Instantaneous Power  P(t)
Instantaneous Power is the rate at which energy is absorbed/delievered
for an element at a specific time.
Instantaneous power varies with time.
P(t) = v(t)*i(t)
P(t) = V_{m}cos(ωt + φ)*I_{m}cos(ωt + θ)
To find the power absorbed in the resistor
we would have to multiply voltage times current.
Since both voltage and current are sinusoidal,
instantaneous power would also be sinusoidal.
To find instantaneous power at a specific time requires
plugging in a specific 't'.


Here are some hypothetical waveforms:
P(t) varies with time 

Summary
Instantaneous Power is a function of time. It is not a constant.
It is not an average, therefore it is not used for billing utility customers.
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Peak Power
Peak Power is the maximum instantaneous power. Peak Power
is an important specification used to characterize a device/element.
If the power absorbed by an element should exceed the Peak Power,
then the device may be permanently damaged.
Thus Peak Power is a spec used during the design phase by engineers.
Here Peak Power is indicated on the waveform: 

Summary
Peak Power is used to characterize a device.
This information is usually found in a Parts Data Book published
by a device manufacturer such as Intel or AMD.
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Average Power
Average Power is the average rate at which energy
is absorbed. Average Power is independent of time.
It is used by utility companies for billing purposes.
Recall Instantaneous Power: P(t) = v(t)*i(t)
P_{average} = (1/T) ∫ P(t) dt
where T is the period and P(t) is integrated over a period.
Average Power is not a function of time: 

Let v(t) = V_{m}cos(ωt + φ)
and Let i(t) = I_{m}cos(ωt + φ  θ)
P_{average} = (ω/2π) ∫ [V_{m}cos(ωt + φ) * I_{m}cos(ωt + φ  θ)]
(Note that T = 2π/ω)
If the above equation is simplifed, one gets the following result:
P_{average} = ½(V_{m}I_{m})cos(θ)
Note that θ = φ  (φ  θ) OR θ = Phase of v(t)  Phase of i(t)
Summary
Average Power is calcuated by multiplying the magnitudes V_{m} and I_{m} and then dividing by 2.
This product is multiplied by the cosine of the the angle at which voltage leads current.
What is the average power for a resistor?
Since voltage and current are in phase across a resistor, then:
P_{average} = ½(V_{m}I_{m})
because cos(0) = 1
What is the average power for a capacitor?
Since voltage lags current by 90^{o} across a capacitor, then:
P_{average} = ½(V_{m}I_{m}) cos(90^{o}) = 0
What is the average power for a inductor?
Since voltage leads current by 90^{o} across an inductor, then:
P_{average} = ½(V_{m}I_{m}) cos(90^{o}) = 0
This should make sense since capacitors and inductors are storage elements.
They store and release energy. They do not absorb energy.
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Calculating Average Power
Alternate forms to P_{average}:
Recall that P_{average} = ½(V_{m}I_{m})cos(θ)
Note that Z = V/I where Z is impedance.
Putting this equation in phasor form:
Z = (V_{m} : Phase of v(t)) / (I_{m} : Phase of i(t))
Z = V_{m}/I_{m} and Angle of Z = Phase of v(t)  Phase of i(t)
We must say angle of Z not phase of Z because Z (unlike voltage and current) is NOT a phasor/sinusoidal!
V_{m}I_{m} = Z I_{m}²
θ = Phase of v(t)  Phase of i(t)
Therefore an alternate to P_{average} is:
P_{average} = ½ Z I_{m}² cos (Angle of Z)
cos (Angle of Z) = Real(Z) / Z
P_{average} = ½ Z I_{m}² Real(Z) / Z
P_{average} = ½ I_{m}² Real(Z)
The real part of Z is the resistive portion of the impedance. Note that
reactive components such as capacitors and inductors do not contribute to
the average power absorbed.
Study Problems
After clicking on the following link enter 111 for the problem and 1 for the step:
Study Problem 111
After clicking on the following link enter 112 for the problem and 1 for the step:
Study Problem 112
After clicking on the following link enter 113 for the problem and 1 for the step:
Study Problem 113
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Root Mean Square (RMS) values
An RMS value is the DC equivalent current or voltage that would deliver
the same average power as the AC current or voltage.
Recall that power in a DC circuit is:
P = VI = I² R
To find I_{RMS}:
I_{RMS} ² R = 1/T ∫ R i(t)² dt = P_{average}
Canceling out the R we get:
I_{RMS} ² = 1/T ∫ i(t)² dt
If this equation is simplified (left to the more advanced student) we get:
I_{RMS} = I_{m}/√2
Similarly we get:
V_{RMS} = V_{m}/√2
We now have other equations for calculating average power:
P_{average} = V_{RMS} I_{RMS} cos (θ)
P_{average} = I_{RMS}² Real(Z)
When we see our household voltage source expressed as:
120 Volts  60 Hz
this is really an RMS voltage value.
To find the peak voltage V
_{m} we must multiply by √2:
V
_{peak} = 120V * √2 = 170 Volts
Study Problems
After clicking on the following link enter 114 for the problem and 1 for the step:
Study Problem 114
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