1 ECE 221 Electric Circuit Analysis I Chapter 10 Circuit Analysis 4 Ways Herbert G. Mayer, PSU...

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ECE 221Electric Circuit Analysis I

Chapter 10Circuit Analysis 4 Ways

Herbert G. Mayer, PSUStatus 11/23/2014

For use at Changchun University of Technology CCUT

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Syllabus

Goal Sample Problem1 Solve by Substitution KCL Solve by Cramer’s Rule Solve by Node-Voltage Method Solve by Mesh-Current Method Conclusion Problem1 Same for Problem2

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Goal We’ll analyze simple circuits, named Sample

Problem1 and Sample Problem2

Both with 2 constant voltage sources, 3 resistors

Computing 3 branch currents i1, i2, and i3

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

Secondly, we use Cramer’s Rule, with these 3 equations, normalizing them to determinant form

Then, we use the Node-Voltage Method

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

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Problem 1

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Circuit for Sample Problem1

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Sample Problem1: 3 Equations

KCL at node n1 states:

(1) i1 = i2 + i3

KVL in the left mesh labeled ia yields:

(2) R1*i1 + R3*i3 - V1 = 0

KVL in the right mesh, labeled ib:

(3) R2*i2 + V2 - R3*i3 = 0

i.e. i3 = (R2*i2)/R3 + V2/R3

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Solve Problem1 by 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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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 Problem1 by Cramer’s RuleUsing 3 Equations from Substitution

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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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): | 10 0 300 |

|-20 200 -300 |

N1 = 0 -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): | 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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| 1 -1 0 |

N(i3): | 100 0 10 || 0 200 -20

|

N3 = 1 | 0 1 | -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 Problem1 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, generating 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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Problem1 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

V300= 60 * 2 / 11

V300= 10.9090 V

i3 = V300 / 300

i3 = 36.36a mA

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Problem1 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 Problem1 by Mesh-Current Method The mesh current is fictitious, one each associated

with each individual mesh

Fictitious in the sense that we act 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 directly measurable with an Ampere meter, when currents from other meshes super-impose

In the Sample Problem 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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Sample Problem1: Two Mesh Currents

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Solve Problem1 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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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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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 Problem1 via Mesh-Current

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

it follows:

i3 = 36.36 mA

We see consistency across 4 different methods of circuit analysis

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Problem 2

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Sample Problem2 We’ll analyze a similar circuit, named Sample

Problem2

With 2 constant voltage sources of 3 V and 4 V

Plus 3 resistors at 100, 200, and 300 Ohm

But now the elements are connected differently

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

Using 4 methods: substitution method, using Kirchhoff’s Laws

Then we use Cramer’s Rule

Thirdly we use the Node-Voltage Method

Finally the Mesh-Current Method

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Circuit for Sample Problem2

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Sample Problem2: 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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-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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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 Problem2 by 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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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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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): | 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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| -1 0 1 |

N(i2): | 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

From this follows i3 = i1-i2

i3 = -13.63 mA

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Node-Voltage With Sample Problem2

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

The circuit is repeated here for convenience

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Mesh-Current With Sample Problem2

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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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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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With i2 = i1 – i3, it follows:

i2 = 14.54 mA

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Which Method is easiest?

• It seems the Mesh-Current Method is simplest for these problems

• With the smallest number of equations

• And less chances for students’ sign confusion, as we follow the same sign (or: direction) through the whole mesh

• But for a large number of unknowns Cramer’s Rule is the only way to compute accurately

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