AC Power Analysis

AC Power Analysis

Power is defined as the time rate of doing work. In an AC circuit, power quantities are continuously varying. Capacitive, resistive, and inductive loads help us identify the amount of power in the circuit and in its elements. Power can be absorbed or supplied by circuit elements. There are formulas for finding the Instantaneous, Average, and Maximum power in an AC circuit.

For Instantaneous Power:

P(t)= ½ Vm Im cos(Ѳv – Ѳi) + ½ Vm Im cos(2ῳt + Ѳv + Ѳi)

-To solve the instantaneous power, you just simply substitute the solved/given voltage, current and their phase angles and also the angular frequency. P is also not time dependent.

For Average Power:

P(t)= ½ Vm Im cos(Ѳv – Ѳi)

-This formula is just similar to instantaneous power, but the ½ Vm Im cos(2ῳt + Ѳv + Ѳi) was removed.

For Maximum Power:

P(t)= ½ (I^2) Rl

But if you substitute other formulas, expand, and use derivation, the final formula would be:

1/8 (Vth^2 / Rth)

-Vth and Rth are only the real part of the obtained voltage and Impedance.

 

Thevenin's theorem/ Norton's Theorem

Thevenin's theorem/ Norton's Theorem

Thevenin's theorem/ Norton's theorem is another method used for circuit analysis and simplification. It involves converting circuit sources and impedances to a thevenin equivalent. The picture below shows the Thevenin's and Norton's equivalent circuit of the original circuit. No current flows in ab because it represents an open circuit.
 

Thevenin is commonly used for voltage analysis while Norton is used in current analysis.
You could use other circuit solving methods to identify Vth, Rth, and In.

Superposition

Superposition

Is a method of circuit analysis and simplification that involves leaving one source in the circuit and killing all the remaining sources, to obtain the unknown value in the circuit. The voltage source would be shorted and the current source would be opened. Here’s a sample problem for this:

First, let’s kill the 12∟0 V source and leave the 6∟0 A source. The resulting circuit would be:

To obtain I, we will use mesh analysis. This loop would create 3 loops but loop 3 is already equal to 6∟0 A and its not necessary for solving I.

Loop 1 would be equal to:

(-2+j1) I1 – (j1)I2=0

Loop 2 would be equal to:

(-j1)I1 + (-1 – j2)I2 = -6

From the loop equations and using matrix, we would obtain:

I= 2.042∟149.78 A

After obtaining I by killing the voltage source, we will now kill the current source and leave the voltage source to solve I. The resulting circuit would be like this:

We will use nodal analysis to solve I. The circuit has a super node, so the resulting equation would be:

(0.5)V1 + (0.1 + j0.7)V2= 0

And for KVL:

V1 – V2 = -12

From the equation and matrix, we would obtain I.

I= 6.51∟40.60 A

To obtain the final I, we will just simply add the 2 I’s we solved.

I= I1 + I2

I= (2.042∟149.78)+(6.51∟40.60)

I= 6.149 ∟58.878 A

Source Transformation

Source Transformation

Is a method used to simplify a circuit by transforming a series voltage source and impedance to a parallel current source and impedance, and vice versa. This process is commonly used to lessen the number of loops in a circuit for easy analysis.

The direction of the arrow of the current source follows the positive sign of the voltage source.

In this circuit, we should identify the voltage at the 4Ω resistor. We will start transforming at the right side of the circuit using simple ohms law.

V= 2A (2Ω)

V= 4 V

 

The resulting circuit would be:

Combine the series impedance's.

Then, transform it again.

I=V/R

I= 4/ 2-j1

I= 1.788∟26.56 A

Combine the parallel impedances.

Z= j3(2-j1)/ j3+(2-j1)

Z= 2.37 ∟18.43 Ω

 

Then transform again so that we can subtract the voltage source and finally obtain V.

V= (1.788∟26.56) (2.25+j0.75)

V= 2.998 + j2.998

V= V1-V2

V= (2.998 +j2.998)-(10)

V= 7.6168 ∟156.82 V

We can now use voltage division to obtain V.

V=(7.616∟156.82)(4)/4+ (2.25+j0.75)

V= 4.84 ∟149.97 V

Mesh Analysis with complex numbers

Mesh Analysis with complex numbers

Mesh analysis is a method that is used to solve planar circuits for the currents at any place in the circuit.

Find I

In solving mesh current you can decide whether you go counter-clockwise or clockwise in applying KCL or creating a loop but I always use the direction of the given current to avoid sign errors.

I1 is already given in the circuit which is 2 A. The only current that we need to solve is I2 to identify current I. 

From I2 we could get the equation:

-5 (I2 –I1) + j3 (I2) – 5 = 0

Transpose 5 and substitute I1 to the equation.

-5 I2 + 5 (2) + j3 I2= 5

                      -5 I2 + 10 + j3 I2= 5 

                      (-5+ j3) I2= 5- 10

                      (-5 + j3) I2 = -5 
                      I2= -5/ -5+ j3 
                      I2= 0.735+ j0.441

To get current I we could use this illustration taken from the middle top node.

(The current entering the node is equal to the current leaving the node.)

Therefore:

    I1= I + I2

    I= I1- I2

I= 2- (0.735+ j0.441)

I= 1.2647- j0.441

Convert it to Polar.

I= 1.339 ∟-19.22 A

Supermesh


A super mesh is created when there are two meshes that have a current source in common. When a super mesh occurs in a circuit it involves opening the given current and combining the formulated equation of the two loops that are connected to the given current. Then applying KCL to the original circuit to get another equation and applying matrix, ohms law or simple calculations to obtain the unknown value in the circuit.

Nodal Analysis with complex numbers

 

Nodal Analysis with complex numbers

 

Nodal analysis is a method of determining the voltage between nodes in an electrical circuit in terms of the branch currents.

Sample problem #2

Find I

The elements in this circuit are already converted to impedances.

First let’s identify the nodes in this circuit.

The node V2 has already its value which is 8 V because of the supply connected under it.

Solving V1 could help identify the current entering the –j3 ohm capacitor. From the node V1, we can get:

V1 (1/j6 + 1/-j3 + 1/8) – V2 (1/8) = 0

(The reciprocal of the nearby impedances of node V1 times V1 minus the neighbor node of V1which is V2 times the reciprocal of the impedance that connects them.)

We could substitute V2= 8.

V1 (0.125+ j0.1666) – 8 (1/8) = 0

              V1 (0.125 +0.1666) = 1 

              V1= 1/ (0.125+ j0.1666)

              V1= 2.881 – j3.840 V

              V1= 4.80 ∟-53.12V

To get I, just use ohms law and divide V1 to the capacitor.

Other problems that involves nodal and mesh analysis requires matrix to solve for the unknown. The more complex the problem, the greater the chance matrix is needed.
Supernode

A super node is contains two nodes, one a non-reference node and another node that may be a second non-reference node or the reference node.

When a super node occurs in a circuit it involves shortening the supply voltage and combining or adding the formulated equation of the two nodes that are connected to the supply voltage. Then applying KVL to the original circuit to get another equation and applying matrix, ohms law or simple calculations to obtain the unknown value in the circuit.