P1.36 (a) Elements C and D are in series. (b) Because elements C and D are in series, the currents are equal in magnitude. However, because the reference directions are opposite, the algebraic signs of the current values are opposite. Thus, we have dc ii −= . (c) At the node joining elements A, B, and C, we can write the KCL equation A 4 1 3 =+=+= cab iii . Also we found earlier that

A. 1− =−= cd ii

P1.37* At the node joining elements A and B, we have .0=+ ba ii Thus, A. 2 −=ai For the node at the top end of element C, we have 3=+ cb ii . Thus,

A 1=ci . Finally, at the top right-hand corner node, we have .d3 e ii =+ Thus, A 4 =di . Elements A and B are in series.

P1.38* find we KCL, Applying A. 4 and A, 5 A, 3 A, 2 given are We =−=== hdba iiii

A 1=−= abc iii A 5=+= hce iii A 3−=+= daf iii A 7−=−= hfg iii

P1.39 A. 1 and A, 5 A, 3 A, 1 given are We ===−= hgca iiii Applying KCL, we find A 2=+= acb iii A 4=+= hce iii A 7=−= afd iii A 6=+= hgf iii

P1.40 If one travels around a closed path adding the voltages for which one enters the positive reference and subtracting the voltages for which one enters the negative reference, the total is zero.

P1.41 (a) Elements A and B are in parallel.

(b) Because elements A and B are in parallel, the voltages are equal in magnitude. However because the reference polarities are opposite, the algebraic signs of the voltage values are opposite. Thus, we have

.ba vv −= (c) Writing a KVL equation while going clockwise around the loop composed of elements A, C and D, we obtain .0=−− cda vvv Solving for

cv and substituting values, we find V. 7=cv Also we have V. 2−=−= ab vv

10

P1.42* Summing voltages for the lower left-hand loop, we have ,0105 =++− av which yields V. 5−=av Then for the top-most loop, we have

,015 =−− ac vv which yields V. 10=cv Finally, writing KCL around the outside loop, we have ,05 =++− bc vv which yields V. 5−=bv

P1.43 We are given V. 6 and V, 10 V, 7 V, 5 =−=== hfba vvvv Applying KVL, we find V 12=+= bad vvv V 1−=−−−= hfac vvvv V 8=+−−= dcae vvvv V 2=−= heg vvv

V 7=+= ecb vvv P1.44* Applying KCL and KVL, we have

A 1=−= dac iii A 2−=−= ab ii V 6−=−= adb vvv V 4== dc vv The power for each element is W 20−=−= aaA ivP W 12== bbB ivP W 4== ccC ivP W 4== ddD ivP

0 Thus, =+++ DCBA PPPP P1.45 (a) In Figure P1.28, elements C, D, and E are in parallel.

(b) In Figure P1.33, no element is in parallel with another element. (c) In Figure P1.34, elements C and D are in parallel.

P1.46 The points and the voltages specified in the problem statement are:

Applying KVL to the loop abca, substituting values and solving, we obtain: 0=−− accbab vvv 0155 =−− acv V 10−=acv

11

P1.56 The power delivered to the resistor is )4exp(5.2/)()( 2 tRtvtp −== and the energy delivered is

J 625.045.2)4exp(

45.2)4exp(5.2)(

000

==

−−

=−==∞∞∞

∫∫ tdttdttpw

P1.57 The power delivered to the resistor is

)4cos(25.125.1)2(sin5.2/)()( 22 ttRtvtp ππ −=== and the energy delivered is

[ ] J 5.12)4sin(425.125.1)4cos(25.125.1)(

10

0

10

0

10

0

=

−=−== ∫∫ ttdttdttpw π

ππ

P1.58 Equation 1.10 gives the resistance as

ALR ρ

=

(a) Thus, if the length of the wire is doubled, the resistance doubles to 1 . Ω(b) If the diameter of the wire is doubled, the cross sectional area A is increased by a factor of four. Thus, the resistance is decreased by a factor of four to 0.125 . Ω

P1.59 (a) The voltage across the voltage source is 10 V independent of the

current. Thus, we have v = 10 which plots as a vertical line in the v−i plane. (b) The current source has i = 2 independent of v, which plots as a horizontal line in the v−i plane. (c) Ohm's law gives i = v/5. (d) Applying Ohm's law and KVL, we obtain 105 += iv which is equivalent to 22.0 −= vi . (e) Applying KCL and Ohm's law, we obtain obtain 105 += iv which is equivalent to 22.0 −= vi .

14

(b) Rx and the 2-Ω resistor are in parallel. Also, the 6-Ω resistor and the the 4-Ω resistor are in parallel. Thus, the voltages across the parallel elements are the same as labeled in the figure. (c) V 61 =v

A 5.14/14 ==vi V 410 1 =−= vvx

A 22/2 == xvi A 112 =−= iis

A 5.04 =−= sx iii Ω== 8/ xxx ivR

P1.69 12/12 vi = 6/6 vi = 36/12/612 =+=+ vvii

12=v V 112 =i A 26 =i A P1.70* (a) Applying KVL, we have xx vv 510 += , which yields V 667.16/10 ==xv

(b) A 5556.03/ == xx vi (c) W. 556.510 −=−=− xsourcevoltage iP (This represents power delivered by

the voltage source.) (absorbed) W 926.0)(3 2 == xR iP

(absorbed) W 63.45 ==− xxsourcecontrolled ivP

P1.71

Applying KVL around the periphery of the circuit, we have

19

P1.75*

V 4)A 1() 4( =×Ω=xv A 312/ =+= xs vi

pplying KVL around the outside of the circuit, we have:

A

V 15243 =++= ss iv

P1.76

A 2 V/15 30 −=Ω−=xi Applying KCL for the node at the top end of the controlled current source:

A 12/2/ =−=−= xxxs iiii he source labled is is an independent curren T t source. The source labeled

ix/2 is a current-controlled current source.

P1.77 Applying Ohm's law and KVL, we have

.51020 xx ii =+ Solving, we obtain A. 4−=xi

The source labeled 20 V is an independent voltage source. The source labeled 5ix is a current-controlled voltage source.

21

and have an equivalent resistance of 6 Ω. Finally, the two parallel combinations are in series, and we have

1064 =+=abR Ω

P2.11 nnReq

1000

1000

1

...1000

11000

11000

11

==+++

=

P2.12*

P2.13 In the lowest power mode, the power is 33.8312021

2

=+

=RR

Plowest W.

For the highest power mode, the two elements should be in parallel with an applied voltage of 240 V. The resulting power is

150050010002402402

2

1

2

=+=+=RR

Phighest W.

Some other modes and resulting powers are: 1R operated separately from 240 V yielding 1000 W

2R operated separately from 240 V yielding 500 W 1R in series with 2R operated from 240 V yielding 333.3 W 1R operated separately from 120 V yielding 250 W P2.14 For operation at the lowest power, we have

21

2120300RR +

=P =

At the high power setting, we have

2

2

1

2 1201201200RR

+=P =

39

Solving these equations we find . 2421 Ω== RR

The intermediate power setting is obtained by operating one of the elements from 120 V resulting in a power of 600 W.

P2.15 By symmetry, we find the currents in the resistors as shown below:

Then, the voltage between terminals a and b is 65316131 =++=eq=ab Rv

P2.16* The 20-Ω and 30-Ω resistances are in parallel and have an equivalent

resistance of Req1 = 12 Ω. Also the 40-Ω and 60-Ω resistances are in parallel with an equivalent resistance of Req2 = 24 Ω. Next we see that Req1 and the 4-Ω resistor are in series and have an equivalent resistance of Req3 = 4 + Req1 = 16 Ω. Finally Req3 and Req2 are in parallel and the overall equivalent resistance is

Ω 6.9/1/1

121

=+

=eqeq RRabR

P2.17 The 20-Ω and 30-Ω resistances are in parallel and have an equivalent

resistance of Req1 = 12 Ω which in turn is in series with the 8-Ω resistance resulting in an equivalent resistance of Req2 = Req1 + 8 = 20 Ω. Next Req2 is in parallel with the 60-Ω resistance resulting in an equivalent resistance Req3 = 15 Ω which in turn is in series with the 7-Ω resistance resulting in an overall equivalent resistance of Rab = 22 Ω.

40

P2.40 (a) Ω==+ 120A 1.0V 12

21 RR 51221

2 =×+ RRR

Solving, we find Ω= 502R and Ω= 701R . (b)

The equivalent resistance for the parallel combination of R2 and the load is

Ω 402001501

1=

+=eqR

Then using the voltage division principle, we have

V 364.4V 121

=×+

=eq

eqo RR

Rv

P2.41* mV 50mA 1.0 =×= wRv

Ω=−

= m 25mA 1.0A 2

mV 50gR

P2.42 The circuit diagram is:

With 0=Li and V 5=Lv , we must have V 59

21

2 =×+ RRR . Rearranging,

this gives

8.02

1 =RR (1)

With mA 25=Li and V 5.4=Lv , we have ( ) 5.4mA 255.4 21 =9 +− RR .

46

,0218 =++− xx vv which yields 6=xv V. Then we have 12212 == xvv V. Using Ohm’s law we obtain 112 12/12 ==vi A and 32 =/= xx vi A. Then KCL applied to the node at the top of the 12-Ω resistor gives yx iii += 12 which yields 2=yi A.

P1.72 Consider the series combination shown below on the left. Because the current for series elements must be the same and the current for the current source is 2 A by definition, the current flowing from a to b is 2 A. Notice that the current is not affected by the 10-V source in series. Thus, the series combination is equivalent to a simple current source as far as anything connected to terminals a and b is concerned.

P1.73 Consider the parallel combination shown below. Because the voltage for parallel elements must be the same, the voltage vab must be 10 V. Notice that vab is not affected by the current source. Thus, the parallel combination is equivalent to a simple voltage source as far as anything connected to terminals a and b is concerned.

P1.74 (a) 10 21 vv +=

(b) iv 151 = iv 52 = (c) ii 515 +=10 A 5.0=i (d)

(absorbed) W 25.15(absorbed) W 75.315

source.) the by delivered (Power W. 510

25

215

==

==

−=−=−

iPiP

iP sourcevoltage

20

006667.006667.03333.0 321 =−− vvv 406667.01333.006667.0 321 =−+− vvv 01733.006667.006667.0 321 =+−− vvv

Solving, we find 0.151 =v 0.502 =v 0.253 =v

P2.50 Writing KCL equations at nodes 1, 2, and 3, we have

022010

211 =+−

+vvv

0542023212 =+

−+

− vvvvv

245

233 =−

+vvv

In standard form, we have: 205.015.0 21 −=− vv

025.05.005.0 321 =−+− vvv 245.025.0 32 =+− vv

Solving, we find 9.121 −=v V 29.12 =v V 16.53 =v V

P2.51 Writing KCL equations at nodes 1 and 2, we have

392821

2111 =−

++vvvv

369212 −=+

− vvv

In standard form, we have: 31111.01944.0 21 =− vv

32778.01111.0 21 −=+− vv Solving, we find 0.121 =v V and V 00.62 −=v . If the source is reversed, the algebraic signs of the node voltages are reversed.

P2.52 To minimize the number of unknowns, we select the reference node at

one end of the voltage source. Then, we define the node voltages and write a KCL equation at each node.

49

225

20 211 =−

+− vvv 3

1020

2212 −=−

+− vvv

Solving, we find V 24.181 =v and V 53.132 =v .

Then, we have A 647.010

20 21 =

−=

vi .

P2.53 We must not use all of the nodes (including those that are inside

supernodes) in writing KCL equations. Otherwise, dependent equations result.

P2.54 The circuit with a 1-A source connected is:

1

20103121 =

−+

− vvvv

0101010

32122 =−

+−

+vvvvv

50

Finally, the power delivered by the source is 1010 1 == iP W.

P2.66

16)(284 =−+ BAA iii 0147)(28 =++− BBAB iiii

Solving we find 5714.0 and A 1 == BA ii A. Then we have 11 == Aii A and 4286.02 =−= BA iii A.

P2.67 Mesh A: 10 0)(105 =−++ BAAA iiii

By inspection: 2=Bi

Solving, we find . A 8.0=Ai Then we have 8.01 == Aii A and

2.12 =−= AB iii A. P2.68 First we select mesh-current variables as shown.

Then, we can write

55