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Question about voltage divider

9905mi

Jul 5, 2018
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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
 

hevans1944

Hop - AC8NS
Jun 21, 2012
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if they all connected to the ground shouldn't they all be equal to the load's voltage monus zero?
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

Jul 5, 2018
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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.
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.
 

Harald Kapp

Moderator
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Nov 17, 2011
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the circuit scheme show's that they are conected to the ground and not to each other?
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.
why finding the resistance in R3,R2 and R1 requires subtraction of the voltages in other loads
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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