Question

1. In single-phase circuits test, consider the instantaneous power waveforms. How is the value of the real power obtained in the software? What is the significance of the instantaneous power going negative (consider direction of power flow and reactive power)?

2.  For each of the resistive, inductive and capacitive loads (loads 1-3), comment on whether the current waveform is smoother, noisier or the same shape as the voltage waveform. Explain the results using the following, for a resistor i = v/R, for an inductor, the current is proportional to the integral of the voltage , while in a capacitor, the current is proportional to the derivative of the voltage.

3.

Device	Acquired Parameters			Calculated Parameters
	Vrms (V)	Irms (A)	P (W)	S (VA)	Q (VAr)	cos q	R (W)	X (W)
1. Resistor R, (50W, D.C. Board)	62.7	1.17	73.5	73.359	j4.55	1.0019	53.69	j3.324
2. Inductor L, 10A	61.8	0.87	4	53.766	53.617	0.0744	5.285	70.834
3. Capacitor C, 18mF	62.5	0.36	-0.1	22.5	-22.50	-0.004	-0.772	-173.6

Calculate the inductance and capacitance (showing your working) corresponding to the calculated reactances in rows 2 and 3 of Table 3. Compare the calculated capacitance to its nominal value.

4. Show the formulas (e.g. Z₁//Z₃, you do not need to show your working) you would use to calculate the impedance for loads 4 to 6 in terms of the impedance of loads 1 to 3.

Answer 1

1)
In AC circuit analysis, what is this power that we talk about. The main problem is that the AC voltage and current varies sinusoidally with time. Moreover the presence of circuit reactive elements like Inductor and capacitor shift the current wave with respect to voltage wave (angle of phase difference).

Power is rate at which energy is consumed by load or produced by generator. Whether it is DC circuit or AC circuit, the value of instantaneous power is obtained by multiplying instantaneous voltage with instantaneous current. If at any instant of time t the voltage and current values are represented by sine functions as

   v = V_m  sin t

i = I_m  sin ( t- )

V_m and I_m  are the maximum values of the sinusoidal voltage and current. Here =2 f
f is the frequency and is the angular frequency of rotating voltage or current phasors. It should be clear that for a power system f is usually 50 or 60 Hz
is the phase difference between the voltage and current.

As we said the instantaneous power is the product of instantaneous voltage and current, if we name instantaneous power as p then

p = v.i = V_m  sin t . I_m  sin (t-)
or p = V_m I_m sin t sin (t-)

Applying trigonometric formula 2.sin A.sin B = cos(A-B) - cos (A+B) we get

It can be written as

This is the equation of instantaneous power

In the Fig-C is drawn all the three waves corresponding to v, i and p. Graphically also we can get the value of instantaneous power (p) at any instant of time t by simply multiplying the value of current i and voltage v at that particular instant t. (You can verify that in the diagram p is negative when either v or i is negative otherwise p is positive. See the points where p is zero). In the graph we have shown horizontal axis as angle ? instead of time t for easy visualization. It should be clear that both way it is correct.

Clearly the instantaneous power p is composed of two terms. The first term is constant because for a given load the phase angle is fixed. It does not change unless the load is changed.  The second term is varying with time sinusoidally due to the presence of the term cos (2t-). Look that the instantaneous power frequency is twice the frequency of voltage or current.

So the instantaneous power in a single phase circuit varies sinusoidally.

The instantaneous power, p = constant term + sinusoidal oscillating term.

In one complete period the average of oscillating term is zero.

Then what is the average power within a given time, say one Time Period of the wave?

It is the constant term.

Here is another way to think about the average power.
Just observe that the instantaneous power is negative for a small time. For any time interval you just find the total +ve area A+ (above horizontal-axis (blue line) and below p curve) and total -ve area A- (below horizontal axis and above p curve). The net area is obtained by subtracting A- from A+. By dividing this net area ( by the time interval T_i we get the average power(P). You can do this using calculus. What you will ultimately get is only the first term in the above formula for instantaneous power p.

In still another way it is easier to realize that the formula for instantaneous power p has a constant term (V_m.I_m / 2) cos and the other sinusoidal term (V_m.I_m / 2) cos (2 wt - ). Actually p is the oscillating power which oscillates about the average constant term (V_m.I_m / 2) cos .

So the average power is

The above formula can be written as

Or,

here,
     

     

V and I are the phasor representation of RMS values* of voltage and current sinusoids. The symbols |V| and |I| are the magnitudes of phasors V and I. (See at the buttom for definition of RMS value).

This above formula is your favorite formula for useful power that we are most concerned about. This average power formula is used to find the power consumed by the load. The monthly electric energy bill at home is based on this power. The engineers and technicians in power or electrical industry simply use the term power instead of average power. So whenever we simply call power it means average power.

Of course the instantaneous power is oscillating in nature. As we already said it does not oscillates about the horizontal-axis rather about the average power P (cyan color horizontal line).

P will be zero when cos =0 or = 90 degree, that is when the phase angle between voltage and current waves is 90 degrees. It is only when the load is pure inductive or capacitive. In this case the second term only remains in the instantaneous power formula.

From the above figure for some time the power becomes negative that means the load supply energy to source for this period. This is due to the presence of reactive element in load.

The above formula for instantaneous power can be written in another form. This form actually is an attempt to distinguish the oscillating reactive power from the instantaneous power formula. Rearranging the terms in equation for instantaneous power above we get

  p = |V| | I | cos (1-cow2t) - |V| | I | sin sin2t

In this equation the first term |V| | I | cos (1-cow2t) is oscillatory whose average value is |V| | I | cos . We already talked about this average power.

The second term |V| | I | sin sin2t which is also oscillatory but with zero average value. The maximum value of this term is |V| | I | sin . This is the so called Reactive power. So Reactive power is the maximum value of a oscillatory power that is repeatedly drawn from the source and again returned to the source within each cycle. So the average of this reactive power is zero.

The average power P is called as Real Power. It is also sometimes called active power.

   Real power = P = |V| | I | cos

It is usually written as P = VI cos . But it should be remembered that V and I are the rms values of voltage and current. For example when we say single phase 220 volt AC it means the rms value of voltage is 220 volts ( it is not maximum value of voltage sinusoid)

   Reactive power = Q = |V| | I | sin

Real power is measured in Watt and the reactive power is measured in VAR (VoltAmpereReactive). In power sector these units are too small so real power is measured in Megawatt (MW) and reactive power in Megavar (MVAR). The letter R at the end denotes reactive power.