Understanding Voltage Division in Circuits: A Superposition Approach

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Discussion Overview

The discussion revolves around the calculation of voltage in a circuit using voltage division and the potential application of superposition. Participants are attempting to solve a homework problem involving multiple voltage sources and resistors, exploring different methods to arrive at the correct voltage output.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a solution using voltage division but finds their answer incorrect, prompting a request for assistance.
  • Another participant suggests using nodal analysis as an alternative method to solve the problem, indicating that all currents can be expressed in terms of voltage and resistance.
  • A participant clarifies that they are required to use voltage division and describes the flow of current in the circuit, detailing how it splits at various points.
  • A repeated post reiterates the initial attempt at a solution, expressing confusion about applying voltage division with multiple inputs and hinting at the possibility of using superposition.
  • A later reply proposes a superposition approach, calculating contributions from different voltage sources and arriving at a combined voltage output, while also mentioning a resource for further reading on voltage dividers.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct method to solve the problem, with some advocating for voltage division and others suggesting nodal analysis or superposition. The discussion remains unresolved regarding the best approach to take.

Contextual Notes

There are limitations in the participants' understanding of how to apply voltage division in circuits with multiple inputs, and the discussion includes various assumptions about current flow and circuit behavior that are not fully explored.

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Homework Statement



Calculate v (refer to attachment).

Homework Equations



Voltage division: Vout = Vin x R2 / R1 + R2

The Attempt at a Solution



V1 = V1 x R2 / R1 + R2
= 15 x 5 / 10 + 5
= 5V

V2 = V2 x R4 / R3 + R4
= 5 x 2 / 10 + 2
= 0.83V

V = 5 + 0.83
= 5.83V

But my answer is wrong. Could someone please tell me what's wrong?
 

Attachments

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Have you tried nodal analysis, or are you supposed to use "voltage division" to solve it for academic perfection of the technique?

[tex]I_1 + I_2 + I_3 + I_4 = 0[/tex]
where I_1 through i_4 refer to the four currents going through the four resistors. Every single current can be written in terms of the voltage across the resistor and the resistance.
 
I am supposed to use voltage division.

In connection with how the current flows (if this was part of an actual circuit), would it be like this:

*The current from the +15V supply (I1) spilts into 2 currents: the one flows from the midpoint to Earth (I2), and the other one flows to the right of the mid point (I3). I1 = I2 + I3.
*The current from the +5V supply (I4) spilts into 2 currents: the one flows from the midpoint to Earth (I5), and the other one flows to the right of midpoint (I6). I4 = I5 + I6.
*I3 would have joined with I6 then traveled through the same line as v and onwards.
 
TsAmE said:

Homework Statement



Calculate v (refer to attachment).

Homework Equations



Voltage division: Vout = Vin x R2 / R1 + R2

The Attempt at a Solution



V1 = V1 x R2 / R1 + R2
= 15 x 5 / 10 + 5
= 5V

V2 = V2 x R4 / R3 + R4
= 5 x 2 / 10 + 2
= 0.83V

V = 5 + 0.83
= 5.83V

But my answer is wrong. Could someone please tell me what's wrong?

I don't know how to use "voltage division" when you have multiple inputs like that. Doesn't make sense to me... Maybe there's some superposition trick, but I'm not seeing it.
 
berkeman said:
... Maybe there's some superposition trick, ...

I think you are right. Superposition with voltage division can solve this.

1) Contribution of +15V:

V=(2k||10k||5k)*15/(10k+2k||10k||5k)=1.25k*15/(10k+1.25k)=1.667V

2) Contribution of -15V:

V=(2k||10k||10k)*(-15V)/(5k+2k||10k||10k)=1.429k*(-15)/(6.429)=-3.333V

3) Contribution of 5V:

V=(2k||10k||5k)*5V/(10k+2k||10k||5k)=1.25k*5V/(10k+1.25k)=0.556V

Superposition:
V=1.667-3.333+0.556=-1.11V

I have a http://www.solved-problems.com/tag/voltage-divider/". I need to add one similar to this.
 
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