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Irwin, Engineering Circuit Analysis, 11e, ISVSOLUTION:Irwin, Engineering Circuit Analysis, 11e, ISVSOLUTION:Irwin, Engineering Circuit Analysis, 11e, ISVSOLUTION:If 퐼 0 = 1퐴 then 푉 1 = 3 + 5 퐼 0 = 8 푉 and 퐼 1 =푉 41 = 2 퐴.ApplyingKCL at node 1 gives퐼 2 =퐼 1 +퐼 0 = 3 퐴푉 2 =푉 1 + 2퐼 2 = 8 + 6 = 14 푉, 퐼...

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1 Irwin, Engineering Circuit Analysis, 11e, ISV

SOLUTION:

Chapter 5: Additional Analysis Techniques

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Chapter 5: Additional Analysis Techniques

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Chapter 5: Additional Analysis Techniques

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SOLUTION: If 𝐼0 = 1𝐴 then 𝑉1 = 3 + 5 𝐼0 = 8 𝑉 and 𝐼1 =

𝑉1 4

= 2 𝐴.ApplyingKCL at node 1 gives

𝐼2 = 𝐼1 + 𝐼0 = 3 𝐴 𝑉2 = 𝑉1 + 2𝐼2 = 8 + 6 = 14 𝑉,

𝐼2 =

𝑉2 7

=2𝐴

Applying KCL at node 2 gives 𝐼4 = 𝐼3 + 𝐼2 = 5 𝐴 Therefore,𝐼𝑠 = 5 𝐴.This shows that assuming𝐼0 = 1 𝐴 gives𝐼𝑠 = 5 𝐴.The actual source current of 15 A will give 𝐼0 = 3 𝐴 as the actual value.

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SOLUTION:

Open circuit the 4 A source. Then, (7 + 2) || (5 + 5) = 4.737 Ω, we can calculate v1’ =1.4.737 = 4.737V To find the total current flowing through the 7 Ω resistor, we first determine the total voltage v1 by continuing our superposition procedure. The contribution to v1 from the 4 A source is found by first open-circuiting the 1 A source, then noting that current division yields: 5 20 4 = = 1.053A 5 + (5 + 7 + 2) 19 V1’’ = (1.053)(9) = 9.477V Hence v1 = 14.21V We may now find the total current flowing downward through the 7 Ω resistor as 14.21/7 = 2.03 A.

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SOLUTION:

Short-circuit the 10 V source Note that 6 || 4 = 2.4 Ω. By voltage division, the voltage across the 6 Ω resistor is then 2.4 4× = 1.778V 3 + 2.4 So that i1’ = 0.2693A Short-circuit the 4 V source Note that 3 || 6 = 2 Ω. By voltage division, the voltage across the 6 Ω resistor is then 2 −10 × = −3.33V 6 i1’’= -5.556mA a. I = i1’ + i1’’ = -259.3mA

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SOLUTION:

Taking one source at a time: The contribution from the 24-V source may be found by shorting the 45-V source and opencircuiting the 2-A source. Applying voltage division vx‘ = 24

20 10 +20+45 ||30

20

= 24 10 +20+18 = 10V

We find the contribution of the 2-A source by shorting both voltage sources and applying current division: 20

vx’’ = 24 10 +20+18 = 8.333V Finally, the contribution from the 45-V source is found by open-circuiting the 2-A source and shorting the 24-V source. Defining v across the 30-Ω resistor with the “+” reference on top: 30

0=

v30 v30 v30 − 45 + + 20 10 + 20 45

solving, v = 11.25 V and hence v ”’ = -11.25(20)/(10 + 20) = -7.5 V 30

x

Adding the individual contributions, we find that v = v ’ + v ” + v ”’ = 10.83 V x

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x

x

x

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SOLUTION:

We find the contribution of the 4-A source by shorting out the 100V source and analysing the resulting circuit:

4=

V1′

+

20 V ′

V 1′ −V′ 10 V ′ −V′

1 4i1’ = 1 + 10 30 Where i1 ' = V '/ 20

1

Simplifying & collecting terms, we obtain

30 V ' – 20 V' = 800 1

-7.2 V ' + 8 V' = 0 1

Solving, we find that V' = 60 V. Proceeding to the contribution of the 60-V source, we analyse the following circuit after defining a clockwise mesh current i flowing in the left a

mesh and a clockwise mesh current i flowing in the right mesh. b

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30 ia – 60 + 30 ia – 30 ib = 0 ib = -0.4i1" = +0.4ia Solving, we find that ia = 1.25 A and so V" = 30(ia – ib) = 22.5 V Thus, V = V' + V" = 82.5 V.

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SOLUTION: Since there are two sources, let 𝑣 = 𝑣1 + 𝑣2 Where 𝑣1 and𝑣2 are the contributions due to the 6-V voltage sourceand the 3-A current source, respectively. To obtain 𝑣1 we set the currentsource to zero, as shown in the figure below.

Applying KVL to the loop, we get, 12𝑖1 − 6 = 0 Thus,

=>

𝑖1 = 0.5 𝐴

𝑣1 = 4𝑖1 = 2 𝑉

To get 𝑣2 we set the voltage source to zero, as shown.

Using current division, 𝑖3 =

8 4+8

3 =2𝐴

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𝑣2 = 4𝑖3 = 8 𝑉

And we find,

𝑣 = 𝑣1 + 𝑣2 = 10 𝑉

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SOLUTION: The circuit involves a dependent source, which must be leftintact. We let, 𝑖0 = 𝑖0𝑎 + 𝑖0𝑏 Where𝑖0𝑎 and 𝑖0𝑏 are due to the 4-A current source and 20-V voltagesource respectively. To obtain𝑖0𝑎 we turn off the 20-V source so that we have the circuit shown below.

We apply mesh analysis in order toobtain 𝑖0𝑎 . For loop 1, 𝑖1 = 4 𝐴

(i)

−3𝑖1 + 6𝑖2 − 𝑖3 − 5𝑖0𝑎 = 0

(ii)

For loop 2,

For loop 3,

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22 Irwin, Engineering Circuit Analysis, 11e, ISV −5𝑖1 − 𝑖2 + 10𝑖3 + 5𝑖𝑜𝑎 = 0

(iii)

𝑖3 = 𝑖1 − 𝑖0𝑎 = 4 − 𝑖0𝑎

(iv)

But at node 0,

Solving equations (i)-(iv), we get,

𝑖0𝑎 =

52 17

𝐴

To obtain 𝑖𝑜𝑏 we turn off the 4-A current source so that the circuitbecomes as that shown below.

For loop 4, KVL gives, 6𝑖4 − 𝑖5 − 5𝑖0𝑏 = 0

(v)

−𝑖4 + 10𝑖5 − 20 + 5𝑖0𝑏 = 0

(vi)

And for loop 5,

But, 𝑖5 = −𝑖0𝑏 . Thus, using this and equations (v) and (vi) we get, Hence,

𝑖0 = 𝑖𝑜𝑎 + 𝑖0𝑏 = −0.4706 𝐴

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𝑖0𝑏 = − 17 𝐴 60

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SOLUTION: In this case, we have three sources. Let

𝑖 = 𝑖1 + 𝑖2 + 𝑖3

Where, 𝑖1 , 𝑖2 and 𝑖3 are due to the 12-V, 24-V, and 3-A sources respectively. To get 𝑖1 , consider the circuit shown below.

Combining the 4 Ω (on the right-hand side) in series with 8 Ω gives 12 Ω. The 12 Ω inparallel with 3 Ω gives4Ω as shown in the figure above. Thus, To get 𝑖2 , consider the circuit shown below,

Chapter 5: Additional Analysis Techniques

𝑖1 =

12 6

=2𝐴

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Applying mesh analysisgives, 16𝑖𝑎 − 4𝑖𝑏 + 24 = 0

=>

4𝑖𝑎 − 𝑖𝑏 = −6

(i)

7𝑖𝑏 − 4𝑖𝑎 = 0

=>

𝑖𝑎 = 𝑖𝑏 4

(ii)

7

𝑖2 = 𝑖𝑏 = −1 𝐴

From (i) and (ii), we get,

To get𝑖3 , consider the circuit shown below.

Using nodal analysis gives, 𝑣2 8

+

𝑣2 −𝑣1 4

𝑣2 −𝑣1 4

=

𝑣1 4

=3

=>

3𝑣2 − 2𝑣1 = 24

(iii)

𝑣1 3

=>

𝑣2 =

10 𝑣 3 1

(iv)

+

Substituting (iv) into (iii) leads to 𝑣1 = 3 and 𝑖3 = Thus,

𝑣1 3

=1𝐴

𝑖 = 𝑖1 + 𝑖2 + 𝑖3 = 2 𝐴

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SOLUTION:

One approach to this problem is to write a set of mesh equations, leaving the voltage source and current source as variables which can be set to zero. We first rename the voltage source as V . We next define three clockwise mesh currents in x

the bottom three meshes: i , i and i . Finally, we define a clockwise mesh current i in the 1

y

4

top mesh, noting that it is equal to –4 A Our general mesh equations are then: -V + 18i – 10i = 0 x

1

y

–10i + 15i – 3i = 0 1

y

4

–3i + 16i – 5i = 0 y

4

3

Set V = 10 V, i = 0. Our mesh equations then become x

3

18i – 10iy’ = 10 1

–10i + 15iy’ – 3i = 0 1

4

– 3iy + 16i = 0 4

Solving, iy’ = 0.6255 A. Set V = 0 V, i = – 4 A. Our mesh equations then become x

3

18i – 10iy’’ = 0 1

–10i + 15iy’’ – 3i = 0 1

4

-3iy’’ + 16i4 = -20 Solving,iy’’ = –0.4222 A Thus, iy = iy’ + iy’’ = 203.3mA Chapter 5: Additional Analysis Techniques

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SOLUTION:

Shorting the 14 V source, we find that RTH = 10 || 20 + 10 = 16.67 Ω. Next, we find V by determining V (recognising that the right-most 10 Ω resistor carries no TH

OC

current, hence we have a simple voltage divider): 10 + 10 = 9.33V VTH = Voc = 14 10 + 10 + 10 Thus, our Thevenin equivalent is a 9.333 V source in series with a 16.67 Ω resistor, which is in series with the 5 Ω resistor of interest. 9.333

Now, I5Ω = 5+16.67 = 0.4307A Thus, P5Ω = (0.4307)2.5 = 927.5mW

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SOLUTION:

First R = 10 mV/ 400 μA = 25 Ω TH

Then R = 110 V/ 363.6×10-3 A = 302.5 Ω TH

Increased current leads to increased filament temperature, which results in a higher resistance (as measured). This means the Thévenin equivalent must apply to the specific current of a particular circuit – one model is not suitable for all operating conditions (the light bulb is nonlinear).

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SOLUTION:

a. Shorting out the 88-V source and open-circuiting the 1-A source, we see looking into the terminals x and x' a 50-Ω resistor in parallel with 10 Ω in parallel with (20 Ω + 40 Ω), so R = 50 || 10 || (20 + 40) = 7.317 Ω TH

Using superposition to determine the voltage V across the 50-Ω resistor, we find xx'

Vxx ′ = VTH

50||(20 + 40) 40 = 88 ] + 1 (50||10)[ 40 + 20 + (50| 10 10 + [50|| 20 + 40 ]

= 69.27V b. Shorting out the 88-V source and open-circuiting the 1-A source, we see looking into the terminals y and y' a 40-Ω resistor in parallel with [20 Ω + (10 Ω || 50 Ω)]: R = 40 || [20 + (10 || 50)] = 16.59 Ω TH

Using superposition to determine the voltage V across the 1-A source, we find yy' 27.27 40 V𝑦𝑦 ′ = VTH = 1. R TH + 88 ] [ 20 + 40 10 + 27.27 = 59.52V

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SOLUTION:

a. Removing terminal c, we need write only one nodal equation 𝑣𝑦 𝑣𝑦 − 𝑣𝑥 3= + 3 2 which may be solved to yield V = 4 V. Therefore, Vab = VTH = 2 – 4 = 2V b

R

TH

= 12 || 15 = 6.667 Ω. We may then calculate I as I = V / R N

= -300 mA (arrow pointing upwards) b. Removing terminal a, we again find R

TH

N

TH

TH

= 6.667 Ω, and only need write a single

nodal equation; in fact, it is identical to that written for the circuit above, and we once again find that V = 4 V. In this case, V = V = 4 – 5 = -1 V, so I = -1/ 6.667 b

TH

= –150 mA (arrow pointing upwards)

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bc

N

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