The sinusoidal steady state response

heir respective nodes, remembering that each node voltage is understood to be measured with respect to the chosen reference.
If the circuit contains only current sources, apply Kirchhoff’s current law at each non-reference node. To obtain the conductance matrix if a circuit has only independent current sources, equate the total current leaving each node through all conductance to the total source current entering that node, and order the term form v1 to vN-1. For each dependent current source present, relate the source current and the controlling quantity to the variables v1, v2,, vN-1, if they are not already in that form.
If the current contains voltage sources, form a super-node about each one by enclosing the source and its two terminals within a broken-line enclosure thus reducing the number of nodes by 1 for each voltage source that is present. The assigned node voltages should not be changed. Using these assigned node-to-references voltages, apply KCL at each of the nodes or super-nodes(that do not contain the references node) in this modified circuit. Relate each source voltage to the variables v1, v2,, vN-1, if it is not already in that form.
Example:
In this circuit, each passive element is specified by its impedance, 
although the analysis might be simplified slightly by using admittance values. Two current sources are given as phasors, and phasor node voltages 
Vi and V2 are indicated. At the left node we apply KCL and I = V/Z: 
At the right node, 
Combining terms, we have 
Using determinants to solve Eqs. (1) and (2), we obtain 
The time-domain solutions are therefore obtained by expressing Vi and 
V2 in polar form: 
2. Mesh analysis
	Method:
Make certain that the network is a planar network. If it is non-planar, mesh analysis is not applicable.
Make a neat, simple circuit diagram. Indicate all element and source values. Resistance values are preferable to conductance values. Each source should have its reference symbol.
Assuming that the circuit has M meshes, assign a clockwise mesh current in each mesh, i1, i2,, iM.
If the circuit contains only voltage sources, apply KVL around each mesh. To obtain the resistance matrix if a circuit has only independent voltage source, equate the clockwise sum of all the resistor voltages to the counterclockwise sum of all the source voltages, and order the term, from i1 to iM. For each dependent voltage source present, relate the source voltage and the controlling quantity to the variables i1, i2,, iM, if they are not already in that form.
If the circuit contains current source, create a super-mesh for each current source that is common to two meshes by applying KVL around the larger loop formed by the branches that are not common to the two meshes; KVL need not be applied to a mesh containing a current source that lies on the perimeter of the entire circuit. The assigned mesh currents should not be changed. Relate each source current to the variables i1,i2,, iM, if it is not already in that form. 
Example:
Solution: Noting from the left source that a> = 103 rad/s, we draw the 
frequency-domain circuit of Fig. 9-26 and assign mesh currents Ii and I2. 
Around mesh 1,
while mesh 2 leads to
Solving
9-3: Superposition, source transformations,Thevenin’s theorem
Superposition
If circuit has two or more independent sources, there several ways to determine the value of a specific variable(voltage,current):
-Use nodal or mesh analysis.
-Superposition approach: Determine the contribution of each independent source to the variable, and then add them up.
 The superpositionprinciple states that the voltage across (or current through) an element in a linear circuitis the algebraic sum of the voltages across (or currents though) that element due to each independent source acting alone.
Superposition is not limited to circuit analysis but is applicable in many field where cause and effect bear a linear relation ship to one another.
Step to apply superposition principle:
 -Turn off all independent sources except one source (dependent sources are left intact):
-Replace voltage source by short circuit
-Replace current source by open circuit
Find the output (voltage or/and current) due to that active source (using nodal or mesh analysis)
-Repeat step 1&2 for each of the other independent sources.
Find the total contribution by adding algebraically all the contributions due to the independent sources.
Example:
Solution: First we redraw the circuit of Fig. 9-1 as Fig. 9-5, where each 
pair of parallel impedances is replaced by a single equivalent impedance. That is, 5 and -j'10 in parallel yield 4 - j2 il; jlO in parallel with -j'5 gives -7IO il; and 10 in parallel with j"5 provides 2 + jA il. To find Vi, we first 
activate only the left source and find the partial response:
With only the right source active, current division helps us to obtain
Source transformation
Similar to the series-parallel combination and wye-delta transformation, source transformation is using to simplify circuits that bases on the concept of equivalence.
An equivalentcircuitis one whose v-I characteristics are identical with theoriginal circuit.
A source transformationis the process of replacing a voltage source vsin series with a resistor R by a current source isin parallel with a resistor R, or vice versa.
Notes: 
-The arrow of the current source is directed toward the positive terminal of the voltage source. 
- The source transformation is not possible when R = 0 (ideal voltage source) or R = ∞ (ideal current source). 
Thevenin’s theorem
A linear two terminal circuit can be replaced by an equivalent circuit consisting of a voltage source VTh in series with a resistor R
Statement of theorem:
Given any linear circuit, rearrange it in the form of two networks A and B that are connected by two wires. A is the network to be simplified, B will be left untouched .
Disconnect network B. Define a voltage Voc as the voltage now appering across the terminals of network A
Turn off or “zero out” every independent source in network A to form an inactive network. Leave dependent sources unchanged
Connect an independent voltage source with value Voc in series with the inactive network. Do not complete the circuit; leave the two terminals disconnected
Connect network B to the terminals of the new network A. All currents and voltages in B will remain unchanged
9-4: The Phasor Diagram
we saw that sinusoidal waveforms of the same frequency can have a Phase Difference  between themselves which represents the angular difference of the two sinusoidal waveforms. Also the terms "lead" and "lag" as well as "in-phase" and "out-of-phase" were used to indicate the relationship of one waveform to the other with the generalized sinusoidal expression given as: A(t) = Am sin(ωt ± Φ) representing the sinusoid in the time-domain form. sinusoidal waveforms be represented by a Phasor Diagram by using the rotating vector method.
Basically a rotating vector, simply called a "Phasor" is a scaled line whose length represents an AC quantity that has both magnitude ("peak amplitude") and direction ("phase") which is "frozen" at some point in time. A phasor is a vector that has an arrow head at one end which signifies partly the maximum value of the vector quantity ( V or I ) and partly the end of the vector that rotates. Generally, vectors are assumed to pivot at one end around a fixed zero point known as the "point of origin" while the arrowed end representing the quantity, freely rotates in an anti-clockwise direction at an angular velocity, ( ω ) of one full revolution for every cycle. This anti-clockwise rotation of the vector is considered to be a positive rotation. Likewise, a clockwise rotation is considered to be a negative rotation.
A sine wave can be constructed by a single vector rotating at an angular velocity of ω = 2πƒ, where ƒ is the frequency of the waveform.
 Phasor Diagram of a Sinusoidal Waveform
Phase Difference of a Sinusoidal Waveform
The generalised mathematical expression to define these two sinusoidal quantities will be written as:
The current, i is lagging the voltage, v by angle Φ and in our example above this is 30o. So the difference between the two phasors representing the two sinusoidal quantities is angle Φ and the resulting phasor diagram will be.
Phasor Diagram of a Sinusoidal Waveform
Phasor Addition
Sometimes it is necessary when studying sinusoids to add together two alternating waveforms, for example in an AC series circuit, that are not in-phase with each other. determine Resultant Phasor or Vector Sum by using the parallelogram law.
Consider two AC voltages, V1 having a peak voltage of 20 volts, and V2 having a peak voltage of 30 volts where V1 leads V2 by 60o. The total voltage, VT of the two voltages can be found by firstly drawing a phasor diagram representing the two vectors and then constructing a parallelogram in which two of the sides are the voltages, V1 and V2 as shown below.
Phasor Addition of two Phasors
9-5: response as function of ω
We will now consider methods of obtaining and presenting the response of a circuit with sinusoidal excitation as a function of the radian frequency w. With the possible exception of the 60-Hz power area in which frequency is a constant and the load is the variable, sinusoidal frequency response is extremely important in almost every branch of electrical engineering as well as in related areas, such as theory of mechanical vibrations or automatic control.
For the first example, we select the series RL circuit. The phasor voltage Vs is therefore applied to this simple circuit, and the phasor current I (leaving the positively marked end of the voltage source) is selected as the desired response. We are dealing with the forced response only, and the familiar phasor 
methods enable the current to be obtained:
the current produced by a source voltage \/Jf_ V. The magnitude of the response is