Power and Power
Measurement
ENGR 10 – Intro to Engineering
College of Engineering
San Jose State University
(c) P.Hsu 2009
Electricity is a convenient way of
transmitting energy -- by wires
Sourcing
Energy
Energy
transmission
Receiving
Energy
Voltage, Current
Rate of energy transmission from left to right
Power (J/s or Watt) = Voltage (Volts) x Current (Amperes)
(c) P.Hsu 2009
In this circuit, energy is carried by the wires from the
battery to the light bulb.
Current (A)
+
Energy
flow
-
Sourcing
energy
Voltage
(V)
Receiving
energy
(c) P.Hsu 2009
A hydraulic analogy
High
pressure
Hydraulic Motor
Handpump
Power
flow
Sourcing
power
Low
pressure
Receiving power
(c) P.Hsu 2009
`
Think about the hydraulic analogy
Voltage is the force (or pressure) that forces the
positive electrical charge to flow (current)
through a circuit.
• Voltage is measured in ‘Volts’..
+
+
1.5v
-
(c) P.Hsu 2009
Light
bulb
Voltage is a relative quantity. The “+” and “-”
symbols in a circuit diagram denote the relative
voltage (pressure) between two wires.
+
+
1.5v
-
Light
bulb
-
In this circuit, the voltage on the wire marked by ‘+’ is
1.5v higher than that the wire marked by ‘-’.
(c) P.Hsu 2009
Think about the hydraulic analogy
Current is the flow rate of positive electrical charge.
• Current is measured in Ampere (or Coulomb per second)
+
+
Light
bulb
-
Current always flows in complete loop.
(c) P.Hsu 2009
Analogy
Electrical Circuit
Voltage (V)
Current (A)
(Charge flow rate)
+
-
Hydraulic system
 Pressure (psi)
 Fluid flow rate (gal/sec)
Pump
+
Light
bulb
Hydraulic
Motor
(c) P.Hsu 2009
Ohm’s Law
I
+
+
V
-
R
-
If the ‘circuit’ is a simple resistor, the voltage, current, and
the resistance of the resistor is related by Ohm’s Law:
V
I
R
Resistance is measured in Ohm (Ω)
(c) P.Hsu 2009
I
+
+
V
-
R
-
In the above circuit, V = 12 volts, R=3Ω.
What is the current I =?
I = V/R = 12/3 = 4 amps
(c) P.Hsu 2009
Question
Voltage in an electrical circuit is
analogous to what physical quantity in a
hydraulic system?
(a) Fluid flow rate
(b) Speed of the hydraulic motor
(c) Volume of the hydraulic fluid
(d) Pressure
(e) Speed of the pump
(c) P.Hsu 2009
Clicker Question
From the values given in the diagram below, what is
the resistance (R) of an Ipod (I=V/R)?
0.1A
+
+
3v
-
(a)
(b)
(c)
(d)
(e)
-
0.3 Ω
3.1 Ω
0.03 Ω
0.9 Ω
30 Ω
(c) P.Hsu 2009
IPod
“Open Circuit” Case
I=0
+
+
Light
bulb
V
-
Power = VI = V0 = 0
(c) P.Hsu 2009
High
pressure
No hydraulic
fluid flow
Hydraulic Motor
Pump
`
No
Power
Power = Pressure  0 = 0
(c) P.Hsu 2009
Please put all cells phones
away
Announcements
• HW #1 will be posted on E10 web site
today –due at BEGINNING of class
Wednesday, September 18.
• Check in Canvas in “Solar Module” for
additional information on series and
parallel circuits.
16
Energy source in a circuit
Output voltage from a typical energy source stays the same
regardless of its output current (I) so long as it is less than the
source’s rated current value. Within this range, the output
current I only depends on the resistance (R) of the equipment
(i.e., the load) connected to the source.
I
I = V/R
+
+
V
R
-
and the power is P = V*I = V2/R
-
The smaller the resistance, the higher the output current (and
power), but V stays the same as long as the current is less
than the rated current of the source.
(c) P.Hsu 2009
Load-down effect
When equipment tries to draw higher than the rated current
of the source, the voltage will sag. This condition is called
“Load down”.
+
+
V=9
-
-
+
V=9
-
No load
+
Normal load
+
+
V=8.2
-
-
Overload (Voltage is loaded down.)
(c) P.Hsu 2009
The maximum output power of a source can
be determined by varying load resistance
from high to low.
Normal
operating
range.
The voltage
is loaded
down.
Maximum
output
power of
the source
Maximum
output
current
Voltage
(Volt)
Current
(Amp)
Power
(W)
Equivalent loading resistance
(R vary from high to low)
10
0
0
infinite (open circuit)
10
2
20
5 Ω (Light load)
9.8
4
39.2
2.5 Ω (Normal)
9.4
5
47
1.9 Ω (Normal)
8.5
6
51
1.4 Ω (MAX POWER)
7.2
7
50.4
1 Ω Overload
5.2
8
41.6
0.65 Ω (Overload)
3.0
9
27
0.3 Ω (Overload)
0
10
0
0 (shorted)
(c) P.Hsu 2009
The table shows power output at different loading
conditions. At what loading resistance can the source
provide the largest amount of energy?
(a) RL=0
(b) RL=1.4 Ω
(c) RL= 5 Ω
(d) RL=infinite
(e) Insufficient information
Voltage
(Volt)
Current
(Amp)
Power
(W)
10
0
0
infinite
10
2
20
5Ω
9.8
4
39.2
2.5 Ω
9.4
5
47
1.9 Ω
8.5
6
51
1.4 Ω
7.2
7
50.4
1Ω
5.2
8
41.6
0.65 Ω
3.0
9
27
0.3 Ω
0
10
0
0 (shorted)
(c) P.Hsu 2009
Equivalent
loading
resistance
(RL)
Voltage vs. current and Power vs. current
of a source
Voltage
Voltage is “loaded down”
for high load current.
light load
Power
(W)
Max power
heavy load
Load Current (Amp)
(Depending on the load)
Max power point
power curve
I (amp)
Max power load current
(c) P.Hsu 2009
Key Concepts
 A source’s output voltage drops when ‘too much’ current
is drawn from it (namely, the “load down” effect.)
 While it is possible to increase output current even when
the voltage is loaded down, the output power (voltage x
current) stops increasing due to decreasing voltage.
 A source outputs its maximum power at a specific
output voltage and current. (A very important parameter
for characterizing solar cell).
based on notes of P. Hsu 2007
Question
Which of the following statements is false?
A) A source’s voltage varies with output current.
B) A source’s current varies with output power.
C) A source’s output current varies with load
D) A source’s maximum output power is its maximum
output voltage times its maximum output current.
Pmax = Vmax x I max
(c) P.Hsu 2009
Solar Cell Lab
I
+
V
_
Smaller
R
POT
To the circuit, the solar cell is an energy source.
A variable resistor (potentiometer or POT) is used as the
load in experimentally determining the V vs. I curve of a solar
cell.
The same procedure is used in the wind turbine experiment.
24
Solar Cell lab basic setup
Review the manual before coming to the Lab
Voltage Current Power
Power meter
Solar Cell
POT
By changing the POT resistance, the current drawn from the
solar cells can be varied.
The output V, I, & power is recorded at each current level, from
high to low.
You will plot V vs. I and Power vs. I from the data.
(c) P.Hsu 2009
Solar cell voltage, current, and power output
V vs. I curve
Voltage
Power
(W)
Max power point
I (amp)
Current
(Amp)
• For most applications (for example, for a laptop
computer,), we only draw as much current (or power)
as needed from the source.
• For solar energy generation, however, we want to
draw as much power as it can generate. Therefore,
it is important to find this maximum power point.
(c) P.Hsu 2009
Data
Collection
Form
27
Solar Cells in Series
Two or more cells can be connected in series.
• The combined output voltage is the sum of cell’s output voltages.
• The combined output current (and the current rating) is the same
as for a single cell.
• The combined power (and the power rating) is the sum of the
individual cell’s power P = (V1+V2)I1 = (V1+V2)I2 = (V1+V2)I .
I2
Cell 2
I = I1 = I2
V2
motor
V =V1+V2
I1
Cell 1
V1
28
Solar Cells in Parallel
Two or more cells can be connected in parallel.
• The combined output current (and current rating) is the sum of cell’s
output currents
• The combined output voltage is the same for each cell.
• The combined power (and power rating) is the sum of the individual
cell’s power P = V(I1+I2) = VI .
I = I1 + I2
I1
I2
Same V
Cell 1
Cell 2
motor
29
Series and Parallel Connection
Series --- Voltage is the sum and current (and current
capacity) is the same.
I2
I = I1 = I2
Cell 2
V2
motor V =V1+V2
I1
Cell 1
V1
Parallel --- Voltage is the same and current (and current
capacity) is the sum.
I1
I = I1 + I2
I2
Same V
Cell 1
Cell 2
motor
30
Solar Cells in parallel
Example: A solar cell is rated 3 volts @ 3 amp. How many cells are
required to power a circuit that need 3 volts @ 5 amp?
Answer: Two cells in parallel. This circuit provides 3v and has a rating of
6 amps (therefore, it can certainly supply 5 amps).
I = I1 + I2
I1
I2
Same V
Cell 1
Cell 2
motor
Note: A source rated 3v @ 3 amp outputs 3v regardless of how much
current is drawn from it by the load as long as it is below 3
amp.
31
Clicker Question
There are 2 solar cells. Each one is
rated 1 volt voltage @ 2 amp. To power
a load that needs 1 volt @ 3 amp, how
should these two cells be connected?
•
•
•
•
•
a) In parallel
b) In series
c) back-to-back
d) by glue
e) there is no way
What is the output voltage in this case?
(a) 1V (b) 2V
(c) 3V (d) 4V
(d) 5V
Clicker Question
There are 2 solar cells. Each one is
rated 3 volt @ 2 Amp.
To obtain output of 4 volt and 4 Amp,
how should these two cells be
connected?
•
•
•
•
•
(c) P.Hsu 2009
a) In parallel
b) In series
c) back-to-back
d) by glue
e) there is no way
Group Problem – 4 Cells
•Work in groups of 3 people
•Write your name and student ID in the paper.
•Draw a picture of your solution
•Turn in at the end of the class
There are 4 solar cells. Each one is capable of output of 1
volt voltage and 2 Amp current.
a) To obtain output 2 volts and 8W, how should these 4 cells
be connected ?
b) What is the current output from the configuration you
made?
(c) P.Hsu 2009
Pros and Cons of Electrical Power
Pros: Convenience for transmission and distribution,
clean, easy to control, easily transformed into many
forms of power (mechanical, heat, light, etc.)
Cons: Requiring power conversion equipment (solar
panels, heaters, motors, etc.). There is always some
conversion loss.
39% of the power used in the US is converted into electric
form first.
The pros clearly outweighs the cons.
(c) P.Hsu 2009
Power Conversion
Rate of energy input
= Psun (J/S)
+
-
Solar Panel
Current (I)
Voltage (V)
Pe =VI
The solar panel converts the power from sunlight to
electric power. If 100% of the power from the sun
is converted, the following equality holds.
Psun = Pe = V*I
In reality, however, only a fraction of the sun power
(typically 15%) can be converted.
(c) P.Hsu 2009
Power Conversion Efficiency
Input
Power
Power Conversion
equipment
Output
Power
Wasted
Power
In an energy conversion process, the ratio between the
desired output power and the input power is the efficiency
of this process.
Efficiency 
Energy output that we care about
Total Input Energy
In the process, some power is inevitably converted to a form
that we don’t care about (such as heating of the panel).
(c) P.Hsu 2009
Rate of energy input
= Psun (J/S)
+
-
Solar Panel
Current (I)
Voltage (V)
Pe =VI
Power from the sun on earth at noon is about 1350W per
m2. For a 2 m2 solar panel, with an efficiency of 15%, the
output power is
1350 W/m2 x 2 m2 x 0.15 = 405 W.
This amount can be achieved if the loading condition is
such that the output voltage and current is at the maximum
power output condition. (c) P.Hsu 2009