hw01_sol

Fundamentals of ECE EE292: Fall 12

Homework #1Due Th. 9/06

1. (Hambley P1.13)The current through a given circuit element is given by i(t) = 10 sin(200pit)A in which t is inseconds and the angle is in radians.

(a) Sketch i(t) to scale versus time for t ranging from 0 to 15 ms.

(b) Determine the net charge that passes through the element between t = 0 and t = 10 ms.

(c) Repeat for the interval from t = 0 to t = 15 ms.

Solution

(a)

0 5 10 1510

5

0

5

10

time [msec]

i(t) [A

]

(b)

Q =

t2t1

i(t)dt

=

0.010

10 sin(200pit)dt = 10200pi

[cos(200pit)]0.010

= 10200pi

[cos(2pi) cos(0)] = 10200pi

[1 1] = 0 C

(c)

Q =

0.0150

10 sin(200pit)dt = 10200pi

[cos(200pit)]0.150

= 120pi

[cos(3pi) cos(0)] = 120pi

[1 1] = 110pi

= 0.0318 C

2. (Hambley P1.18)We have a circuit element with terminals a and b. Furthermore, the element has vab = 5Vand iab = 2A.

(a) Over a period of 10 seconds, how much charge moves through the element?

(b) If electrons carry the charge, which terminal do they enter?

(c) How much energy is transfered?

(d) Is it delivered to the element or taken from it?

Solution

(a)Q = I time = 2 10 = 20 C

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(b) Since iab is positive, positive charge flows from terminal a to b. However, electrons arethe charge carrier and they will flow the opposite direction and instead enter terminal b.

(c)w = Q V = 20 5 = 100 J

(d) Since iab is positive, current flows from terminal a to b. Given the voltage across theterminals, vab, is positive, terminal a is the positive polarity which means this is inpassive reference configuration. Since energy is positive, energy is absorbed by theelement (energy is delivered to the element).

3. (Hambley P1.25)The element shown in Figure P1.25 has v(t) = 10 V, and i(t) = 3etA.

(a) Compute the power for the circuit element.

(b) Find the energy transferred between t = 0 and t =.(c) Is this energy absorbed or supplied by the element?

Solution

(a) Notice the circuit is not in passive reference configuration.

p(t) = v(t)i(t) = 10 3et = 30et W

(b)

w =

t2t1

p(t)dt

=

030etdt = 30 [et]

0= 30[0 1] = 30 J

(c) Since the energy is negative, the energy is supplied by the element.

4. (Hambley P1.26)The current and voltage of an electrical device are iab(t) = 5A. and vab = 10 sin(200pit)V inwhich the angle is in radians.

(a) Find the power delivered to the device and sketch it to scale vs. time for t [0, 15] ms.(b) Determine the energy delivered to the device for the interval 0 t 2.5 ms.(c) Repeat for the interval 0 t 10 ms.Solution

i(t) v(t)

+

Figure P1.25

3 A

+

15 V5W

iR

Figure P1.66

6W

3W

+

Vx12W

2W

i1

2 A

i2v1

+

Figure P1.68

2


(a) p(t) = vabiab = 50 sin(200pit) W

0 5 10 1550

0

50

time [msec]

p(t) [W

]

(b)

w =

0.00250

50 sin(200pit)dt = 50200pi

[cos(200pit)]0.00250

= 14pi

[cos(pi/2) cos(0)] = 120pi

[0 1] = 14pi

= 0.0796 J

(c)

w =

0.010

50 sin(200pit)dt = 50200pi

[cos(200pit)]0.010

= 14pi

[cos(2pi) cos(0)] = 14pi

[1 1] = 0 J

5. (Hambley P1.66)Consider the circuit shown in Figure P1.66

(a) Find the current iR flowing through the resistor.

(b) Find the power for each element in the circuit.

(c) Which elements are absorbing energy?

Solution

(a) Since this is a parallel connection, the voltage across the current source and resitor are15 V. The current flowing through the resistor is

iR =vRR

=15

5= 3 A.

(b) Applying KCL at the bottom node, the current flowing into the voltage source is thesum of current from the current source and from resistor which is 6 A. The power canbe computed for each element as

pvs = vi = 15 6 = 90 Wpcs = vi = 15 3 = 45 WpR = vi = 15 3 = 45 W

(c) Energy is absorbed by elements with positive power. Therefore energy is absorbed bythe current source and resistor and supplied by the voltage source.


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3W

6W 8W

Rx

+

10 V

2 A

isi6

i8

ix+

vy

+vy

+vx

+

vx

Figure P1.72

+

16 V

2W

12W

+

avx

vx+ iy

a = 3 V/V

i12

ix

Figure P1.74

(a) Which elements are in series?

(b) Which elements are in parallel?

(c) Apply Ohms and Kirchoffs laws to solve for Vx.

Solution

(a) The 3W resistor, voltage source Vx, and 2W are in series.

(b) The 6W and 12W resistors are in parallel.

(c) Using Ohms Law the voltage across the parallel resistors is caluclated as

v1 = iR = (2)(6) = 12 V.

This voltage can be use to find the current through the 12W resistor

i1 =v1R

=12

12= 1 A.

The current through the voltage source is found by applying KCL at the bottom nodeof the parallel resistors

i2 = 2A+ i1 = 3 A.

Finally using KVL around the right hand loop Vx can be calculated as

Vx 3i2 v1 2i2Vx = 3(3) + 12 + 2(3) = 9 + 12 + 6 = 27 V


I1 R1 I2 R2

b

a

vab

Figure T1.3

+

Vs

R

avx

vx+ isc

a = 0.3 S

ix

Figure T1.5

4


(a) Which elements are in series?

(b) Which elements are in parallel?

(c) Apply Ohms and Kirchoffs laws to solve for Rx.

Solution

(a) There are no elements in series.

(b) Notice the wire connecting the top and bottom nodes of the diamond. This connectioncauses the 8W resistor and Rx to be in parallel and the 6W and 3W resistors to be inanother parallel connection.

(c) Since there are parallel elements, the voltages across them are the same.

vy = iR = (2)(3) = 6 V

i6 = vy/R6 = 6/6 = 1 A

Using KVL around the top loop

vx = 10 vy = 10 6 = 4 Vi8 = vx/R8 = 4/8 = 0.5 A

Using KCL at the top node

is = i6 i8 = 1 0.5 = 0.5 AUsing KCL at the bottom node

ix = 2 + is = 2.5 A

Rx = vx/ix = 4/2.5 = 8/5 = 1.6W

8. (Hambley P1.74)What type of controlled source appears in the circuit of Figure P1.74? Determine the valuesof vx and iy.

SolutionThe circuit has a voltage-controlled voltage source. Applying KVL around the entire circuitgives

16 vx avx = 0

vx =16

1 + a=

16

1 + 3= 4 V.

The current down through the 12W resister is found using Ohms Law as

i12 =v

R=

3vx12

=3(4)

12= 1 A

Similarly application of Ohms Law across the vx resistor gives

ix =vx2

=4

2= 2 A.

Finally iy is found by applying KCL at the top node

iy = ix i12 = 2 1 = 1 A.

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9. (Hambley T1.3)The circuit of Figure T1.3 has I1 = 3A, I2 = 1A, R1 = 12, and R2 = 6.

(a) Determine the value of vab.

(b) Determine the power for each current source and state whether it is absorbing energyor delivering it?

(c) Compute the power absorbed by R1 and by R2.

Solution

(a) Applying KCL at the node a and defining the currents throught the resistances leaving,

I1 +vabR1

+vabR2

= I2

Solving for vab,

vab = (I2 I1)(

R1R2R1 +R2

)= (1 3)

((12)(6)

12 + 6

)=

(2)(72)18

= 8 V

(b) The power for each source is

pI1 = I1 vab = (3)(8) = 24 W Energy suppliedpI2 = I2 vab = (1)(8) = 8 W Energy aborbed

(c) The power absorbed by each resistor is

pR1 =v2abR1

=64

12=

16

3W

pR2 =v2abR2

=64

6=

32

3W

Notice that the power supplied by I1 is equal to the amount aborbed by the resistorsand I2.

10. (Hambley T1.5)We are given Vs = 15V, R = 10, and a = 0.3S = 0.3

1 for the circuit of Figure T1.5. Findthe value of the current isc flowing through the short circuit.

SolutionSince both ends of the current source are connected by a wire, there is no voltage drop acrossit and KVL around the left loop results in

vx = Vs = 15 V.

Using Ohms Law

ix =vxR

=15

10= 1.5 A.

Finally, using KCL at the top node results in

isc = ix avx = 1.5 0.3(15) = 1.5 4.5 = 3 A.

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hw01_sol

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