Transcript Optimum Power

Ideal Source
In an ideal source, all the internal energy or
power, e.g., the chemical reaction in a
battery, is available to the load without any
“losses” in the form of heat.
An ideal battery is shown symbolically as
• An ideal voltage source (battery) in
series with a load (light bulb):
• Find the following:
– The current through the bulb whose
resistance is Rb = 2.5 Ω.
– The power consumed by the light bulb in W.
– The power supplied by the ideal battery in W.
Solutions
The current drawn by the bulb is
I = Vsource / Rbulb = 1.5 / 2.5 = 0.6 A
The power consumed by the bulb is
Pbulb = I2 Rbulb = (0.6 A)2 2.5 = 0.9 W
The power supplied by the ideal battery is
Psource = V I = 1.5 (0.6) = 0.9 W
From the law of the conservation of energy,
Psource = Pbulb
Real Source
• Real sources have an internal resistance.
• This is not a visible resistor like you used in
the lab, but it is a “model resistor” added to
the circuit to analyze the source realistically.
• The internal resistance causes energy loss
(dissipated heat for example) and causes
the source to provide less energy to the
load.
• Ideal source
(battery) in series
with a load (light
bulb)
• Real source (battery
with internal resistance)
in series with a load
(light bulb)
Repeat the previous exercise with a real 1.5 V
battery source. Assume the battery’s internal
resistance to be 0.3Ω.
Find the following:
•Current through the bulb
•Power consumed by the bulb
•Power supplied by the ideal portion of the battery
•Power consumed by the battery’s internal resistance.
Req = Rinternal + Rb
= 0.3 + 2.5 = 2.8 Ω
• Current: I = 1.5 V / 2.8 Ω = 0.536 A
• Power consumed by the bulb:
Pbulb = I2 Rbulb = (0.5357 A)2 2.5 Ω = 0.717 W
• Power provided by the ideal portion of the battery:
Psource = V I = 1.5V(0.5357 A) = 0.803 W
• Power consumed by the battery’s internal
resistance:
Pinternal = I2 Rinternal = (0.5357 A)2 0.3 Ω = 0.086 W
Comparison with Previous
Results
Ideal Source:
• I = 0.6 A
• Pbulb = 0.9 W
• Psource = 0.9 W
Balance of power:
Psource = Pbulb
0.9 W = 0.9 W
Real Source:
• I = 0.536 A
• Pbulb = 0.717W
• Psource = 0.803W
• Pinternal = 0.086 W
(dissipated as heat)
Balance of power:
Psource = Pinternal + Pbulb
0.803W = 0.086W + 0.717W
Connection to Real Life
• As batteries age, their internal
resistance increases.
• That means, even if you measure 1.5 V
across the terminals with a meter, when
you connect it to a load, it will provide a
lower net voltage
In-Class Activity
• Given this circuit where the
real battery has been
connected for a while.
• Assume the bulb needs at least 0.4 A to light.
• What will the internal resistance be when the
bulb goes dark?
• What voltage will be measured across the bulb
at that time?
Question
Can the bulb be replaced by a different bulb to
maximize the power delivered to the bulb for a
given battery voltage and internal resistance (Ri) ?
In other words, if you replace the bulb with a
symbolic resistance Rb as shown here, can you find
the value of Rb for maximum bulb power?
Start by expressing the
power absorbed by the
bulb in terms of V and both
R’s.
Power Absorbed by Bulb
• Pb = Ib * Vb
• Ib =
(in terms of Rb, Ri and Vs)
• Vb =
(in terms of Rb, Ri and Vs)
• Therefore, Pb =
The power absorbed by the light bulb is given
by
Rb
2
Pb 
V
2 s
Rb  Ri 
where
Pb = bulb power (W)
Rb = bulb resistance (Ω)
Ri = internal resistance (Ω)
Vs = battery voltage (V)
Is there a value of Rb that maximizes Pb?
Graphical Approach
In-Class Activity
• Use Excel to generate a table of Pb
as a function of Rb for V = 1.5 V and
Ri = 0.3 Ω.
• Choose values of Rb from 0 Ω to
0.8 Ω in 0.05 Ω steps.
• Plot your data using the “Scatter
Plot”
Names:
Team:
Date:
R_bulb
0
P_bulb
Vs = 1.5 V
Ri = 0.3 Ω
0.05
0.1
0.15
0.2
What is Pb for Rb = 0?
What is Pb for Rb → ∞ ?
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
From the curve, identify
what value of Rb makes
Pb maximum.
Analytical Approach
• In calculus you’ve learned about
derivatives.
• The first derivative of a function is a
measure of its slope (rise/run).
• When we take the derivative of our
power function with respect to Rb, we
find that the slope is zero at the
maximum.
• The value of Rb at that maximum will be
R b = Ri