Download εmf = I(R+r)

March 04, 2013
Emf and internal resistance
Emf and internal resistance
ΔVof battery =
or
ε
mf
ε - Ir
= I(R+r)
(for this particular series circuit)
As the current in the circuit
increases the voltage, supplied to
the external circuit, by the battery
goes down.
Application
· temporary dimming of lights with
large current draw
Sources of electromotive force?
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Itotal = 3.14 A
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Which Bulb is Brighter - Concepts
IA
What determines brightness?
IB
If all bulbs have the same
resistance, order the bulbs in
order of brightest to least bright?
Can resistance of a bulb change?
explain
ID
IC
What happens to order of
brightness if bulb B is removed?
Will bulb D glow? why or why
not? explain
page 598
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Adjustable or Variable Resistor in a circuit
adjustable or variable resistor (dimmer switch)
ΔVbattery = 120 volts
Variable resistor = Rv
Resistance of Light Bulb = 220 Ω
Light Bulb - that emits light of varying brightness, due to the
variable resistor (that limits current flow)
What current would be required in the circuit to give the light bulb an output of 60 watts?
What resistance would the variable resistor have to be?
If the variable resistance doubled, what would happen to the current flow through the circuit? and the power output of
the light bulb? (brightness)
If I want to add a stereo to this circuit that required 120 volts all the time (even if I adjusted the variable resistor), how
would I do it? explain?
Adjustable or Variable Resistor in a circuit
adjustable or variable resistor (dimmer switch)
ΔVbattery = 120 volts
Variable resistor = R
v
Resistance of Light Bulb = 220 Ω
Light Bulb - that emits light of varying brightness, due to the
variable resistor (that limits current flow)
What current would be required in the circuit to give the light bulb an output of 60 watts?
p = I2R ( of light only)
find I
What resistance would the variable resistor have to be?
V = IR eq , find Req for full circuit, it is larger than 220,
If the variable resistance doubled, what would happen to the current flow through the circuit? and the power output of
the light bulb? (brightness)
If I want to add a stereo to this circuit that required 120 volts all the time (even if I adjusted the variable resistor), how
would I do it? explain?
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Electrical Power and Mechanical Power
P = W/Δt
Rate at which work is done
or energy is used,
measured in watts (J/s)
Power = Work/time
Power can also be: P = F*v
Work = F*d
P = W/t = F*d/t or F*v
Work is done in raising an elevator upward. The power
consumption of the motor will be related to the rate at which
work is done or, in this case, the rate at which the elevator is
lifted.
P = N*m/s or J/s or w (watt)
Simple Resistor and Capacitor in Combination Circuits
If ΔV = 10 volts, R= 5 ohms and C = 4 Farad
switch
What is charged stored on capacitor? (while the
switch is closed)
R
V
C
What is the current through the resistor when
the circuit is closed?
What is the power of the bulb? (resistor)
When the switch is opened
How long would the light bulb "run" with the
energy stored in the capacitor at designed
power rating of the light bulb?
R
C
What about this orientation? It is different than your initial logical
thoughts.
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Power Loss in high voltage lines
17.34 (modified)
A high-voltage transmission line with a resistance of 0.31Ω/km carries a current of 1000 A. The line is at a
potential of 700kV at the power station and carries the current to a city located 160 km from the station.
a. What is the power loss due to the resistance of the line?
b. What total power is transmitted?
c. What is the fraction of the transmitted power does this loss represent? fractional loss = powerloss/ Powerin
d. What is the actual potential at the city (adjusting for loss)?
e. If the distribution cost (cost related to this energy loss above) is .026396 dollars/ kwhr, how much would it
cost to run this line for a month, (just for energy loss)?
f. How much energy is lost (distribution costs) in transporting the energy each month in kwhr and Joules?
Power Loss in high voltage lines
17.34 (with additional)
A high-voltage transmission line with a resistance of 0.31Ω/km carries a current of 1000 A. The line is at a
potential of 700kV at the power station and carries the current to a city located 160 km from the station.
a. What is the power loss due to the resistance of the line?
2
P = I R = 49,500,000 watts
b. What total power is transmitted?
P = IV = 1000A * 700,000 Volts = 700,000,000 watts
c. What is the fraction of the transmitted power does this loss represent? fractional loss = powerloss/ Powerin
50 MW/ 700 MW = 7.1 % (loss)
d. What is the actual potential at the city (adjusting for loss)?
current remains the same, energy per charge goes down P = IV, 650,000 W = 1000A *V, V = 650 kVolts
e. If the distribution cost (cost related to this energy loss above) is .026396 dollars/ kwhr, how much would it
cost to run this line for a month, (just for energy loss)?
.026396 dollars/kwhr * 50,000 kw*24hr*30 days/month = .95 million dollars/month (lost)
f. How much energy is lost (distribution costs) in transporting the energy each month in kwhr and Joules?
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Kirchoff's Laws
Junction rule demonstrates
conservation of ?
The loop rule demonstrates
conservation of ?
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