elg2138hw1soln11sep28

8/3/2019 elg2138HW1soln11sep28

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Elg2138 HW#1 Solution

P1) P2.5-2

(a) By KCL, current through the resistor is 3A.From Ohms law s 5 3 15 Vv R i . (The resistor voltage does not depend on thevoltage source voltage.)

Next2 215

45 W5

vP

R .

By KVL using sum of voltage rise = sum of drop, voltage across the current source is

10+15=25 V (with +ve at the end where current is coming out)

By passive sign convention,i) power developed at the voltage source is (3)(10)=30W. Therefore power absorbed

by the voltage source.

ii) power developed at the current source is (-3)(25)=-75W. Therefore power deliveredby the current source

In summary, we have power delivered by the current source = sum of power absorbedby the resistor and by the voltage source

(b) Since and do not depend ons

v P v the values ofv and P are 15 V and 45 W

s sboth when v =10V and when v =5V.

P2) P2.7-5

4 22

R

(by Ohms Law) a

Since A i a = A(-0.5) = 2, we have2 V

40.5 A

A

P3) P3.2-9

KVL: 56 24 0 80 V

KCL: 8 0 8 A

8010

8

R R

R R

R

R

v v

i i

vR

i

by Ohms Law

AAt node A:


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P4) P3.2-13We can label the circuit as follows:

The subscripts suggest a numbering of the circuit elements. Apply KCL at node b to get

4 40.25 0.75 0 1.0 Ai i

Next, apply KCL at node dto get

3 4 0.25 1.0 0.25 0.75 Ai i

Next, apply KVL to the loop consisting of the voltage source and the 60 resistor to get

2 215 0 15 Vv v

Apply Ohms law to each of the resistors to get

22 15 0.25 A

60 60

vi ,

3 310 10 0.75 7.5 Vv i

and

4 420 20 1 20 Vv i

Next, apply KCL at node c to get

1 2 3 1 3 2 0.75 0.25 1.0 Ai i i i i i

Next, apply KVL to the loop consisting of the 0.75 A current source and three resistors to get

6 4 3 2 6 4 3 20 20 ( 7.5) 15 12.5 Vv v v v v v v v

Finally, apply KVL to the loop consisting of the 0.25 A current source and the 20 resistor toget

5 4 5 40 20 20 Vv v v v


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Using passive sign convention on the assigned directions and polarities, the power of each

component is as follows:i) 15V-voltage source: 15(i1)= 15(-1) = -15W therefore power generated

ii) 0.75A-current source : v6(0.75)=(-12.5)(0.75)= -9.375W. therefore power generated

iii) 0.25A-current source : v5(-0.25)= (20)(-0.25) = -5W therefore power generated

iv) 60 resistor: v2i2 = (15)(0.25) = 3.75W therefore power absorbedv) 20 resistor: v4i4 = (-20)(-1) = 20W; therefore power absorbed

vi) 10 resistor: v3i3 = (-7.5)(-0.75) = 5.625W; therefore power absorbed

One can check that total power absorbed = 29.375W = total power generated

P5) P3.2-19We assign current and voltage as shown. Then, usingpassive sign convention on the power generated at the

voltage source we have 3.6 0.3 A

12i

Similarly, v(-0.5) = -4.8 for the current source

4.89.6 V

0.5v

Using KVL on the left loop in clockwise direction, sum

of voltage rise = 12= sum of voltage drop = iR1 + v

(Ohms Law on the resistor)

1

12 9.68

0.3R

By KCL, current (moving down ward) on R2 =0.3+0.5.

Since voltage across R2 = v (KVL) on right loop.

Then by Ohms Law2

9.612

0.3 0.5R

P6) P3.2-20Apply KCL at node a to determine the current

in the horizontal resistor as shown.

Apply KVL to the loop consisting of thevoltages source and the two resistors to get

-4(2-i) + 4(i) - 24 = 0 i = 4 A


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P7) P3.8-5

By KCL, the current through the 2 ohm resister is in the direction shown in thediagram.

Top mesh: we would like to check that First, we have by Ohms Law.

Proceeding, +1-2 Substituting in the given values of , we have

Similarly, we want to check if lower left mesh gives

10 2 0.5 10 2 2 14 Vs a bv i i by Ohms law

Lower right mesh, check that

Checking, we have . Therefore, KVL is satisfied.

Summarizing the analysis and observations from the 3 meshes, we conclude that the analysis is

correct.

Note: ia and ib are in Amperes in the above solution from the author. The conclusion is the

opposite if mA is used as given in the text.

elg2138hw1soln11sep28

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