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ECE 221Electric Circuit Analysis I

Chapter 10Circuit Analysis 3 Ways

Herbert G. Mayer, PSUStatus 10/21/2015

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Syllabus

Goal Sample Problem 1 Solve by Substitution KCL Using Cramer’s Rule Solve by Node Voltage Method Solve by Mesh Current Method Conclusion Problem 1 Same for Problem 2

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Goal We’ll analyze simple circuits, named Sample

Problem 1 and Sample Problem 2

With various constant voltage sources and resistors

Goal is to compute branch currents i1, i2, and i3

First by using conventional algebraic substitution, applying Kirchhoff’s Laws; we’ll need 3 equations

Secondly, we use the Node Voltage Method

Thirdly we compute fictitious currents ia and ib, using the Mesh Current Method

Any method may apply Cramer’s Rule to conduct the arithmetic computations, once the equations exist

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Problem 1

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Circuit for Sample Problem 1

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Solve Solve Problem 1Problem 1

Via KCL, KVLVia KCL, KVL

Using Arithmetic SubstitutionUsing Arithmetic Substitution

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Sample Problem 1: 3 Equations

KCL at node n1:

(1) i1 = i2 + i3

KVL in the left mesh, labeled ia:

(2) R1*i1 + R3*i3 - v1 = 0

KVL in the right mesh, labeled ib:

(3) R2*i2 + v2 - R3*i3 = 0

(3)’ i3 = (R2*i2)/R3 + v2/v3

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Solve Problem 1 Arithmetic Substitution

(1) in (2)

R1*(i2+i3) + R3*i3 = v1

R1*i2 + R1*i3 + R3*i3 = v1

R1*i2 + i3*(R1+R3) = v1

R1*i2 + (R2*i2 + v2)*(R1+R3)/R3 = v1

. . .

i2*(R1+R2*(R1+R3)/R3) = v1-v2*(R1+R3)/R3

. . .

i2*(100+2*400/3) = 10 - 20*(400/300)

i2 = -45.45 mA

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Solve Problem 1 Arithmetic Substitution

i3 = i2 * R2/R3 + v2/R3 = -0.0303+0.066667

i3 = 0.03636 A

i3 = 36.36 mA

i1 = i2 + i3

i1 = -9.09 mA

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Solve Solve Problem 1Problem 1

Via KCL, KVLVia KCL, KVL

Using Cramer’s RuleUsing Cramer’s Rule

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Solve Problem 1 Using Cramer’s Rule

i1 = i2 + i3

R1*i1 + R3*i3 - v1 = 0

R2*i2 + v2 - R3*i3 = 0

Normalized:

i1 – i2 - i3 = 0

R1*i1 + 0 + R3*i3 = v1

0 + R2*i2 - R3*i3 = -v2

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Cramer’s Characteristic Determinant

Normalize i1, i2, i3 positions in matrix

| 1 -1 -1 | | 0 |Δ = | R1 0 R3 |, R = | v1 |

| 0 R2 -R3 | |-v2 |

| 1 -1 -1 |Δ = |100 0 300 |

| 0 200 -300 |

| 1 -1 1 |S = | -1 1 -1 |

| 1 -1 1 |

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Cramer’s Characteristic Determinant

Δ = 1 | 0 300 | -100 | -1 -1 | + 0| 200 -300 | | 200 -300|

Δ = 1*( 0 – 60,000 ) - 100*( 300 + 200 )

Δ = -60k - 50k

Δ = -110,000

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Numerator Determinant N1, and i1| 0 -1 -1 |

N(i1) = N1 = | 10 0 300 ||-20 200 -300|

N1 = -10 | -1 -1 | -20|-1 -1|| 200 -300| | 0 300|

N1 = -10 * (300+200) -20 * (-300 )

N1 = -10*500 + 6,000

N1 = 1,000

i1= 1,000 / -110,000

i1 = -0.00909 A = -9.09 mA

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Numerator Determinant N2, and i2

| 1 0 -1 |N(i2) = N2 = | 100 10 300 |

| 0 -20 -300 |

N2 = 1 | 10 300 | -100 | 0 -1 ||-20 -300 | | -20 -300|

N2 = -3,000 + 6,000 -100 * ( 0 - 20 )

N2 = 3,000 + 2,000 = 5,000

i2 = 5,000 / -110,000

i2 = -0.04545 A = -45.45 mA

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Numerator Determinant N3, and i3

| 1 -1 0 |N(i3) = N3 = | 100 0 10 |

| 0 200 -20 |

N3 = 1 | 0 10 | -100 | -1 0 || 200 -20 | | 200 -20 |

N3 = -2,000 - 100 * (20 ) = -4,000

i3= -4,000 / -110,000

i3 = 0.0363636 A = 36.36 mA

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Solve Solve Problem 1Problem 1

Using NoVoMoUsing NoVoMo

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Solve Problem 1 by Node Voltage Method

Ignoring the current or voltage directions from the substitution method, we use the Node Voltage Method at node n1, currents flowing toward reference node n2

We generate 1 equation with unknown V300, voltage at the 300 Ω resistor, yielding i3

Once known, we can compute the voltages at R1 and R2, and thus compute the currents i1 and i2, using Ohm’s law

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Solve Problem 1 by Node Voltage Method

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Solve Problem 1 by Node Voltage Method

3 currents flowing from n1 toward reference node n2:

V300/300 + (V300-10)/100 + (V300-20)/200 = 0

V300 + 3*V300 + V300*2/3 = 30 + 3*20/2

V300*( 1 + 3 + 2/3 ) = 60

Students Compute V300

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Solve Problem 1 by Node Voltage Method

3 currents flowing from n1 toward reference node n2:

V300/300 + (V300-10)/100 + (V300-20)/200 = 0

V300 + 3*V300 + V300 * 3/2 = 30 + 3*20/2

V300*( 1 + 3 + 3/2 ) = 60

V300 = 60 * 2 / 11

V300 = 10.9090 V

Students Compute i3

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Solve Problem 1 by Node Voltage Method

3 currents flowing from n1 toward reference node n2:

V300/300 + (V300-10)/100 + (V300-20)/200 = 0

V300 + 3*V300 + V300 * 3/2 = 30 + 3*20/2

V300*( 1 + 3 + 3/2 ) = 60

V300 = 60 * 2 / 11

V300 = 10.9090 V

i3 = V300 / 300

i3 = 36.363 mA

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Solve Problem 1 by Node Voltage Method

V(R1) = v1 - V300

V(R1) = 10 - 10.9090 = -0.9090 V

i1 = V(R1) / R1

i1 = -0.9090 / 100

i1 = -9.09 mA

Students Compute i2

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Solve Problem 1 by Node Voltage Method

V(R1) = v1 - V300

V(R1) = 10 - 10.9090 = -0.9090 V

i1 = V(R1) / R1

i1 = -0.9090 / 100

i1 = -9.09 mA

From this follows i2 using KCL:

i2 = i1 - i3

i2 = -9.0909 – 36.3636

i2 = -45.45 mA

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Solve Solve Problem 1Problem 1

Using MeCuMoUsing MeCuMo

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Solve Problem 1 by Mesh Current Method The mesh current is fictitious, one such current

associated with its own individual mesh

Fictitious in the sense as if it were uniquely tied to a mesh; yet depending on the branch of the mesh, mesh currents from other parts flow though that very mesh as well

Kirchhoff’s current law is trivially satisfied, but mesh currents are not everywhere measurable with an Ampere meter: not measurable, when currents from other meshes super-impose

In Sample Problem 1 we have 2 meshes, with mesh currents indicated as ia and ib

But we must track that, R3 for example, has both flowing though it in opposing directions

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Solve Problem 1 by Mesh Current Method

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Solve Problem 1 by Mesh Current Method

KVL for mesh with ia yields:

(1) R1*ia + R3*(ia-ib) = v1

KVL for mesh with ib yields:

(2) R3*(ib-ia) + R2*ib = -v2

Students Compute (1) for ib

Then substitute ib in (2)

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Solve Problem 1 by Mesh Current Method

KVL for mesh with ia yields:

(1) R1*ia + R3*(ia–ib) = v1

KVL for mesh with ib yields:

(2) R3*(ib-ia) + R2*ib = -v2

From (1) follows:

(1) ib = ( R1*ia + R3*ia - v1 ) / R3

Substitute ib in (2):

(2) -v2 = ib*(R2+R3) - R3*ia

-v2 = ia*(R1+R3)*(R2+R3)/R3 -

v1*(R2+R3)/R3 - R3*ia

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Solve Problem 1 by Mesh Current Method

v1*(R2+R3)/R3 - v2 =

ia*( (R1+R3)*(R2+R3)/R3 – R3)

-20 + 10*5/3 = ia*(400*500/300 – 300)

ia = -10 / 1100

ia = -0.00909 A = -9.09 mA

Since ia = i1:

i1 = -9.09 mA

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Solve Problem 1 by Mesh Current Method

Recall (1):

(1) R1*ia + R3*(ia–ib) = v1

R3*ib = ia*(R1+R3) - v1

ib = ia*(R1+R3)/R3 - v1/R3

ib = -10*400/(1,100*300) - 10/300

ib = -0.04545 A = -45.45 mA

since i2 = ib:

i2 = -45.45 mA

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Conclusion Problem 1 via Mesh Current

Since i3 = i1 - i2, i3 = -9.09 mA - -45.45 mA

it follows:

i3 = 36.36 mA

We see consistency across 3 different approaches to circuit analysis

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Problem 2Problem 2

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Sample Problem 2 We’ll analyze another, similar circuit, named Sample

Problem 2

With 2 constant voltage sources of 3 V and 4 V

Plus 3 resistors at 100, 200, and 300 Ohm

Again we compute 3 branch currents i1, i2, and i3

Using 3 methods:

First we use substitution, applying Kirchhoff’s Laws

Then we use the Node Voltage Method

Thirdly the Mesh Current Method

Any of these methods may use Cramer’s Rule

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Circuit for Sample Problem 2

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Sample Problem 2: Three Equations

KCL states:

(1) i1 = i2 + i3

KVL in the upper mesh labeled ia yields:

(2) i1*100 + i2*200 -3 = 0

KVL in the lower mesh, labeled ib yields:

(3) -i2*200 + i3*300 + 4 + 3 = 0

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Solve Problem 2 by Substitution

-200*i2 + (i1-i2)*300 = -7 // (1)in(3)

-500*i2 + 300*i1 = -7 // (3’)

100*i1 + 200*i2 = 3 // (2)*3

300*i1 + 600*i2 = 9 // (2’)

(3’)-(2’)

-500*i2 - 600*i2 = -7 -9 = -16

i2*1,100 = 16

i2 = 16 / 1,100

i2 = 14.54 mA

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Solve Problem 2 by Substitution

i1*100 + i2*200 = 3

i1*100 = 3-200*(16/1,100)

i1*100 = 100/1,100

i1 = 1 / 1,100

i1 = 0.91 mA

i3 = i1 - i2

i3 = -15 / 1,100

i3 = -13.63 mA

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Solve Solve Problem 2Problem 2

Via KCL, KVLVia KCL, KVL

Using Cramer’s RuleUsing Cramer’s Rule

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Solve Problem 2 Using Cramer’s Rule

i1 = i2 + i3

i1*100 + i2*200 -3 = 0

-i2*200 + i3*300 +4 +3 = 0

Normalized:

i1 - i2 - i3 = 0

100*i1 + 200*i2 + 0 = 3

0 - 200*i2 + 300*i3 = -7

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Cramer’s Characteristic Determinant

Normalize i1, i2, i3 positions

| -1 1 1 | | 0 |D = | 100 200 0 |, R = | 3 |

| 0 -200 300 | | -7|

| 1 -1 1 |S = | -1 1 -1 |

| 1 -1 1 |

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Cramer’s Characteristic Determinant

Δ = -1 | 200 0 | -100 | 1 1 | + 0| 200 -300| |-200 300 |

Δ = -60,000 – 50,000 = -110,000

Δ = -110 k

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Numerator Determinant N1, and i1

| 0 1 1 |N(i1) = N1 = | 3 200 0 |

| -7 -200 300 |

N1 = 0 - 3| 1 1 | -7 | 1 1 ||-200 300 | |200 0 |

Students Compute N1,

Given Δ = -110 k

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Numerator Determinant N1, and i1| 0 1 1 |

N(i1) = N1 = | 3 200 0 || -7 -200 300 |

N1 = 0 - 3| 1 1 | -7 | 1 1 ||-200 300 | |200 0 |

N1 = -3*(300+200) -7*(-200) =

N1 = -1,500 + 1,400

N1 = -10

Now Students Compute i1

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Numerator Determinant N1, and i1| 0 1 1 |

N(i1) = N1 = | 3 200 0 || -7 -200 300 |

N1 = 0 - 3| 1 1 | -7 | 1 1 ||-200 300 | |200 0 |

N1= -3*(300+200) -7*(-200) =

N1= -1,500 + 1,400

N1= -100

i1 = -100 / -110,000

i1 = 0.000909 A

i1 = 0.91 mA

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Numerator Determinant N2, and i2

| -1 0 1 |N(i2) = N2 = |100 3 0 |

| 0 -7 300 |

N2 = -1 | 3 0 | -100 | 0 1 | + 0| -7 300 | | -7 300|

N2 = -(900) - 100* (7) = -1,600

i2 = -1,600 / -110,000

i2 = 14.54 mA

With i3 = i1 - i2 it follows:

i3 = -13.63 mA

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Solve Solve Problem 2Problem 2

Using NoVoMoUsing NoVoMo

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Solve Problem 2 by Node Voltage Method

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Solve Problem 2 by Node Voltage Method

There are 2 essential nodes, n1 and n2

One will be selected as reference node: pick n2

Compute 3 currents from n1 to n2, express as function of v200

Students compose single KCL equation

For node n1, using single unknown v200

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Solve Problem 2 by Node Voltage Method

Use KCL to compute 3 current from n1 toward reference node n2:

V200/200 + (V200-3)/100 + (V200-3-4)/300 = 0

Students compute v200, and i2

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Solve Problem 2 by Node Voltage Method

Use KCL to compute 3 current from n1 toward reference node n2:

V200/200 + (V200-3)/100 + (V200-3-4)/300 = 0

V200*(3/2 + 3 + 1 ) = 9 + 7

V200*11/2 = 16

V200 = 2.9090 V

i2 = V200 / 200

i2 = 14.54 mA

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Solve Problem 2 by Node Voltage Method

KVL in the lower mesh, with V300 being the voltage drop across the 300 resistor, yields:

V300 = -7 + V200 = -7 + 2.9090 = -4.091 V

i.e. i3 = V300/300 = -0.013637 mA

i3 = -13.63 mA

i1 = i2 + i3 = 14.54 - 13.63

i1 = 0.91 mA

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Solve Solve Problem 2Problem 2

Using MeCuMoUsing MeCuMo

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Solve Problem 2 by Mesh Current Method

Again we analyze 2 meshes, with fictitious currents ia and ib

Circuit is repeated below for convenience

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Mesh Current In Sample Problem 2

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Solve Problem 2 by Mesh Current Method

KVL for mesh with ia yields:

(1) 100*ia + 200*( ia-ib ) = 3

(1) 300*ia - 200*ib = 3

(1) ib = (300*ia-3)/200

KVL for mesh with ib yields:

(2) 300*ib+200*( ib – ia ) = -7

(2) 500*ib-200*ia = -7

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Solve Problem 2 by Mesh Current Method

Substitute ib from (1) in (2):

500*(300*ia - 3)/200 - 200*ia = -7

ia = 1/1,100 = 0.91 mA

i1 = ia, hence:

i1 = 0.91 mA

ib = 3*ia/2-3/200 = 3/(1,100 * 2) - 3/200

ib = -13.63 mA

i3 = ib, hence

i3 = -13.63 mA

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Solve Problem 2 by Mesh Current Method

With i2 = i1 - i3, it follows:

i2 = 14.54 mA

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Which Method is easiest?

• It seems the Node Voltage Method is simplest for these problems

• With the smallest number of equations

• Mesh Current method has smaller number of equations than pure KCL and KVL

• Small number of equations yields less chances for sign confusion

• But for a large number of unknowns Cramer’s Rule is THE methodical way to compute