Transcript electrical energy & power

Electrical Energy and Power
AP PHYSICS C
MONTWOOD HIGH SCHOOL
R. CASAO

For a simple circuit consisting of a battery connected to a
resistor R, the positive terminal of the battery (the longer
plate) is at the higher potential while the negative terminal
(the shorter plate) is at the lower potential.


For a positive charge dQ moving
around the circuit from point a
thru the battery and resistor
and back to point a, point a is a
reference point that is
grounded and its potential is
0 V.
As the charge moves from a to
b thru the battery, its electrical
potential energy increases by
an amount V·dQ while the
chemical potential energy
decreases by the same amount.
 As
the charge moves from c to
d thru the resistor, it loses this
electrical potential energy as it
undergoes collisions with
atoms in the resistor,
producing thermal energy.
 Neglect the resistance of the
connecting wires, so there is
no energy loss for paths bc
and da.
 When the charge returns to
point a, it must have the same
potential energy (O) as it had
at the start.

The figure on the right
represents a circuit
element with potential
difference Va – Vb = Vab
between its terminals and current I passing
thru it in the direction from a to b.
 The
circuit element could be a battery, a resistor, or
something else.
 As charge passes thru the circuit element, the
electric field does work on the charge.
 As an amount of charge q passes thru the circuit
element, there is a change in potential energy
equal to q·Vab.
If q > 0 and Vab = Va – Vb is positive, potential energy
decreases as the charge falls from potential Va to lower
potential Vb.
 The moving charges don’t gain kinetic energy because the
rate of charge flow (current) out of the circuit element
must be the same as the rate of charge flow into the
element.
 Instead, the quantity q·Vab represents electrical energy
transferred into the circuit element. This energy is often
converted into another form of energy (thermal, EM).


If the potential at b is higher than the potential at a,
Vab is negative and a net transfer of energy out of
the circuit element occurs.

The element acts as a source of energy (a battery).
 The
circuit element delivers electrical energy into the
circuit to which it is attached.
 A battery usually converts chemical energy into
electrical energy and delivers it to the external circuit.
 q·Vab.can denote either a quantity of energy delivered to
a circuit element or a quantity of energy extracted from
that element.

In electric circuits, the rate at which energy is
delivered to or extracted from a circuit element is
important.
 If
the current thru the element is I, then in a time
interval dt an amount of charge dQ = I·dt passes thru the
element.
 The
potential energy change for this amount of
charge is Vab·dQ = Vab·I·dt.
 Dividing by dt gives us the rate at which energy is
transferred either into or out of the circuit element.
 The time rate of energy transfer is power P: P = Vab·I.
 Unit: Watt (J/s = V·A)

Power Input to a Pure Resistance:
 If
the circuit element is a resistor,
the potential difference is
Vab = I·R.
 The electrical power delivered to
the resistor by the circuit is:
Vab
P  Vab  I  I  R 
R
2
2
 The
potential at a (where the current enters the
resistor) is always higher than that at b (where the
current exits).
 Current enters the higher potential terminal of the
device and transfers electric potential energy into
the circuit element.
 What happens to the energy?
 The
moving charges collide with atoms in the resistor and
transfer some of their energy to these atoms, increasing
the internal energy of the material.
 Either the temperature of the resistor increases or there is
a flow of heat out of it, or both.
 Energy
is dissipated in the resistor at the rate I2·R.
 Every resistor has a power rating, the maximum power the
resistor can dissipate without becoming overheated and
damaged.

Power Output of a Source:
 For
the car battery connected to a headlight, point a
is at higher potential than point
b, so Va > Vb and Vab is
positive.
 The current I is leaving the
source at the higher
potential terminal and
energy is being delivered to
the external circuit.
 The
rate of energy delivery to the circuit is: P = Vab·I.
 For a source that can be described by EMF and an
internal resistance r: Vab = EMF - I·r
 Multiplying the equation by I: P = Vab·I = EMF·I - I2·r
 EMF represents the work per unit charge the battery
performs on the charges by the nonelectrical force in
the battery that pushes the charges from point b to
point a in the battery.
 In a time dt, a charge dQ = I·dt flows thru the battery;
the work done on it by the nonelectrostatic force is
EMF·dQ = EMF·I·dt.
 EMF·I is the rate at which work is done on the
charges by the battery and represents the rate of
conversion of nonelectrical energy to electrical energy
within the battery.
 I2·r
is the rate at which
electrical energy is
dissipated in the internal
resistance of the source.
 The difference EMF·I - I2·r
is the net electrical power
output of the battery, the
rate at which the battery
delivers electrical energy
to the external circuit.

Power Input to a Source:
In a car, the alternator is
connected to the battery.
 The battery is charged by
the alternator.

 The
EMF of the alternator is larger than the EMF of
the battery.
 The current I in the circuit travels from the alternator
to the battery, from the larger EMF to the smaller
EMF.
 Because of the reversal of current, for the battery:
Vab = EMF + I·r and P = Vab·I = EMF·I + I2·R
 Work is being done on the battery and there is a
conversion of electrical energy into nonelectrical
energy in the battery at a rate EMF·I.
 The I2·R term is the rate of dissipation of energy in
the internal resistance of the battery.
 The sum EMF·I + I2·R is the total electrical power
input to the battery.
 This
also happens when a rechargeable battery (a
storage battery)is connected to a charger.
 The charger supplies electrical energy to the battery;
part of it is converted to chemical energy, to be
reconverted later, and the rest is dissipated (wasted)
in the battery’s internal resistance, warming the
battery and causing heat to flow out of it.

Power companies sell energy in kW·hr;
Energy = power·time; E = P·t
 Energy
is usually expressed in Joules, but electric
companies use kW·hr.
1 kW·hr = 3.6 x 106 J
 The
cost of operating an electrical device is:
$


cost  E (kW  hr)  

 kW  hr 

Energy Transmission thru Power Lines:
 When
transporting electrical energy thru power
lines, power companies want to minimize power
transformed into internal energy in the lines and
maximize the energy delivered to the other end of
the transmission line.
 Because
P = I·V, the same amount of power can be
moved either at high currents and low potential
differences or at low currents and high potential
differences.
 Power companies choose to move electrical energy at
low currents and high potential differences for
economic reasons:
 Copper
wire is very expensive and so it is cheaper to use
high resistance wire (wire having a small cross-sectional
area).
 Power delivered to a resistor is I2·R, so a high resistance
wire will result in a small current, reducing the I2·R loss
(called Joule heating) in the wire.
 Transformers are used to step-up and step-down the
potential difference of the power in the transmission line.
 Efficiency of transmission:
power delivered by line
power delivered by line
e

power delivered to line power delivered by line  power lost in line
 Power
delivered to the line = power lost in line +
power delivered by the line