Bleeder Current Circuit Help.

Discussion in 'Homework Help' started by JustDave, Feb 1, 2011.

  1. JustDave

    Thread Starter New Member

    Feb 1, 2011
    I'm on my second semester in EG and was wondering if anyone can help me with these two circuit problems. I've been doing great so far with little need for help outside the class, but it's just these two circuits that are bothering me.

    The problems are 7-27 (bottom left) and 7-29 (bottom right). It wants me to find certain values for both circuits and both have a bleeder current and that's what gets me. Sorry in advance if the scans are a little dark or hard to read, but my scanner is small compared to my textbook (along with some added editing on my part).


    For 7-27, I'm pretty sure I already have the IR Voltage drops. I did the fallowing to find them.

    V1 = VT-VB = 25V-15V=10V
    V2 = VB-VC=15V-6V=9V
    V3 = VT-(V1+V2)=25V-(10V+9V)=6V

    The rest, however, I'm not sure how to get the current and resistor values. Power will be easy once I have R and I values. Just use Ohm's Law. I attempted to find the current values but I'm pretty sure they are wrong. The proses I used came from another student who thought he figured it out. It just looks very odd to me.

    Below is a copy of my work so far for 7-27 (excuse my lefty writing).

    I'm not sure what formula to use next. Any help would be greatly appreciated. Some starting-steps on where to go next. What formula(s) to use. So on and so forth. After that I can fallow up with my results for the current and resistors for 7-27 and even 7-29. I don't want answers but instead I want to know how it works.

    Thanks in advance.

    Note: I may be away from my computer for a few hours after posting and updating this post. However, I will return.
    Last edited: Feb 1, 2011
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    I'm assuming IB is the current in R3.

    You have IB=6.667mA

    Voltage across R3 is 6V.

    So R3=6/6.667mA=899.96Ω

    The current in R2 will be IB + Load C current or 11.667mA.

    The drop across R2 is 15V-6V=9V

    So you could then find R2. And so on ....

    Does that help?
  3. JustDave

    Thread Starter New Member

    Feb 1, 2011
    Thanks for the reply. However, my question is if the current value I found for IB and the process I used to get it was correct or not. I'm just not sure on how to find the current values for this type of circuit. I'll be back in EG class tomorrow and I'll try to get my teacher to go over it a little more. I would still like to get some other help and opinions from this site though.

    The answers for just 7-27 are in the book. However, he scores you on both your answers and how you got them. I can post the right answers for 7-27 but I would still like to know how to get them

    IR Voltage Drops (which I already know how to find).

    Current Values.

    Resistor values.

    Power Dissipated by.

    So, there are the answers. I just need to know how to get them.
    Last edited: Feb 2, 2011
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    So where did your value of 6.667mA for IB fit with the solution?

    IB is the current in R3. It must be 10% of the total LOAD (A+B+C) current. The total LOAD current is 60mA.

    So IB must be 6mA.

    The resistor R3 therefore drops 6V at 6mA, requiring


    The current in R2 is 6mA + 5mA or 11mA. R2 drops 9V, requiring


    The current in R1 is 11mA + 45mA or 56mA. R1 drops 10V, requiring


    That's it.
  5. JustDave

    Thread Starter New Member

    Feb 1, 2011
    So, I am to understand that your current bleed, IB, will always be the current for the last(?) resistor of the loads? In this case it’s R3. I hope I said that right.

    And basically I could of just done the below to find IB and such.

    ITL (current total load) = IA+IB+IC
    IB = ITL/IB%
    IB = I3
    R3 = V3/I3
    I2 = I3 + IC
    R2 = V2/I2
    I1 = I2 + IB
    R1 = V1/I1

    ITL = 10mA + 45mA + 5mA = 60mA
    IB = 0.06A / .10 or 6mA / 10 = 6mA
    R3 = 6V / 6mA = 1k Ω
    I2 = 6mA + 5mA = 11mA
    R2 = 9V/11mA = 818.2Ω
    I1 = 11mA + 45mA = 56mA.
    R1 = 10V/56mA = 178.6Ω
    Finish with Ohm’s Law for power values.

    I know a lot of the above was repeat information that you gave me but re-typing it helps me understand it a little more.

    Also, thank you very much for all your help. The 7-29 circuit should be a breeze now. Again… thanks.
    Last edited: Feb 3, 2011