pp_chap1_6.doc

Engineering Circuit Analysis, Sixth Edition Practice Problem SolutionsChapters One through Six

2.1 (p9)

= 248 nm = 0.248 m (c)

2.2 (p9)

1 ns = 1000 ps (d)

2.3 (p11)

electrons moving left result in a positive current flowing right. Thus, I2 = + 1 mA, and I1 = I 2 = 1 mA .

2.4 (p13)

v1 = v2 so v2 = v1 = 17 V

2.5 (p15)

We have + 4.6 A flowing into the positive reference terminal of a voltage 220 mV. Using the passive sign convention, the absorbed power is Pabs = 4.6 0.220 = 1.012 W.

2.6 (p15)

We have a current 1.75 A flowing out of our positive voltage reference terminal, or + 1.75 A flowing into the positive reference terminal.

Thus, applying the passive sign convention we find

Pabs = 1.75 3.8 = 6.65 W or a generated power of Pabs = + 6.65 W.

2.7 (p15)

We see that a current of + 3.2 A flows out of our positive voltage reference terminal, so a current 3.2 A flows into that positive reference terminal.

Applying the passive sign convention,

Pabs = 8e 100t 3.2 = 25.6 e 100t W

= 15.53 W

Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved


2.8 (p18)

Moving from left to right and applying the passive sign convention,

Pabs = 7 8 = 56 W Pabs = 2 8 = + 16 WPabs = 5 12 = 60 W Pabs = 8 20 = + 160 WPabs = 0.25vx 20 = [0.25 12] 20 = 60 W

(Check: 56 + 16 60 + 160 60 = 0 )

2.9 (p23)

3.938 k

2.10 (p23)

= 1.89 W

2.11 (p23)

30 mA



3.1 (p32)

5 branches3 nodes

Define out of the “+” terminal of the source; downward through resistor RA.

By KCL,

so

By Ohm’s Law, 1

3.2 (p34)

By KVL, + 3 + 1 + vx = 0 so vx = 4 V

By Ohm’s Law, 400 mA

3.3 (p36)

We begin by writing a simple KVL equation:

where v10 has a “+” reference on the top terminal of the 10 resistor.

The current down through the 10 resistor is

Thus, 2 1.4 = 0.6 A flows left to right through the top 2 resistor.

KVL yields 12.8 V

3.4 (p37)

A single KVL equation yields

or



240 W

120 W

60 W

60 W

(Check: 240 + 120 + 60 + 60 = 0)

3.5 (p38)

Writing a single KVL equation, with a clockwise current i

and

leads to

768 mW

1.92 W

204.8 mW

179.2 mW

3.072 W

3.6 (p39)

Summing the currents entering the top node,

leads to

v = 50 V



3.7 (p40)

Summing the currents entering the top node,

leads to vx = 54 V

3 A, 5.4 A , 6 A

3.8 (p43)

leads to v = 50 V

3.9 (p45)

We first simplify the circuit:

where Vs = 5 + 5 + 5 =15 VR = 15 + 25 +5 = 45

333.3 mA

3.10 (p46)

3.11 (p48)

We first simplify the circuit:

where R = 4 // (2 + 2) // (10 + 10)= 4 // 4 // 20= 1.818

V = 7R = 7 (1.818) = 12.73 V


i

+

v

-


3.12 (p50)

Noting that we can combine the two 10 resistors into a single 5 resistor,

2 V

3.13 (p52)

(1) The resistors to the right of the 125 resistor combine into R = 50 // [2 + 240 // (40 + 20)] = 25

By current division, 100 mA

(2) The resistors to the right of the 50 resistor combine into Rx = 2 + 240 // (40 + 20) = 50

By current division, 50 mA

(3) The current through the 20 resistor is

Thus, v3 = 20 0.04 = 0.8 V



4.1 (p71)

The bottom node is the obvious choice for a reference node.

At the top left node

and at the top right node,

Simplifying, we obtain

30 = 2.5v1 v2 [1]and 60 = v1 + 4v2 [2]

Solving, we obtain v1 = 20 V and v2 = 20 V

4.2 (p74)

we number the top nodes 1, 2 and 3 moving left to right.

At node 1,

At node 2,

At node 3,

Simplifying, we obtain

Solving,

v1 = 5.235 Vand v3 = 11.47 V



4.3 (p75)

Define the top left node 1, and the top right node 2.

Treating these two nodes as a single supernode,

or

The remaining equation is simply v1 v2 = 5 [2]

Solving, v1 = 5.375 V and v2 = 375.0 mV

4.4 (p77)

(a)

Simplifying, 120 = 4.2vx 1.2vy [1] 150 = vx + 2.25 vy [2]

Solving, vx = 10.91 V

(b)

vx = 10 V

(c) Form a supernode between nodes x and y. Then

[1]

and [2]

Simplifying,

and




(d) At node x,

(vx = vy due to short circuit)


4.5 (p80)

mesh 1:

mesh 2:

Simplifying,

Solving, i1 = 184.2 mAi2 = 157.9 mA

4.6 (p80)

define clockwise i3 in third mesh.

mesh 1:

mesh 2:

mesh 3:

Simplifying,

Solving, i1 = 2.220 A and i2 = 469.9 mA

4.7 (p82)

define clockwise current i2 in top right mesh and clockwise current i3 in bottom right mesh. form supermesh with meshes 1 and 3.

Supermesh: [1]

mesh 2: [2]

remaining equation: [3]



Simplifying, [1]

[2] i1 + i3 = 3 [3]

Solving, i1 = 1.933 A

4.8 (p83)

define two clockwise currents: ia in top right mesh, ib in bottom right mesh.

mesh 1: [1]

mesh a: [2]

mesh b: [3]

no additional equation required as the dependent source depends on a mesh circuit.

Simplifying, [1]

[2]

[3]

Solving, ib = 1.98 A

Thus, 79.20 V



5.1 (p105)

Shorting the 3.5-V source,

Open circuiting the 2-A source,

660 mA

5.2 (p107)

(1) Shorting the 3-V source, we apply nodal analysis to find:

Node 1: [1]

Node 2: [2]

The presence of the dependent source has introduced a new variable, requiring an additional equation:

[3]

Simplifying and reducing to two equations,

[4]

[5]

Solving, v1=9.180 V and v2 = 1.148 V

(2) Open-circuiting the 2-A source, we apply mesh analysis after defining two clockwise mesh currents i1 and i2.

mesh 1: [1]

mesh 2: [2]

where [3]



Simplifying,

[1]

[4]

Solving,

and

so, and

Thus, v1 = v 1 + v 1 = 11.15 V and v2 = v 2 + v 2 = 1.394 V

5.3 (p114)

5 V in series with 5 k 1 mA in parallel with 5 k (arrow pointing up).

192.3 A

5.4 (p115)

75 A in parallel with 4 M 300 V in series with 4 M (“+” on top). 40 A in parallel with 200 k 8 V in series with 200 k (“+” on right).

This circuit may be further reduced to

By voltage division, 27.23 V

* The circuit could by transformed to have a current source, but no reduction in complexity would result.


300 + 8 – 3 =10.2 M 1 M

+ V -

1 M4 M

6 M

0.2 M

+ V -


5.5 (p118)

5 A in parallel with 2 10 V in series with 2 (“+” on top) we now have 2 in series with 8 10 10 V in series with 1 1 A in parallel with 10 (arrow pointing up) we now have 10 // 10 = 5 1 A in parallel with 5 This is the Norton equivalent

5.6 (p119)

(1) Removing the 2- resistor and leaving the terminals open-circuited, no current flows through the 5- resistor; hence, there is no voltage drop across it.

So,

(2) Shorting the 9-V source with the 2- resistor removed, we see

(3) We may now find I2 by

260.8 mA

5.7 (p121)

(1) First, short the 3-V source and open-circuit the 7-mA source. Looking in from the terminals, we see a resistance 1 + 5 // 2 = 2.429 k .

(2) Realising that no current flows through the 1-k resistor, we quickly apply nodal analysis to find:

Solving, v = 7.857 V = V TH (“+” on top)

(3) 3.235 A (source oriented with arrow pointing upwards).



5.8 (p122)

Transform the dependent source to a dependent voltage source:

A simple KVL equation yields or V1 = 502.5 mV = V TH

Shorting the terminals, we then find

or

100.5

5.9 (p123)

There are no independent sources so VTH = 0. install a 1-V test source, “+” on top, across the open terminals. define two clockwise mesh currents iL and iR.

left mesh: [1]

right mesh: [2]

and: [3]

Simplifying,

Solving,

iR flows out of the 1-V source. The Thévenin equivalent resistance, therefore, is

20


0

+-

20 k

200V1

+

V1

-


5.10 (p127)

First find the Thévenin equivalent connected to Rout

Removing Rout, we are left with a single mesh. Define a clockwise mesh current i1.

Then (2000 + 2000) i1 + 20 + 30 = 0

and i1 = 12.5 mA.

Shorting the sources,

(a) With

229.7 mW

(b) 306.3 mW

(c)

so

Solving, Rout = 16.88 or 59.23 k


Rout

1 k


5.11 (p129)

convert RV3, RV4, RV5 to a T-network RT1, RT2, RT3.

RH1 + RT1 = RS1 = 13.33 RS2 = RH2 + RT2 = 13.33 RS3 = RH3 + RT3 = 13.33

convert T-network RS1, RS2, RS3 to a -connected network RD1, RD2, RD3

= 40

RD3 = RD2 = RD1 = 40

11.43



6.1 (p151)

Rule #1 offers no assistance here. Involving Rule #2, vout = vin

6.2 (p152)

By Rule #2, [1]

A KVL equation starting at v1 yields:[2] (making use of Rule #1).

To obtain an expression for vout in terms of the input, we need to eliminate i1.

[3]

Substituting into [2],

or vout = v 2 v 1

6.3 (p157)

[1]

[2]

Simplifying [2] yields


vin

vout

-

vd

+

Avd+-


Using [1], we then obtain

6.4 (p163)

The circuit from Fig. 6.5 (p. 150) cannot be simulated without performing a more sophisticated simulation than we have discussed so far. Replacing the sinusoidal source with a 1-V dc source, however, verifies a gain of 11:

The circuit from Fig. 6.7 (p. 151) may be simulated in a similar fashion:



As a final example, we simulate the circuit in Fig. 6.8 (p. 151), which is a basic summing amplifier. Choosing all resistors to have equal value and neglecting the output resistor, we expect the output to be the negative of the sum of the three input voltages, or –(1 + 2 + 3) = -6. The simulation agrees to within 3 significant figures. The slightly nonideal behaviour of the op amp in this particular circuit is also evident in comparing the voltages at the input, which are not precisely equal, leading to the observed discrepancy with the ideal op amp prediction.


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