Monday, December 15, 2014
Saturday, October 11, 2014
Inductor
Inductor is an electrical component that stores energy in magnetic field.
The inductor is made of a coil of conducting wire.
In an electrical circuit schematics, the inductor marked with the letter L.
The inductance is measured in units of Henry [L].
Inductor reduce current in AC circuits and short circuit in DC circuits.
Inductor symbols
Inductor |  |
Iron core inductor |  |
Variable inductor |  |
Inductors in series
For several inductors in series the total equivalent inductance is:
LTotal = L1+L2+L3+...
Inductors in parallel
For several inductors in parallel the total equivalent inductance is:
Inductor's voltage
Inductor's current
Energy of inductor
CAPACITORS AND INDUCTORS
Capacitor:
In both digital and analog electronic circuits a capacitor is a fundamental element. It
enables the filtering of signals and it provides a fundamental memory element.
The capacitor is an element that stores energy in an electric field.
The circuit symbol and associated electrical variables for the capacitor is shown on
Figure 1.
The capacitor may be modeled as two conducting plates separated by a dielectric as
shown on Figure 2.
When a voltage v is applied across the plates, a charge +q accumulates on one plate and a
charge –q on the other.
Capacitor:
In both digital and analog electronic circuits a capacitor is a fundamental element. It
enables the filtering of signals and it provides a fundamental memory element.
The capacitor is an element that stores energy in an electric field.
The circuit symbol and associated electrical variables for the capacitor is shown on
Figure 1.
The capacitor may be modeled as two conducting plates separated by a dielectric as
shown on Figure 2.
When a voltage v is applied across the plates, a charge +q accumulates on one plate and a
charge –q on the other.
Capacitance
The capacitance (C) of the capacitor is equal to the electric charge (Q) divided by the voltage (V):
C is the capacitance in farad (F)
Q is the electric charge in coulombs (C), that is stored on the capacitor
V is the voltage between the capacitor's plates in volts (V)
Capacitance of plates capacitor
The capacitance (C) of the plates capacitor is equal to the permittivity (ε) times the plate area (A) divided by the gap or distance between the plates (d):
C is the capacitance of the capacitor, in farad (F).
ε is the permittivity of the capacitor's dialectic material, in farad per meter (F/m).
A is the area of the capacitor's plate in square meters (m2].
d is the distance between the capacitor's plates, in meters (m).
Capacitors in series
The total capacitance of capacitors in series, C1,C2,C3,.. :
Capacitors in parallel
The total capacitance of capacitors in parallel, C1,C2,C3,.. :
CTotal = C1+C2+C3+...
Capacitor's current
The capacitor's momentary current ic(t) is equal to the capacitance of the capacitor,
times the derivative of the momentary capacitor's voltage vc(t):
Capacitor's voltage
The capacitor's momentary voltage vc(t) is equal to the initial voltage of the capacitor,
plus 1/C times the integral of the momentary capacitor's current ic(t) over time t:
MAXIMUM POWER TRANFER
The Maximum Power Transfer Theorem is another useful Circuit Analysis method to ensure that the maximum amount of power will be dissipated in the load resistance when the value of the load resistance is exactly equal to the resistance of the power source. The relationship between the load impedance and the internal impedance of the energy source will give the power in the load.
Then if the 8Ω speaker is attached to the amplifiers output, the amplifier will see the speaker as an 8Ω load. Connecting two 8Ωspeakers in parallel is equivalent to the amplifier driving one 4Ωspeaker and both configurations are within the output specifications of the amplifier.One good example of impedance matching is between an audio amplifier and a loudspeaker. The output impedance, ZOUT of the amplifier may be given as between 4Ω and 8Ω, while the nominal input impedance, ZIN of the loudspeaker may be given as 8Ω only.
The Maximum Power Transfer Theorem is another useful Circuit Analysis method to ensure that the maximum amount of power will be dissipated in the load resistance when the value of the load resistance is exactly equal to the resistance of the power source. The relationship between the load impedance and the internal impedance of the energy source will give the power in the load.
Maximum Power Transfer Example No1.
 |
Where:
RS = 25Ω RL is variable between 0 – 100Ω VS = 100v |
Then by using the following Ohm’s Law equations:
We can now complete the following table to determine the current and power in the circuit for different values of load resistance.
Table of Current against Power
|
|
Using the data from the table above, we can plot a graph of load resistance, RL against power, P for different values of load resistance. Also notice that power is zero for an open-circuit (zero current condition) and also for a short-circuit (zero voltage condition).
Graph of Power against Load Resistance
From the above table and graph we can see that the Maximum Power Transfer occurs in the load when the load resistance, RL is equal in value to the source resistance, RS that is: RS = RL = 25Ω. This is called a “matched condition” and as a general rule, maximum power is transferred from an active device such as a power supply or battery to an external device when the impedance of the external device exactly matches the impedance of the source.
Then if the 8Ω speaker is attached to the amplifiers output, the amplifier will see the speaker as an 8Ω load. Connecting two 8Ωspeakers in parallel is equivalent to the amplifier driving one 4Ωspeaker and both configurations are within the output specifications of the amplifier.One good example of impedance matching is between an audio amplifier and a loudspeaker. The output impedance, ZOUT of the amplifier may be given as between 4Ω and 8Ω, while the nominal input impedance, ZIN of the loudspeaker may be given as 8Ω only.
Improper impedance matching can lead to excessive power loss and heat dissipation. But how could you impedance match an amplifier and loudspeaker which have very different impedances. Well, there are loudspeaker impedance matching transformers available that can change impedances from 4Ω to 8Ω, or to 16Ω’s to allow impedance matching of many loudspeakers connected together in various combinations such as in PA (public address) systems.
Monday, October 6, 2014
Thevenins Theorem Case 2
thevenin's theorem states that :
A linear two-terminal circuit can be replaced by an equivalent circuit consisting of avoltage source Vth in series with a resistor Rth.
CASE 2: If the network has dependent sources, turn off all independent sources. Apply a voltage source Vo at the terminals a-b and determine the resulting current Io. Alternatively, insert a current Io and determine Vo. Where Rth=Vo/Io
Assume any value of Vo and Io.
Maximum Power Transfer
The Maximum Power Transfer Theorem is not so much a means of analysis as it is an aid to system design. Simply stated, the maximum amount of power will be dissipated by a load resistance when that load resistance is equal to the Thevenin/Norton resistance of the network supplying the power. If the load resistance is lower or higher than the Thevenin/Norton resistance of the source network, its dissipated power will be less than maximum.
This is essentially what is aimed for in radio transmitter design , where the antenna or transmission line “impedance” is matched to final power amplifier “impedance” for maximum radio frequency power output. Impedance, the overall opposition to AC and DC current, is very similar to resistance, and must be equal between source and load for the greatest amount of power to be transferred to the load. A load impedance that is too high will result in low power output. A load impedance that is too low will not only result in low power output, but possibly overheating of the amplifier due to the power dissipated in its internal (Thevenin or Norton) impedance.
Taking our Thevenin equivalent example circuit, the Maximum Power Transfer Theorem tells us that the load resistance resulting in greatest power dissipation is equal in value to the Thevenin resistance (in this case, 0.8 Ω):
With this value of load resistance, the dissipated power will be 39.2 watts:
If you were designing a circuit for maximum power dissipation at the load resistance, this theorem would be very useful. Having reduced a network down to a Thevenin voltage and resistance (or Norton current and resistance), you simply set the load resistance equal to that Thevenin or Norton equivalent (or vice versa) to ensure maximum power dissipation at the load. Practical applications of this might include radio transmitter final amplifier stage design (seeking to maximize power delivered to the antenna or transmission line), a grid tied inverterloading a solar array, or electric vehicle design (seeking to maximize power delivered to drive motor).
The Maximum Power Transfer Theorem is not: Maximum power transfer does not coincide with maximum efficiency. Application of The Maximum Power Transfer theorem to AC power distribution will not result in maximum or even high efficiency. The goal of high efficiency is more important for AC power distribution, which dictates a relatively low generator impedance compared to load impedance.
Sunday, September 21, 2014
NORTON'S THEOREM
Norton's Theorem states that a linear two terminal circuit can be replaced by an equivalent circuit consisting of a current source In in parallel with a resistor Rn where In is the short circuit current through the terminals and Rn is the input or equivalent resistance at the terminals when the independent sources are turned off.
the process of finding Rn is the same way on finding Rth by the use of source transformantion, thevenins and norton's resistances are equal:
Norton's Theorem states that a linear two terminal circuit can be replaced by an equivalent circuit consisting of a current source In in parallel with a resistor Rn where In is the short circuit current through the terminals and Rn is the input or equivalent resistance at the terminals when the independent sources are turned off.
the process of finding Rn is the same way on finding Rth by the use of source transformantion, thevenins and norton's resistances are equal:
Rn=Rth
To find the norton current In, we determine the short circuit current flowing from the terminal a to b in both circuit. It is evident that the short circuit current is In. It must be the same short circuit current from terminal a to b. since the two circuits are equivalent.
In=isc
In=Vth/Rth
DESIGN EXPERIMENT
Objective: Design a circuit that will apply the use series resistive method
Problem : Design a series resistive circuit which contains the desired voltage output of 3.7V in a resistor with the given voltage source of 12V.
R23=5.72kΩ
by using ration and proportion we can find the R3
8.7 / 5.27 = 3.3 / R3
8.7(r3)=3.3(5.27)
R3=2k
R2=Rorig-R3
r2= 5.27-2k
R2=3.27k
this is the equivalent circuit
Reflection:
In order to find the value of each resistor we use voltage divider to get the value of each resistors and also by using the ratio and proportion method. based on our laboratory experiment the value of calculated and measured are not exactly the same because of the tolerance of 5% in the resistors.
Objective: Design a circuit that will apply the use series resistive method
Problem : Design a series resistive circuit which contains the desired voltage output of 3.7V in a resistor with the given voltage source of 12V.
the first thing we do is we assume resistors for R1 with a given voltage of 3.7V,
then because the the total voltage of each all resistors is 12v because of the battery of 12V we deduct 12v by 3.7V and we an answer 8.7v, this 8.7V is the summation R2 and R3 but we do not know yet the value of resistor of R2 and R3.
by the assumed resistor R1 = 2kΩ
we use voltage divider to find the Rtotal of R2 and R3
Vo=V1 (R1) / R1 + R2
8.7V = 12V ( R2) / 2 + R2
8.7V (2+R2) = 12(r2)
17.4 = 12r2 - 8.7r2
17.4=3.3r2R23=5.72kΩ
by using ration and proportion we can find the R3
8.7 / 5.27 = 3.3 / R3
8.7(r3)=3.3(5.27)
R3=2k
R2=Rorig-R3
r2= 5.27-2k
R2=3.27k
this is the equivalent circuit
In order to find the value of each resistor we use voltage divider to get the value of each resistors and also by using the ratio and proportion method. based on our laboratory experiment the value of calculated and measured are not exactly the same because of the tolerance of 5% in the resistors.
Thevenin's Theorem
Thevenin's Theorem States that it is possible to simplify any circuit, no mattter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load. The qualification of linear is identical to that found in Superposition theorem, where all the underlying equation must be linear.
Thevenin's theorem is useful in analyzing power system and other circuits where one particular resistor in the circuit ( called the load resistor) is subjected to change, and re construct the circuit with calculation depend on what method you are gonna use.
example:
In this example I decide to choose R2 as the LOAD resistance in the circuit. We temporarily remove R2(load resistance) from the circuit and reducing what's left to an equivalent circuit composed of a single voltage source and series resitance. The load resistance can be reconnected to this Thevenins equivalent circuit and calculation carried out as if the whole network were nothing but a simple series circuit.
This should be the picture of a Thevenin's equivalent circuit
Next, the voltage between the two points where the load resistor determined. Use whatever analysis methods are at your disposal to do this. In this case, the orignal circuit with the load resistor removed is nothing more than a simple series circuit with the voltage Vth, and so we can also determine the voltage across the open load terminals by applying the OHM's law and KVL.
there are the results:
the voltage between the two load connection points can be figured from the voltage and one of the resistor's voltage drops, and come out to 11.2 V which is our thevenins voltage.
to find the Thevenins series resistance for our equivalent circuit, we need the original circuit, then we remove the power source (independent sources) which is the same method of superposition theorem
The voltage source will be SHORTED and current sources will be OPEN circuit. And then we can figure out the resistance from one load terminal to other.
with the load resistor (2Ω) attached between the connection points, we can determine voltage acroos it and current through it as though the whole network were nothing more than a simple series circuit:
in the example notice that the voltage and current of R2 (8V and 4A) are identical to other method of analysis. The voltage and current figures for the thevenin series resistance and the thevenin source(total) do not apply to any component in the original circuit. Thevenins theorem is only useful for determining what happens to a single resistor in a network: the load.
REFLECTION:
The Thevenin's theorem is a way to reduce a network to an equivalent circuit composed of a single voltage source (Vth),series resistance (Rth), and the series load.
In using the thevenins theorem :
-find thevenins source voltage by removing the load resistor from the original circuit and calculate the voltage across the open connection points where load is connected.
-find the thevenins resistance by removing all power sources in the original circuit by shorting voltage sources and current sources open circuit.
-analyze the voltage and current for the load resistor following the rules for series circuits.
the other case in using the thevenin theorem is a circuit the contains a DEPENDENT source. which our professor did not tackle yet.. thanks for reading :)
Thevenin's Theorem States that it is possible to simplify any circuit, no mattter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load. The qualification of linear is identical to that found in Superposition theorem, where all the underlying equation must be linear.
Thevenin's theorem is useful in analyzing power system and other circuits where one particular resistor in the circuit ( called the load resistor) is subjected to change, and re construct the circuit with calculation depend on what method you are gonna use.
example:
In this example I decide to choose R2 as the LOAD resistance in the circuit. We temporarily remove R2(load resistance) from the circuit and reducing what's left to an equivalent circuit composed of a single voltage source and series resitance. The load resistance can be reconnected to this Thevenins equivalent circuit and calculation carried out as if the whole network were nothing but a simple series circuit.
This should be the picture of a Thevenin's equivalent circuit
The Thevenin equivalent circuit, if correctly derived, will behave exactly the same as the original circuit formed by B1, R1, R3, and B2. In other words, the load resistor (R2) voltage and current should be exactly the same for the same value of load resistance in the two circuits. The load resistor R2 cannot “tell the difference” between the original network of B1, R1, R3, and B2, and the Thevenin equivalent circuit of EThevenin, and RThevenin, provided that the values for EThevenin and RThevenin have been calculated correctly.
The advantage in performing the “Thevenin conversion” to the simpler circuit, of course, is that it makes load voltage and load current so much easier to solve than in the original network. Calculating the equivalent Thevenin source voltage and series resistance is actually quite easy. First, the chosen load resistor is removed from the original circuit, replaced with a break (open circuit):
Next, the voltage between the two points where the load resistor determined. Use whatever analysis methods are at your disposal to do this. In this case, the orignal circuit with the load resistor removed is nothing more than a simple series circuit with the voltage Vth, and so we can also determine the voltage across the open load terminals by applying the OHM's law and KVL.
there are the results:
the voltage between the two load connection points can be figured from the voltage and one of the resistor's voltage drops, and come out to 11.2 V which is our thevenins voltage.
to find the Thevenins series resistance for our equivalent circuit, we need the original circuit, then we remove the power source (independent sources) which is the same method of superposition theorem
The voltage source will be SHORTED and current sources will be OPEN circuit. And then we can figure out the resistance from one load terminal to other.
with the load resistor (2Ω) attached between the connection points, we can determine voltage acroos it and current through it as though the whole network were nothing more than a simple series circuit:
in the example notice that the voltage and current of R2 (8V and 4A) are identical to other method of analysis. The voltage and current figures for the thevenin series resistance and the thevenin source(total) do not apply to any component in the original circuit. Thevenins theorem is only useful for determining what happens to a single resistor in a network: the load.
REFLECTION:
The Thevenin's theorem is a way to reduce a network to an equivalent circuit composed of a single voltage source (Vth),series resistance (Rth), and the series load.
In using the thevenins theorem :
-find thevenins source voltage by removing the load resistor from the original circuit and calculate the voltage across the open connection points where load is connected.
-find the thevenins resistance by removing all power sources in the original circuit by shorting voltage sources and current sources open circuit.
-analyze the voltage and current for the load resistor following the rules for series circuits.
the other case in using the thevenin theorem is a circuit the contains a DEPENDENT source. which our professor did not tackle yet.. thanks for reading :)
Monday, August 25, 2014
WEEK 9 : SUPERPOSITION THEOREM
SUPERPOSITION is an algorithmic circuit analysis method with a prescribed procedure like the node voltage or loop current method.
The Superposition is applied only in circuits with 2 or more independent source in which a final solution can be built as the additive of two or more partial solutions
Superposition only applies to a linear circuit, which are made of circuit elements. It does not apply in non linear circuits.
The superposition principle states that the voltage across the voltage or current in a element in an linear circuit is the voltage across (current through) that element due to each independent source acting alone.
The Superposition helps us to solve the circuit with many independent source but we just have to remember the things
to use superposition.
1. Consider one independent source at a time while all other independent source are turned off. For voltage source, the source will have a short circuit while the current sources are opened.
2. Dependent Sources are left the same because they are control by circuits variables.
Steps in using superposition theorem:
1. Turn off all independent source except for one source. Then find the output depending on what you will use to simplify the circuit, it could be nodal or mesh analysis .
2. Do the in steps for the other sources.
3. Find the total contribution by assuming up the partial results (voltage or current) to find the unknown.
Recollection About:
in superposition theorem, there is only a simple solution find the unknowns, you can use the Voltage divider and current divider.
and by summing the partial results you will come up with the result.
You can also get directly the unknown by using the voltage or current divider with in that loop as long there is the given independent source and a two resistors given.
SUPERPOSITION is an algorithmic circuit analysis method with a prescribed procedure like the node voltage or loop current method.
The Superposition is applied only in circuits with 2 or more independent source in which a final solution can be built as the additive of two or more partial solutions
Superposition only applies to a linear circuit, which are made of circuit elements. It does not apply in non linear circuits.
The superposition principle states that the voltage across the voltage or current in a element in an linear circuit is the voltage across (current through) that element due to each independent source acting alone.
The Superposition helps us to solve the circuit with many independent source but we just have to remember the things
to use superposition.
1. Consider one independent source at a time while all other independent source are turned off. For voltage source, the source will have a short circuit while the current sources are opened.
2. Dependent Sources are left the same because they are control by circuits variables.
Steps in using superposition theorem:
1. Turn off all independent source except for one source. Then find the output depending on what you will use to simplify the circuit, it could be nodal or mesh analysis .
2. Do the in steps for the other sources.
3. Find the total contribution by assuming up the partial results (voltage or current) to find the unknown.
Recollection About:
in superposition theorem, there is only a simple solution find the unknowns, you can use the Voltage divider and current divider.
and by summing the partial results you will come up with the result.
You can also get directly the unknown by using the voltage or current divider with in that loop as long there is the given independent source and a two resistors given.
Sunday, August 17, 2014
LINEARITY PROPERTY
Linear property is the linear
relationship between cause and effect of an element. This property gives
linear and nonlinear circuit definition. The property can be applied in
various circuit elements. The homogeneity (scaling) property and the additivity property are both the combination of linearity property.
The
homogeneity property is that if the input is multiplied by a constant k
then the output is also multiplied by the constant k. Input is called
excitation and output is called response here. As an example if we
consider ohm’s law. Here the law relates the input i to the output v.
mathematically ,
V= IR
If we multiply the input current i by a constant k then the output voltage also increases correspondingly by the constant k. The equation stands,
kiR = kv
The additivity property is that the response to a sum of inputs is the sum of the responses to each input applied separately.
Using voltage-current relationship of a resistor if
v1 = i1R and v2 = i2R
Applying (i1 + i2)gives
V = (i1 + i2)R = i1R+ i2R = v1 + v2
We
can say that a resistor is a linear element. Because the
voltage-current relationship satisfies both the additivity and the
homogeneity properties.
We can tell a circuit is linear if the circuit both the additive and the homogeneous. A linear circuit always consists of linear elements, linear independent and dependent sources.
What is linear circuit?
A circuit is linear if the output is linearly related with its input.
The relation between power and voltage is nonlinear. So this theorem cannot be applied in power.
See a circuit in figure 1. The box is linear circuit. We cannot see any independent source inside the linear circuit.
The linear circuit is excited by another outer voltage source vs. Here the voltage source vs acts as input. The circuit ends with a load resistance R. we can take the current I through R as the output.
Suppose vs = 5V and i = 1A. According to linearity property if the voltage is multiplied by 2 then the voltage vs = 10V and then the current also will be multiplied by 2 hence i = 2A.
The power relation is nonlinear. For example, if the current i1 flows through the resistor R, the power p1 = i12R and when current i2 flows through the resistor R then power p2 = i22R.
If the current (i1 + i2) flows through R resistor the power absorbed
P3 = R(i1 + i2)2 = Ri12 + Ri22 + 2Ri1i2 ≠ p1 + p2
So the power relation is nonlinear. Circuit solution method superposition is based on linearity property.
Saturday, August 9, 2014
MESH ANALYSIS
The Mesh Current Method, also know as the Loop Current Method, is quite similar to the Branch current method in that it uses simultaneous equations, Kirchhoff Voltage Law, and Ohms Law to determine unknown currents in a network. It differs from the Branch current method in that it does not use Kirchhoff Current Law, and it is usually able to solve a circuit with less unknown variables and less simultaneous equation, which is especially nice if you're forced to solve without a calculator
-28 + 2 (I1 - I2) + 4I1 = 0 Original Form of Equation
-28 + 6I1 + 2I2 = 0 Simplified Form of Equation
At this time we have one equation with two unknowns. To be able to solve for two unknown mesh currents, we must have two equations. If we trace the other loop of the circuit, we can obtain another KVL equation and have enough data to solve for the two currents. Creature of habit that I am, I'll start at the upper-left hand corner of the right loop and trace counter-clockwise:
-2I1 - 3I2 +7 = 0
I1 = 5 A
I2 = -1 A
CASES TO BE CONSIDERED FOR MESH ANALYSIS
CASE 1
A current source exists only in one mesh
The Mesh Current Method, also know as the Loop Current Method, is quite similar to the Branch current method in that it uses simultaneous equations, Kirchhoff Voltage Law, and Ohms Law to determine unknown currents in a network. It differs from the Branch current method in that it does not use Kirchhoff Current Law, and it is usually able to solve a circuit with less unknown variables and less simultaneous equation, which is especially nice if you're forced to solve without a calculator
The first step in the Mesh Current method is to
identify “loops” within the circuit encompassing all components. In our
example circuit, the loop formed by B1, R1, and R2 will be the first
while the loop formed by B2, R2, and R3 will be the second. The
strangest part of the Mesh Current method is envisioning circulating
currents in each of the loops. In fact, this method gets its name from
the idea of these currents meshing together between loops like sets of
spinning gears:
The choice of each current's direction is entirely arbitrary,
just as in the Branch Current method, but the resulting equations are easier to
solve if the currents are going the same direction through intersecting
components (note how currents I1 and I2 are both going “up” through
resistor R2, where they “mesh,” or intersect). If the assumed direction of a
mesh current is wrong, the answer for that current will have a negative value.
The next step is to label all
voltage drop polarities across resistors according to the assumed directions of
the mesh currents. Remember that the “upstream” end of a resistor will always
be negative, and the “downstream” end of a resistor positive with respect to
each other, since electrons are negatively charged. The battery polarities, of
course, are dictated by their symbol orientations in the diagram, and may or
may not “agree” with the resistor polarities (assumed current directions):
Using Kirchhoff's
Voltage Law, we can now step around each of these loops, generating
equations representative of the component voltage drops and polarities.
As with the Branch Current method, we will denote a resistor's voltage
drop as the product of the resistance (in ohms) and its respective mesh
current (that quantity being unknown at this point). Where two currents
mesh together, we will write that term in the equation with resistor
current being the sum of the two meshing currents.
Tracing the left loop of
the circuit, starting from the upper-left corner and moving
counter-clockwise (the choice of starting points and directions is
ultimately irrelevant), counting polarity as if we had a voltmeter in
hand, red lead on the point ahead and black lead on the point behind, we
get this equation:
-28 + 2 (I1 - I2) + 4I1 = 0
Notice that the middle
term of the equation uses the sum of mesh currents I1 and I2 as the
current through resistor R2. This is because mesh currents I1 and I2 are
going the same direction through R2, and thus complement each other.
Distributing the coefficient of 2 to the I1 and I2 terms, and then
combining I1 terms in the equation, we can simplify as such:
-28 + 2 (I1 - I2) + 4I1 = 0 Original Form of Equation
..Distributing to terms within parenthesis..
-28 + 2I1+ 2I2 + 4I1 = 0
-28 + 2I1+ 2I2 + 4I1 = 0
..Combining like terms..
-28 + 6I1 + 2I2 = 0 Simplified Form of Equation
At this time we have one equation with two unknowns. To be able to solve for two unknown mesh currents, we must have two equations. If we trace the other loop of the circuit, we can obtain another KVL equation and have enough data to solve for the two currents. Creature of habit that I am, I'll start at the upper-left hand corner of the right loop and trace counter-clockwise:
-2 (I1 +I2) +7 - 1I2 = 0
Simplifying the equation as before, we end up with:
-2I1 - 3I2 +7 = 0
Now, with two equations, we can use one of several methods to mathematically solve for the unknown currents I1 and I2:
-28 + 6I1 + 2I2 = 0
-2I1 - 3I2 + 7 = 0
..rearranging equation for easier equation..
6I1 + 2I2 = -28
-2I1 - 3I2 = -7
Solutions:
I1 = 5 A
CASES TO BE CONSIDERED FOR MESH ANALYSIS
CASE 1
A current source exists only in one mesh
CASE 2
A current source exist between two meshes
We can write one mesh equation by considering the current source. The current source is related to the mesh currents at by
i1=i2+1.5
In order to write the second mesh equation, we must decide what to do about the current source voltage. Notice that there is no easy way to express the current source voltage in terms of the mesh currents. In this example, illustrate two methods of writing the second mesh equation.
Apply KVL to the supermesh corresponding the current source. Shown in blue, this supermesh is the perimeter of the two meshes that each contain the current source.
Apply KVL to the supermesh to get
9i1+3i2+6i2-12=0 or 9i1+9i2=12
This is the same equation that was obtained using method 1. Applying KVL to the supermesh is a shortcut for doing three things.
1. labeling the current source voltage as v
2. applying KVL to both meshes that contain the current source
3. eliminating v from the KVL equations
In summary, the mesh equation are
i1=i2+1.5
and
9i1+9i2=12
by simplying this, the equation results to
i1=1.4167 A and i2=-83.3 mA
Reflection:
In this topic : Mesh analysis , i have learned that it is use to solve planar circuits for the currents at any place in the circuit. this means that there no wires crossing to each other. The difference between a mesh and a loop is that, the mesh is a loop which does not contain any other loops within it while a loop is any continuity path that is available to a current flow from a given circuit. But so far in answering the mesh analysis, it is still challenging because it is hard for me to get equations in a given circuit especially in using matrix solving. I am troubling in using matrix because of lack of practice. That's why I have practice in order have correct answer to the problem.
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