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Looking for answers to questions in 2nd edition of the art of electronics

Discussion in 'General Electronics Discussion' started by Robert Hill, Apr 22, 2015.

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  1. Robert Hill

    Robert Hill

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    Mar 5, 2015
    Hi there, does anyone know a website or such where someone has posted answers to the self test questions in the art of electronics book? I'm wanting to check the answers I'm coming up with. Thanks in advance.
     
  2. Arouse1973

    Arouse1973 Adam

    5,165
    1,087
    Dec 18, 2013
    Hello Robert.
    I am not sure if they are available, I couldn't find them. Some people are saying the student edition but this doesn't have the answers. It does however have worked examples of its own. Why not post some of the questions on here with your answers stating the page number and the example number.
    Thanks
    Adam
     
    davenn likes this.
  3. Robert Hill

    Robert Hill

    112
    12
    Mar 5, 2015
    Thanks for getting back to me, Yes, I'll post up some of the questions with my answers. Right now though I need to go to bed!
     
  4. Robert Hill

    Robert Hill

    112
    12
    Mar 5, 2015
    Hi again,

    So i'm looking at exercise 1.6 on the right hand column of page 7:
    New york city requires about 10 to the power of 10 watts of electrical power at 110v. A one foot diameter pure copper cable has a resistance of 5x10 to the minus 8 ohms per foot.
    -Calculate the power lost per foot
    -the length of cable over which all 10 to the 10 watts of power would be lost
    - the solution to the preposterous seeming answer to the above question.

    I got:
    - P= V squared / R. This means P=12100 /5x10 to the minus 8 = 2.42 times 10 to the 11 watts per foot.

    So, all 10 to the 10 watts gets used up in less than a foot?

    My solution is to use AC rather than DC power?


    The next question I'm looking at is on page 11, exercise 1.9
    It refers to a picture 1.10.
    It asks if the Vin is 30v and r1 and r2 are 10k, find the output voltage when:
    - the output voltage with no load attached
    - the output with a 10k load attached
    - the Thevenin equivalent
    - The power diappated in each of the resistors.

    I got:
    The output voltage with no load attached is Vin x R2 divided by R1+R2 so 30x(10,000/20,000)= 15v
    With a 10k load attached it is 30x(20/30) = 20v
    the Thenenin equivalent, To calculate the Rth valve I did R1xR2/R1+R2 = 200,000000/30,000=6666.666667 so Rth is 6666.666667 ohms
    To calculate The short circuit current I did Vin/R1 and got 0.003 amps
    To calculate Vth I did Rth x I = Vth and got 6666.666667 x 0.003 = 20 volts
    So I calculated that the power dissipated in each resistor as P=I x V which is P= 0.003 x 20 = 0.06 Watts

    So, How much did I get wrong!
     
  5. Arouse1973

    Arouse1973 Adam

    5,165
    1,087
    Dec 18, 2013
    For the first one I get 110V/10GW = 90x10^6 Amps which is 405 MW per foot so 10 GW will be lost in 24.69 feet, I think. You have the first part of the second question correct. The second line is wrong, how can you have a higher value than without a load? Is the Thevenins value with the load attached?
    Thanks
    Adam
     
  6. Robert Hill

    Robert Hill

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    Mar 5, 2015
    Hm, I didn't do very well then...
    So for the first question you calculated the currrent running through the cable by rearranging the P=IxV formula then calculated the power dissipated by using the current and the resistance of the cable. Then divided the power into the 10GW to work out how many feet.

    So I think I went wrong on that one because I tried to calculate power based on the Voltage and resistance when In fact I already had been told that the power was 10 to the power of 10. Instead I should have calculated the current those values would generate in order to work out the power the cable would dissipate. Does that mean the power value I generated (2.42x 10 to the 11) is the maximum power that could be generated by that voltage and resistance? or was it just a nonsense number I came up with?
    A rule I could create to solve these questions would be - calculate the values I don't know from the ones I do, in this case the current.

    On the second question, Yes I thought it must be wrong because the second figure with the load was higher. Yes, the Thevenin value should include the load.

    So having gone back and read over the section again I've decided to use 15v as the Vin for the calculation when a 10k Load is used. This is because the output voltage from the first 'stage' of the voltage divider (the R1 and R2 stage) is what is available for the second stage. I believe that when the load is added a new voltage divider is created that is between R1 and the combined resistance value of R2 and Rload which are in parallel.

    The Vout value I get for when a load is added is now:
    15x (20k/30k) = 10 volts
     
  7. Robert Hill

    Robert Hill

    112
    12
    Mar 5, 2015
    So thinking about this more, and going a bit mad in the process. The basic idea of the Thevenin theorem is that you reduce most of a circuit to a single voltage source and and single resistor in series so that you can then vary the final part of the circuit (load resistance) without having to recalculate most of the circuit. This allows you to see how much of the initial voltage will end up in the final stage of the circuit.

    So the Thevenin equivalent for the circuit with load on is Vth (15v) in series with Rth (5kohm) and Rload (10kOhm) so it is 15v in series with 15kOhm resistance. So the current would be 15/15000= 0.001amp so the power dissipated by the resistors would be 15 * 0.001= 0 015 watts.

    You can then reduce this also to an equivalent circuit which has 15v x (10/15) = 10v. The Rth for the second equivalent circuit is 3.3Kohms and the Vth is 10. This could then be combined with a further load circuit.

    So maybe I have finally understood it? If so Horray!!!!
     
    Last edited: May 4, 2015
  8. davenn

    davenn Moderator

    13,802
    1,941
    Sep 5, 2009
    holy crap ... that's not a transmission line ... its a dummy load :(
     
    Arouse1973 likes this.
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