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Circuits help

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  • 31-07-2016 1:56am
    #1
    thepikminman
    Registered Users Posts: 178 ✭✭


    I have merged the threads - having five different threads about solving circuit related problems is not necessary.

    I have a life-changing test in two days and have tried everything to try and answer this question but am getting nowhere. If someone can tell me the steps to answer this you will literally be saving my life (if I fail, I fail college and can't repeat again). The question:

    For the circuit shown in Figure 1 (below in attachment):
    (a) Draw the circuit in the form required for phasor analysis [3 marks]
    (b) Determine Iin in terms of Vo [2 marks]
    (c) Determine Vin in terms of Vo [3 marks]
    (d) Find Iin [2 marks]
    (e) Specify the rms output voltage [1 mark]
    (f) By what angle does the output voltage lag the input voltage?
    [1 mark]
    (g) Calculate the rms input current [1 mark]
    (h) By what angle does the input current lag the input voltage?
    [1 mark]
    (i) Calculate the amount of power entering the circuit [2 marks]
    (j) How much power is absorbed by each of the two resistors?[2 marks]
    (k) Does the power absorbed by the two resistors match the circuit input
    power? [2 marks]

    Any amount you can explain would help, even if it's just one part.

    Figure 1:
    Tagged:


Comments

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    #2
    mikhail
    Registered Users Posts: 3,863 ✭✭✭mikhail


    Sadly, it seems your exam was a few days ago. I hope it went well. For the record, and because I hate this feeling: https://xkcd.com/979/

    (a) Replace all the resistors, capacitors and inductors with little rectangles (called impedences). The point of phasor analysis is that they're all basically the same thing. Resistors have impedence R Ohms. Capacitors have impedence 1/(j omega C) Ohms. Inductors have impedence j omega L Ohms. Mark those beside the impedences. omega is 4 here.
    Replace the voltage source with 5 sqrt(2) Volts.
    (b) Iin = (5 sqrt(2) - Vo)/(2 + j(4)(3)). You can simplify with a little algebra if you like.
    (c) The question's unclearly worded. Vin = 5 sqrt(2) - it doesn't depend on Vo. I guess you could do it the other way around. Vo = (4//j(4)(0.125))Vin/(2+j12) = Vin/(1/4 + j2)(2+j12) which should again yield to a little algebra. The symbol // just means "in parallel with" - you add impedences like resistors: 1/Zequivalent = 1/Z1 + 1/Z2.
    (d) Assuming nothing connected to the output, Iin = 5 sqrt(2) / Ztotal = 5 sqrt(2) / (2 + j 12 + (1/(0.25 + j2))) which should simplify.
    (e) VoRMS = Vo/sqrt(2) = 5/(1/4 + j2)(2+j12)
    (f) angle(Vo)-angle(Vi)... angle(x) = arctan(Im(x)/Re(x))
    (g) Iin/sqrt(2)
    (h) angle(Iin)-angle(Vin)
    (i) P = Vin Iin
    (j) 2 Iin^2 and 4 (Iin(j0.5)/(4 + j0.5))^2 (P = I^2 R, current division)
    (k) Just add them up and compare.


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    #3
    thepikminman
    Registered Users Posts: 178 ✭✭thepikminman


    Thank you. For part (a) what happens to Vout, and what do I use for the value of j?


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    #4
    thepikminman
    Registered Users Posts: 178 ✭✭thepikminman


    Hi,

    I tried doing this question with little success. If you can help me in any way, even if you just show me the process needed to get marks, I will literally be eternally grateful. Thank you. Even part of an answer would help.

    Here's the question in the attachment:


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    #5
    thepikminman
    Registered Users Posts: 178 ✭✭thepikminman


    Hi,

    I tried doing this question with little success. If you can help me in any way, even if you just show me the process needed to get marks, I will literally be eternally grateful. Thank you. I understand that it'd be hard to show me the full graph, but that's fine even part of an answer would help.

    Here's the question in the attachment:


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    #6
    thepikminman
    Registered Users Posts: 178 ✭✭thepikminman


    Hi,

    I tried doing this question with little success. If you can help me in any way, even if you just show me the process needed to get marks, I will literally be eternally grateful. Thank you. Even part of an answer would help. Apologies for the poor quality.

    Here's the question in the attachment:


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    #7
    mikhail
    Registered Users Posts: 3,863 ✭✭✭mikhail


    v_o(t) just becomes the complex number V_o.

    j is the square root of -1. Depending on your background, you may call it i.


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