EMT1150 Primer - EMT 1150 PDF

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EMT 1150...

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Primer for EMT 1150 Professor Levy January 27, 2020

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Contents 0.1 Electronics Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.2 Electrical Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.2.1 Current-Voltage Laws. . . . . . . . . . . . . . . . . . . . . . . . . . 0.2.2 Continuity Laws. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.3 Voltage Dividers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.4 Current Dividers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.5 Canonical Circuit Analysis Techniques . . . . . . . . . . . . . . . . . . . . 0.5.1 Mesh Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.5.2 Nodal Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.6 Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.6.1 Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.6.2 Improper to Proper Fraction . . . . . . . . . . . . . . . . . . . . . . 0.6.3 Solution of 2×2 System of Equations . . . . . . . . . . . . . . . . .

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4 4 4 5 6 7 7 7 9 10 10 10 11

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CONTENTS

0.1

Electronics Terminology

Figure 1: Circuit with nodes, branches, meshes and loops.

1. Element: The term elements includes both components (i.e., resistors, capacitors, etc) and sources (i.e, current sources, voltage sources, etc.). 2. Node: A junction where two or more elements connect is called a node. 3. Branch: Branches are the connections between nodes. A branch is an element (resistor, capacitor, source, etc.). The number of branches in a circuit is equal to the number of elements. 4. Loop: A loop is any closed path going through circuit elements. To draw a loop, select any node as a starting point and draw a path through elements and nodes until the path comes back to the node where you started. There is only one rule: a loop can pass through a node only one time. It is ok if loops overlap or contain other loops. 5. Mesh: A mesh is a loop that has no other loops inside it.

0.2

Electrical Laws

We will study five laws in the class. These are current-voltage laws of resistive, inductive, and capacitive components, and two continuity law.

0.2.1

Current-Voltage Laws.

There are three current-voltage laws: once for resistive components, one for capacitive components, andone for inductive components.

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0.2. ELECTRICAL LAWS Table 1: Some very informative caption

Ohm’s Law1

V R = IR R

dVC dt dIL Henry’s Law3 VL = L dt 1 The voltage difference across a resistor, VR , is equal to the product of the current through the resistor, IR , by the resistance of the resistor, R. ; 2 The current through the capacitor, IC , is equal to the product of the time-rate of change of the voltage difference accross the dVC capacitor, , with the capacitance of the dt capacitor, R. 3 The voltage difference across an inductor, VL , is equal to the product of the time-rate of change of the current thorugh the inductor, dIL , with the inductance of the indcuctor, R. dt Farad’s Law2

IC = C

Table 2: Current-Voltage Laws.

0.2.2

Continuity Laws.

There are two continuity laws: one for voltages (i.e., Kirchoff’s voltage law), and one for one for currents (i.e., Kirchoff’s currents law). Kirchoff’s voltage law states that the algebraic sum of the potential rises and drops around any loop or closed surface is zero. Kirchoff’s current law states that the algebraic sum of the currents entering into and out of any node is zero. Table 3: Current-Voltage Laws. P

V =0

Kichoff’s Current Law (KCL)

P

I=0

œ

Kichoff’s Voltage Law (KVL)

J

6

0.3

CONTENTS

Voltage Dividers

A voltage divider is shown in the dashed box in Figure ??. By RT I denote the total resistance of the volage divider, which I write as RT =

n X

Rk

k=1

= R1 + R2 + · · · + Ri−1 + Ri +i+1 + · · · + Rn By Vin , I denote the voltage across the votlage divider. By Ri , I denote the resistance of the ith resistor. By Vi , I denote the votlage accross the ith resistor. The votlage divider rule states that Ri Vi . = Vin RT

0.4

Current Dividers

A current divider is shown in a dashed box in Figure 3. By RT I denote the total resistance of the current divider, which I write as RT = Pn

1

1 k=1 Rk

=

1 R1

+

1 R2

1

+ · · · Ri−1

1 + R1i +

1 Ri+1

+ ··· +

1 Rk

By Iin , I denote the current through the current divider. By Ri , I denote the resistance of the ith resistor. By Ii , I denote the current through the ith resistor. The current divider rule states that RT Ii . = Iin Ri

0.5

Canonical Circuit Analysis Techniques

In the next two subsections we outline the methods of mesh analysis and nodal analysis. In performing these analysis, we will use the circuit of Figure 4 to guide our understanding.

0.5.1

Mesh Analysis

In performing mesh analysis, network equations are obtained by applying Kirchhoff’s voltage law around each mesh. The mesh-analysis method is applied as follows:

0.5. CANONICAL CIRCUIT ANALYSIS TECHNIQUES

Figure 2: Voltage divider is the circuitry given inside the fine dashed line.

Figure 3: Current divider is the circuitry given inside the fine dashed line.

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8

CONTENTS

Figure 4: Circuit with nodes, branches, meshes and loops. 1. Label each mesh with an Roman number and assign a distinct mesh current in the clockwise direction to each closed mesh of the network. The mesh-analysis method is applied as follows: • It is not absolutely necessary to choose the clockwise direction for each mesh current. In fact, any direction can be chosen for each mesh current with no loss in accuracy, as long as the remaining steps are followed properly. 2. Indicate the polarities within each mesh for each resistor as determined by the assumed direction of mesh current for that mest. • Note the requirement that the polarities be placed within each mesh. This requires, as shown in Figure 4, that the resistor R2 has two sets of polarities across it: one for mesh 1 and one for mesh 2. 3. Apply Kirchhoff’s voltage law around each mesh in the clockwise direction. • Again, the clockwise direction was chosen to establish uniformity. (a) When applying KVL to a component, if the component has two or more assumed mesh currents through it, the total current through the component is written as • the assumed current of the mesh in which Kirchhoff’s voltage law is being applied, • plus the assumed mesh currents of the other meshes passing through the elment in the same direction in which KVL is being appled, • minus the assumed mesh currents of the other meshes passing through the elment in the opposite direction in which KVL is being appled.

0.5. CANONICAL CIRCUIT ANALYSIS TECHNIQUES

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(b) When applying KVL to a source, the polarity of a source is unaffected by the direction of the assigned mesh currents. 4. Solve the resulting simultaneous linear equations for the assumed mesh currents (see further Section 0.6.3 herein). 5. Using the schematic and the now solved for mesh currents, solve for the current through any elements of interest.

0.5.2

Nodal Analysis

In performing nodal analysis, network equations are obtained by applying Kirchhoff’s current law at each node. If we define one node of a network as a reference (that is, a point of zero potential or ground), the remaining nodes of the network will all have a fixed potential relative to this reference. For a network of N nodes, therefore, there will exist N − 1 nodes with a fixed potential relative to the assigned reference node. Equations relating these N − 1 nodal voltages re written by applying Kirchhoff’s current law at each of the N − 1 nodes. The nodal-analysis method is applied as follows: 1. Ascribe a letter to each nodes of circuit. 2. Select one node as the ground reference. • The choice does not affect the result and is just a matter of convention. • Choosing the node with the most branches can simplify the analysis. 3. Create a table where each row contains a variable for each node voltage. 4. If the voltage of any nodes are known, write down its value in the second column of the aformentioned table. 5. For each unknown voltage, form an equation based on Kirchhoff’s current law. • For a resistive component, use Ohm’s Law to write the current between two nodes as ‘the voltage of the node with the higher potential, minus the voltage of the node with the lower potential, all divided by the resistance between the two nodes.’ 6. Solve the system of simultaneous equations for each unknown voltage (see further Section 0.6.3 herein). 7. Using the schematic and the now solved for node voltages, solve for the voltage across any elements of interest.

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0.6 0.6.1

CONTENTS

Algebra Operations a b + a c = a (b + c) a

ab b = c c

ac a  = b b c  a b c

=

a bc

a c ad+ bc + = b d bd a b a+b = + c c c ab+ ac = b + c for a 6= 0 a

0.6.2

Improper to Proper Fraction   b b a+ c c = c   c d d

a+

  b c a+ c = (d) c   b ac + c c = dc ac + b = dc

(1)

(2)

(3) (4)

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0.6. ALGEBRA d





 d c = b b c a+ a+ c c (d) c  =  b a+ c c dc  =  b ac + c c dc = ac + b   b b a+ cf c c =  e  cf e d+ d+ f f   b a+ cf c  = e d+ cf f   b c f acf + c  = e dcf + cf f acf + bf = dcf + ec

(5)

(6)

(7)

(8)

a+

0.6.3

(9)

(10)

(11)

(12)

Solution of 2×2 System of Equations

Many of the classroom problems and many of the home work problems are comprised of a system of two equations with two unknown variables. This section is provided as a courtesy for those students who want to better understand this general class of problems. In the study of circuit analysis, there are two instances where the solution of a system of equations becomes necessary. These instances are those in which either mesh analysis or nodal analysis is performed. In one or the other of these cases either Kirchoff’s-voltage law is used to write the sum of all voltages around a mesh is equal to zero, or Kirchoff’s-current law is used to write the sum of all currents into and out of a node is equal to zero. One then has to solve for an number of unknown node voltages or unknown loop currents. In what follows, you can see the general solusion of a system of two equations each with two unknown variables.

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CONTENTS

Theorem 1 (Solution of 2x2 System of Eqaution ). Let, x1 and x2 be variables that are to be solved for and let a11 , a12, a21 , a22 , b1 , and b2 be constants. Let that the following equations hold. 0 = a11 x1 + a12 x2 + b1 0 = a21 x1 + a22 x2 + b2

(13) (14)

Then the solution of x1 and x2 are given as a12 b2 − a22 b1 a11 a22 − a12 a21 −a11 b2 + a21 b1 x2 = a11 a22 − a12 a21

x1 =

(15a) (15b)

Proof. We can rewrite Eqations 13 and 14 as a21 · 0 = a21 [a11 x1 + a12 x2 + b1 ]

(16)

a11 · 0 = a11 [a21 x1 + a22 x2 + b2 ]

(17)

Now subtract bottom from top a21 · 0 − a11 · 0 = a21 [a11 x1 + a12 x2 + b1 ] − a11 [a21 x1 + a22 x2 + b2 ]

(18)

0 = [a21 a12 x2 + a21 b1 ] − a11 [a22 x2 + b2 ] 0 = a21 a12 x2 + a21 b1 − a11 a22 x2 − a11 b2

(19) (20)

0 = [a21 a12 − a11 a22 ] x2 + a21 b1 − a11 b2

(21)

Now we look at three cases 1. a21 a12 6= a11 a22 −a21 b1 + a11 b2 = x2 a21 a12 − a11 a22 −a21 b1 + a11 b2 x2 = a21 a12 − a11 a22 a21 b1 − a11 b2 x2 = a11 a22 − a21 a12 −a11 b2 + a21 b1 x2 = a11 a22 − a12 a21

(22) (23) (24) (25)

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0.6. ALGEBRA Now plug this result into Equation 13 to obtain

−b1 − a12



0 = a11 x1 + a12 [x2 ] + b1   −a21 b1 + a11 b2 0 = a11 x1 + a12 + b1 a21 a12 − a11 a22 

−a21 b1 + a11 b2 a21 a12 − a11 a22 = x1 a11

(26) (27)

(28) (29)

Now get rid of the offensive part of this improper fraction. The denominator is fine, but the numerator has a fraction in it. Since there is a denominator is the numerator, multiple bot top and bottom of the big fraction by the denominator as follows:   −a21 b1 + a11 b2 −b1 − a12 a21 a12 − a11 a22 [a21 a12 − a11 a22 ] = x1 (30) [a21 a12 − a11 a22 ] a  11  −a21 b1 + a11 b2 −b1 [a21 a12 − a11 a22 ] − a12 [a21 a12 − a11 a22 ] a21 a12 − a11 a22 = x1 (31) a11 [a21 a12 − a11 a22 ] −b1 [a21 a12 − a11 a22 ] − a12 [−a21 b1 + a11 b2 ] = x1 (32) a11 [a21 a12 − a11 a22 ] −b1 [−a11 a22 ] − a12 [a11 b2 ] = x1 (33) a11 [a21 a12 − a11 a22 ] b1 a22 − a12 b2 = x1 (34) a21 a12 − a11 a22 2. a21 a12 = a11 a22 and a21 b1 = a11 b2 Returning to Equation 21, we can see that if these to conditions are met that 0 = 0 x + 0. This means that any real value of x satisfies the equation. So that there are infinitely many solutions for x that satisfy the linear system of equations. We can write the solution set as {(x1 , x2 ) | a11 x1 + a12 x2 + b1 = 0} 3. a21 a12 = a11 a22 and a21 b1 6= a11 b2

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CONTENTS Returning to Equation 21, we can see that if these to conditions are met that 0 = [a21 a12 − a11 a22 ] x2 + a21 b1 − a11 b2 0 = 0 x2 + a21 b1 − a11 b2 0 = a21 b1 − a11 b2 0 a21 b1 − a11 b2 = a21 b1 − a11 b2 a21 b1 − a11 b2 0=1

(35)

This is a cotradiction in terms, because zero is not equal to one. So we find that there are exists no solutions to the linear system of equations. We can write the solution set as {(x1 , x2 )} = ∅, In conclusion, if one can first write the electrical system in a form given in Equations 13 and 14, then one can directly obtain the solutions to the unknown circuit varaibles by using the above boxed results....