Voltage Divider: Solving for V in Series Circuits

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SUMMARY

The discussion centers on the application of the voltage divider technique in series circuits, specifically focusing on the formula VRn = Rn/Req * V. Participants clarify that the voltage divider can be applied even when not all elements are in series, particularly when dealing with current sources. The solution involves calculating the equivalent resistance of resistors and deriving voltages based on circuit configurations. Key calculations include determining V1 as 24V using a 0.9mA current through a 40K equivalent resistance.

PREREQUISITES
  • Understanding of series and parallel circuits
  • Familiarity with Ohm's Law and basic circuit analysis
  • Knowledge of voltage and current sources
  • Ability to calculate equivalent resistance in circuits
NEXT STEPS
  • Study the derivation of the voltage divider rule
  • Learn to analyze circuits with both current and voltage sources
  • Practice calculating equivalent resistance in complex circuits
  • Explore the implications of using voltage dividers in non-series configurations
USEFUL FOR

Electrical engineering students, circuit designers, and hobbyists looking to deepen their understanding of voltage dividers and circuit analysis techniques.

Miike012
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The voltage divider technique was derived from a circuit where the resistors and battery were in series.
I want to focus on the the variable V in the following formula.

And I only want to focus only on circuits whose elements are all in series with one another (No current division)

(VRn = Rn/Req*V, where V is the sum of the the constant voltage sources in series.

in the problem they used that voltage divider technique to solve for V0, however they have a multiple of V1 which is the voltage source not in series with V0.

My question is regaurding circuits with only battereis and resistors: It is obvious that the voltage divider works with elements that are not in series. How will I know when I can use volt divider tech.? Because the only way I understand how to use it is if all elements are in series.
 

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The circuit doesn't have a voltage source. It has a current source. And what you claim is obvious isn't actually true.

Instead of blindly applying a formula, you'll probably find it useful to derive the voltage divider rule. If you understand how and why it works, you'll better understand to what situations it applies.
 
Miike012 - They didn't explain the first steps in the solution, perhaps that confused you?

Follow these steps drawing circuits as you go..

First they calculated the equivalent resistance of all the resistors to be 40K.

Then refer to diagram (c)...

The current source forces 0.9mA through that 40K equivalent resistance so you can calculate V1...

V1 = 0.9mA * 40K = 24V

Then back to the original circuit in (a)...

Then they mentally "removed" the current source in the original circuit and replaced it with a voltage source equal to V1.

The 60K resistor is in parallel with V1 so easy to work out the current through that if needed.

The 40 and 80k form a potential divider from V1 so you can use that to calculate Vo.

Note: The 60K in parallel with the 40 & 80K doesn't effect how the voltage divider works because V1 is a voltage source. The 60k doesn't effect the current flowing in the 40 & 80K branch.
 

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