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Question about voltage divider

Discussion in 'Electronics Homework Help' started by 9905mi, Jul 5, 2018.

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  1. 9905mi

    9905mi

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    Jul 5, 2018
    This is a question about building a voltage divider for 3 loads as shown in the image below, the thing that I don't understand is why finding the resistance in R3,R2 and R1 requires subtraction of the voltages in other loads and then dividing by the current, if they all connected to the ground shouldn't they all be equal to the load's voltage monus zero?

    20180706_005248.jpg
     
  2. hevans1944

    hevans1944 Hop - AC8NS

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    Jun 21, 2012
    The four voltage-divider resistors are not ALL connected to ground. Only R4 is connected to ground. The other three resistors are in series with R4 and each other, hence the need to calculate their values based on the current through each of them and the voltage drop across each resistor. Go back and read the text explanation again more carefully.
     
    9905mi likes this.
  3. 9905mi

    9905mi

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    Jul 5, 2018
    Yes, but how can I infer that if it's not mentioned in the example and the circuit scheme show's that they are conected to the ground and not to each other?

    Thanks for answering b.t.w.
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    The schematic you present us shows the resistors in the divider chain connected to each other, not to ground whereas the loads are connected to ground. Hence Hop's argument is valid.
    Resitance is R = V/I.
    You get V by subtracting the voltages at both ends of the resistors. In the example given these voltages are equal to the load voltages,
    e.g.: V(R3) = V(load2) - V(load3) = 50 V - 25 V = 25 V.
    The current through a resistor is the current flowing out from the bottom node (connecting it to the load and the lower resistor), e.g. I(R3) = I(R4) + I(load3) = 15 mA (as given in the description).
     
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