Effective or RMS Value

Maximum Average Power Transfer

Maximum Average Power is transferred from a source to a load, the load impedance should be chosen equal to the conjugate of the Thevenin equivalent impedance representing the reminder of the network.

Maximum power transfer theorem is explained by the figure. Let a network consisting of one or more independent sources and other resistive elements be represented by its Thevenin's equivalent circuit.  We find out now the value of load resistance for which the power transferred to it is the maximum.

The average power dissipated in the load is the square of the current multiplied by the resistive portion (the real part)   of the load impedance.

Instantaneous and Average Power

Instantaneous and Average Power in an AC Circuit

The instantaneous power at any one moment is the same as in a DC circuit - Joules Law

The average power is the time average of the power over one period.

For a current driven circuit we can rewrite this as,

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 ωt  

          i = I_m  sin (ωt-φ)

V_m and I_m  are the maximum values of the sinusoidal voltage and current. Here ω=2 π f
f is the frequency and ω 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
φ 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 ωt  .  I_m  sin (ωt-φ)
         or  p = V_m I_m  sin ωt  sin (ωt-φ)

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

It can be written as

Thevenin's Theorem in AC Analysis

Source Transformation in AC Analysis

• Transform a voltage source in series with an impedance to a current source in     parallel with an impedance for simplification or vice versa.

Source transformations are easy to perform as long as there is a familiarity with Ohm's law. If there is a voltage source in series with an impedance, it is possible to find the value of the equivalent current source in parallel with the impedance by dividing the value of the voltage source by the value of the impedance. The converse also applies here: if a current source in parallel with an impedance is present, multiplying the value of the current source with the value of the impedance will result in the equivalent voltage source in series with the impedance.

Superposition Theorem in AC Analysis

The superposition theorem states that in a linear circuit with several sources, the current and voltage for any element in the circuit is the sum of the currents and voltages produced by each source acting independently. The theorem is valid for any linear circuit. The best way to use superposition with AC circuits is to calculate the complex effective or peak value of the contribution of each source applied one at a time, and then to add the complex values. This is much easier than using superposition with time functions, where one has to add the individual time functions.

To calculate the contribution of each source independently, all the other sources must be removed and replaced without affecting the final result.

When removing a voltage source, its voltage must be set to zero, which is equivalent to replacing the voltage source with a short circuit.

When removing a current source, its current must be set to zero, which is equivalent to replacing the current source with an open circuit. 

Now let's explore an example. 

In the circuit shown below" 

R_i = 100 ohm, R₁ = 20 ohm, R₂ = 12 ohm, L = 10 uH, C = 0.3 nF, v_S(t)=50cos(wt) V, i_S(t)=1cos(wt+30°) A, f=400 kHz. 

Notice that both sources have the same frequency: we will only work in this chapter with sources all having the same frequency. Otherwise, superposition must be handled differently.

Find the currents i(t) and i₁(t) using the superposition theorem.

• REVIEW:
• The Superposition Theorem states that a circuit can be analyzed with only one source of power at a time, the corresponding component voltages and currents algebraically added to find out what they'll do with all power sources in effect.
• To negate all but one power source for analysis, replace any source of voltage (batteries) with a wire; replace any current source with an open (break).

AC Circuits:Nodal Analysis & Mesh Analysis

Ø Since KCL is valid for phasors, we can analyze AC circuits by NODAL analysis.
Ø Determine the number of nodes within the network.
Ø Pick a reference node and label each remaining node with a subscripted value of voltage: V1, V2 and so on.
Ø Apply Kirchhoff’s current law at each node except the reference. Assume that all unknown currents leave the node for each application of Kirhhoff’s current law.
Ø Solve the resulting equations for the nodal voltages.
Ø For dependent current sources: Treat each dependent current source like an independent source when Kirchhoff’s current law is applied to each defined node. However, once the equations are established, substitute the equation for the controlling quantity to ensure that the unknowns are limited solely to the chosen nodal voltages.



ØPractice Problem 10.1: Find v1 and v2 using nodal analysis







Mesh Analysis

Practice Problem 10.4: Calculate the current Io

Please do prefer from this following links.

file:///Z:/downloads/Part2%20AC%20Circuit%20analysis%20examples.pdf]


http://home.chuhai.hk/~wllo/BSC/CSC23/CSC23Ch6.pdf