliances. Each appliance requires the same voltage drop and this is
what happens when they are connected in parallel.
4/29/2016
Physics 214 Summer 2016
30
Ch 13 E 14
24  resistor has voltage difference 3V across leads.
a) What is the current through the resistor?
b) What is the power dissipated in resistor?
3V
24 
a) V = IR
I = V/R = 3/24 = 0.125A
b) P = IV = V2/R = (3)2/24 = 0.375W
4/29/2016
Physics 214 Summer 2016
31
Ch 13 E 16
A toaster draws current = 7A on a 110-V AC line
a) What is the power consumption of the toaster?
b) What is the resistance of the heating element
in the toaster?
→ I = 7A
a) P = IV = 7.110 = 770 W
b) V = IR, R = V/I = 110/7 = 15.7 
4/29/2016
Physics 214 Summer 2016
110 V
R
32
Ch 13 CP 2
Three 30- light bulbs connected in PARALLEL to 1.5
V battery with negligible internal resistance.
a) What is the current through the battery?
b) What is the current through each bulb?
c) If one bulb burns out, does the brightness of the
other bulbs change?
1.5 V
4/29/2016
R
R
R
Physics 214 Summer 2016
R = 30 
33
Ch 13 CP2 cont.
a) 1/Rp = 1/R1 + 1/R2 + 1/R3 = 1/30 + 1/30 + 1/30 = 1/10
Rp = 10 
V = ItRp , It = V/Rp = 1.5/10 = 0.15 A
b) V = IR, I = V/R = 1.5/30 = 0.05 A
Notice that total current, It, through the battery is the sum of
currents through each bulb.
It = 3(I) = 0.15 A
c) Brightness of remaining two bulbs do not change. Instead, load
on the battery is reduced. Each remaining bulb still feels 1.5 V.
So, each remaining bulb still draws 0.05 A of current. Since,
P=IV, each remaining bulb still outputs same power. This is
benefit of hooking circuits in parallel.
4/29/2016
Physics 214 Summer 2016
34
CH 13 CP 4
R
A
R
R
R
R
R
B
R = 3
a) What is the resistance of each parallel combination?
b) What is the total resistance between A and B?
c) 6V voltage difference b/w A and B. What is the current through
the entire circuit?
d) What is the current through each resistor in three-resistor parallel
combination?
a) Two-resistor parallel combination
I/RP2 = 1/R + 1/R = 1/3 + 1/3 = 2/3,
RP2 = 3/2 
b) Three resistor parallel combination
I/RP3 = 1/R + 1/R + 1/R = 1/3 + 1/3 + 1/3 = 1, RP3 = 1 
4/29/2016
Physics 214 Summer 2016
35
CH 13 CP 4 cont
R
R
A
R
R
C
R
B
R
3/2 
3
R = 3
1
C
A
B
9/2 
C
1
B
A
c) V = I RT, I = V/RT = 6/(11/2) = 12/11 = 1.09 A
4/29/2016
Physics 214 Summer 2016
36
CH 13 CP 4
After accounting for AC part of circuit we have left
C
d) VAC = I RAC = (1.09) (9/2) = 4.91 V
VCB = V – VAC = 1.09 V
I = V/R = 1.09/3 = 0.37 A
V=1.09 V
R
R
R
B
This answer could have been easily noted as follows: We calculated 1.09 A
flowing from A to B. That means, at point C, there is 1.09 A. Now this
current must make it to point B, but there are 3 different paths. Since each
path is of equal resistance, the current will equally choose all three paths.
thus any one path has 1.09/3 A = 0.37 A of current
4/29/2016
Physics 214 Summer 2016
37