ECE 202 Solutions - Chapter 1, Probs. 7,8,11,12,13,15,16,18,19,20,23,25

 
1.7) First, we'll find the charge per cubic meter by multiplying the number of electrons by the charge

of a single electron:

1.6022x10^-19C*10²⁹electrons/m³ = 1.6.022x10¹⁰C/m³

Next, we determine the charge in the cross sectional area of the bar (the number of electrons flowing past us at a given time):

1.6022x10¹⁰C/m³*(0.4x10^-2m)(16x10^-2m) = 10.25x10⁶C/m

We need to find average velocity, in m/s. We have charge/meter (above) and current, which is charge/second. , so

1600A/(10.25x10⁶C/m) = (1600C/s)/(10.25x10⁶C/m) = 156.04x10^-6m/s

  1.8)# Megatons = [((845miles)(5280ft/mile)(2.526lb/ft)(4conductors))/(2000x10⁶lbs/megaton)] = 0.02254megatons

  1.11) Assume we are standing ant Box A looking toward Box B, then p=-vi (Note: there

is a misprint here in the solutions manual, if any copies of that are floating around).

  a) p=-vi=-(120)(25)= -3000W, so power is flowing from B to A

b) p= -(240)(-5) = 1200W (from A to B)

c) p= -(-720)(1.5) = 1080W (from A to B)

d) p= -(-20)(-480) = -9600W (from B to A)

  1.12) Power is defined as the change in energy over the change in time. In this problem, we know the change in time and can compute the power from the voltage and current, and are asked to find the change in energy.

P = vi = (12V)(100mA) = 1.2W

Energy Supplied = (Power)(change in time in seconds) = 1.2W(4hours)(3600sec/hr)= 17,280J

    1.13) Given the plots of voltage and current, find:

(a) The total charge transferred to the battery.

          Charge is related to current and time through i = [dq/dt] , so to find the charge we merely need to integrate the current with respect to time. We can do this by simply finding the 3 equations for the 3 straight lines which compose the charge graph:

0<t<5ks;i = -1.2t+20

5<t<15ks;i = -0.6t+17

15<t<20ks;i = -1.6t+32

then integrating each equation separately over the appropriate time range:

ò⁵₀(-1.2t+20)dt = -0.6t²+20t|₀⁵ = -0.6(25)+20(5) = 85kC

repeat with the 2nd function for 5-15ks and the third function for 15-20ks, and your

total should be 215kC=215,000C

Incidentally, this is also the area under the current curve, so you could divide it into rectangles and tringles and sum the area (which is what the solutions manual does).

(b)The total energy transferred to the battery.

           Energy relates to power the same way charge relates to current, so again we need to integrate over the range when enrgy was being transferred. However, we need to integrate power, so we must determine the voltage equations from the graph and multiply by the appropriate current equation in that time range. The voltage equations are:

0<t<20;v(t) = 0.2t+8

So, using p(t)=v(t)i(t)

0<t<5ks;w = ò⁵₀pdt = ò⁵₀(-1.2t+20)(0.2t+8)dt = ò⁵₀(-0.24t²-5.6t+160)dt = -0.12t³-2.8t²+

160t|₀^5ks = 720kJ

Repeat this procedure for 5-15ks and 15-20ks with the appropriate current equation, sum the results, and your answer should be: 2036.67kJ

  1.15) Given equations for v(t) and i(t), find the time when the power delivered is maximum.

For all functions, maxima (and minima) occur when the derivative is equal to zero. So let's

compute p(t)=v(t)i(t), then take dp/dt and set it equal to zero:

p = vi = 80,000te^-500t(15te^-500t); t>0 = 12x10⁵t²e^-1000tW

dp/dt = 12x10⁵[t²(-1000)e^-1000t+e^-1000t(2t)] = 12x10⁵te^-1000t[t(2-1000t)]

dp/dt = 0; t = 0 or t = 2x10^-3s

Since we know v=i=0 at t=0, that must be the minimum, so the maximum occurs at 2ms.

b) At t=2ms, p=649.61mW (plug .002 seconds in equation for power above)

c) Total energy is the integral of the power,

w = 12x10⁵ò^¥₀t²e^-1000tdt = 12x10⁵{[(e^-1000t)/((-1000)³)][10⁶t²+2000t+2]|^¥₀} = 2.4mJ

1.16) a) p(t) = vi = (20e^-400t-20e^-1600t)(30-40e^-400t+10e^-1600t)

p(625ms) = 20.77mW

b) Find the energy delivered between 0 and 625us:

w = ò^625ms₀p(t) = [562.5+1000e^-800t+375e^-1600t+62.5e^-3200t-1500e^-400t-500e^-2000t]mJ

w(625us) = 3.99mJ

c.) w(total)-562.5 mJ (Change upper limit of integral to infinity)

1.18) a) See figure. Since the current is always a contant, the graph should look loke the voltage graph adjusted by constants.

b) @4s, p=vi=2.5t mW; w= 1.25t² =1.25(16)=20 mW (here's another problem where the solutions in the solution manual make no sense).

@12s, w=0

@36s, w=7.2 mJ

@50s, w=0

  1.19) a) Again, to find a maximum we must set the derivative of the power equal to zero. The expresion for power in terms of time is:

p(t) = v(t)i(t) = t(1-0.025t)(4-0.2t) = 4t-0.3t²+0.005t³   0 £ t £ 40

[dp/dt] = 4-0.6t+0.015t²

[dp/dt] = 0 at t=8.45 and t=31.55s (use quadratic formula). These 2times are when the minimum and maximum power was delivered, but we don't know which is which, so plug back into the equation for p(t):

p(8.45) = 15.4W

p(31.55) = -15.4W
so maximum power delivered is at 8.45 seconds.

We've already firgured out (b), (c) and (d) as well:

(b) p_max =15.4 W delivered

(c) 31.55s

(d) p_max =15.4W extracted

(e) Energy is the integral of the power, so

w = ò^t₀pdx = ò^t₀(4x-0.3x²+0.005x³)dx = 2t²-0.1t³+0.00125t⁴

Plugging the various values of t into this equation:

w(0)=0J         w(10)=112.5J            w(20)=200J           w(30)=112.5J         w(40)=0J

    1.20) The measured current is -50A, or +50A if we change the reference so it flows in the direction of car B. Since current is flowing towards the voltage drop in car b (and so is the power- check this) car B

has the ``dead'' battery.

(b)Our power is a constant, p=vi=(12)(50)=600W. Our energy is the integral of this constant, 600t,

so after 2 minutes w=600(2)(60)=72,000J=72kJ

    1.23) Voltage is given as linear function of time, so once again, we must find the equations of this

line. We have 2 points on this line, (0s,1.5V) and (144,000s,1.0V). The slope is -0.5/144,000=

3.47222x10^-6 and the y(or V) intercept is 1.5, so the equation for the voltage is:

v(t) = -3.47222x10^-6t+1.5

p = vi = 9mA = -3.125x10^-8t+0.0135W

w = òpdt = -1.5625x10^-8t²+0.0135t J

w(144,000) = -324+1944 = 1620J

You can also do this by plotting the power vs. time and taking the area under the curve. The official solutions manual is obsessed with that...

      1.25)

OK, remember p=vi when the current flows in the direction of the voltage DROP. In this problem, they play with the way they choose the voltage and current references, so when current flows in the direction of a voltage GAIN, we must use p=-vi. Regardless of which equation we're using, positive power means absorbing, negative means delivering.

P_a = -v_ai_a = -(225)(-36) = 8100W

p_b = -v_bi_b = -(120)(-60) = 7200W

p_c = v_ci_c = (60)(120) = 7200W       etc....