How Is Voltage Divided in a Circuit with 3k and 4k Resistors?

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Discussion Overview

The discussion revolves around the calculation of current and voltage in a circuit containing 3k and 4k resistors. Participants explore the implications of current flow and voltage drop across resistors in a circuit, focusing on the conditions under which voltage is divided.

Discussion Character

  • Homework-related, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant attempts to calculate the current and voltage at points X and Y, arriving at a value for Vx that contradicts their expectation.
  • Another participant emphasizes that current through a resistor is necessary for a voltage drop to occur, questioning the flow of current through the 4k resistor.
  • A subsequent reply suggests that if there is no current through the 4k resistor, it could be treated as an ideal wire, implying no voltage drop across it.
  • A later response agrees with the idea that without current flowing through the 4k resistor, there would be no voltage drop, supporting the notion of it behaving like an ideal wire.

Areas of Agreement / Disagreement

Participants express a shared understanding that current is required for a voltage drop across a resistor. However, there is no consensus on the implications of this for the specific circuit in question, as the initial participant's calculations lead to confusion regarding the expected voltage at point X.

Contextual Notes

The discussion highlights assumptions about current flow and voltage behavior in circuits, particularly in relation to the treatment of resistors with no current. There are unresolved aspects regarding the calculations and the conditions under which voltage is divided.

Who May Find This Useful

This discussion may be useful for students and individuals interested in circuit analysis, particularly those grappling with concepts of voltage division and current flow in resistive circuits.

TsAmE
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Homework Statement



Calculate the current i and the voltage appearing at X and Y

Homework Equations



None.

The Attempt at a Solution



I found I to be 0.002A and Vy = 2V, but I am having trouble with Vx

Attempt:

I tried getting the current that goes through the 3k and 4k resistors:

Ix = V / (R1 + Rx)
= 12 / ((3 + 4) x 10^3)
= 1.71 x 10^-3 A

I then calculated the voltage on the first line after the 3k resistor:

Vx' = V(R2 + R3) / (R1 + R2 + R3)
= (12(2 + 1) x 10^3) / (3 + 2 + 1) x 10^3
= 6V

I then subtracted the Vdrop across the 4k resistor from Vx' to get the voltage at X:

Vx = Vx' - IxRx
= 6 - (1.71 x 10^-3)(4 x 10^3)
= -0.84V

but the correct answer was Vx = 6V. How can this be possible? Why isn't there a voltage drop across the 4k resistor?
 

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Current through a resistor is needed to have a voltage drop. If there is current going through the 4k resistor, where is it going?
 
Tweedle_Dee said:
Current through a resistor is needed to have a voltage drop. If there is current going through the 4k resistor, where is it going?

It would then go to X. So if there is no current through a resistor, would the resistor be treated as an ideal wire (with no resistance) and thus the same voltage as the line?
 
TsAmE said:
It would then go to X. So if there is no current through a resistor, would the resistor be treated as an ideal wire (with no resistance) and thus the same voltage as the line?

Correct. Since there is no where for the current to flow through the 4k resistor, there is no voltage drop.
 

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