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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.