Circuit- Current/Power

Discussion in 'Homework Help' started by J.live, Sep 12, 2010.

  1. J.live

    Thread Starter Member

    Sep 10, 2010
    35
    0
    Find the power absorbed by 60 V and supplied by 10 ohms

    Attempt:

    60+5i+3i+ 10i= 0

    i= 3.3 repeating

    power absorbed is just a bizarre number I dont think it is correct



    power supplied

    (33.33)(3.3)= 111.1 W
     
    Last edited: Sep 12, 2010
  2. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    Be careful of polarity.
    Sometimes it helps if you label the pluses and minuses on the diagram.

    You also seem to have messed up your power equations...
     
  3. J.live

    Thread Starter Member

    Sep 10, 2010
    35
    0
    I have no idea about the polarity.

    current is positive since it is coming about of a positive terminal that's all I could figure out.
     
  4. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    You need to stay with a sign convention.
    The standard one is that positive current enters a positive terminal for positive (absorbed) power.

    I drew this in:
    polarity.png

    Note the direction of the current versus the polarity of the voltage.

    When you write a loop equation you need to keep polarity in mind...

    Go clockwise (direction of current) starting with the resistor. If you enter a plus terminal, you put a plus, if you enter a minus terminal, you put a minus.

    As you go around this is the loop equation you get:

    5i - 3i + 10i - 60 = 0
     
  5. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    This is also the reason the voltage source supplies power while the resistance absorbs power (I take it you did a typo on the first post). In the source, voltage and current have opposite polarity directions. This means this device supplies power. On the contrary, in the resistance, current and voltage have the same polarity.
     
  6. J.live

    Thread Starter Member

    Sep 10, 2010
    35
    0
    Oh, makes sense now. Thank you for drawing this out.

    i= 5

    Using Power absorbed equation for 60 V

    (5)^2 (300) = 7500 W ?


    Using Power supplied equation for 10 V

    (5)(10)= 50 W?
     
  7. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    @Georacer:
    Sort of, that dependent source is a negative resistor.

    You could get the series resistance as 5 + 10 - 3 = 12, which is exactly what the loop equation gives you.
    Overall you still just have a resistor.

    A negative resistance by itself isn't unstable, for example, try a 10V source with a -1 ohm resistor.
    You'll still get 10A but it's the opposite polarity. Turns out the resistor is providing power to the 10V 'source'...

    Negative resistances show up all the time actually, you just rarely notice it.

    The most common one I always think of is the input to a DC-DC converter. When you lower the input voltage the input current increases. [Edit: the incremental input resistance is negative... it's still positive overall]
     
    Last edited: Sep 12, 2010
  8. Ghar

    Active Member

    Mar 8, 2010
    655
    72

    I have no idea where the 300 came from.

    The power equation is always P = IV, absorbed vs supplied doesn't change anything except the polarity.

    With a resistor you get the additional relationship that V = IR or I = V/R
    This gives you these extra two equations:

    P = I^2 * R
    P = V^2 / R

    If the power is plus, it is absorbed.
    Notice that the two special cases for a resistor cannot be negative since V and I are squared, meaning a resistor always absorbs power (ignoring the above discussion of negative resistors...)

    Remember what I said about the convention; positive current entering a positive terminal is power absorbed. If you make one of those negative the power becomes negative, meaning it is being supplied.

    In this circuit you have 60V and the current is leaving the positive terminal, therefore, the power is negative.
    The 60V source power is P = -60*5 = -300W, meaning it is supplying 300W.
     
  9. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    @Ghar

    On my previous post, at first I wrote about an instability issue, mistaking the dependent voltage source for a current source. I understood what a nonsense I wrote right after reading it, so I changed it. Now, I don't remeber if I actually posted it (there isn't an edit date and time), so if you are referring to this, I already understood my mistake.

    If you are referring to my current post, isn't it true that the dependent source supplies power? Where do you disagree with me?
     
  10. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    It's about your old post about instability.
    I need to quote more often, I just rarely do it when it's a slow thread.

    I agree, the dependent source is supplying power.
    Since it's a negative resistance it's always going to supply power.
     
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