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

[TsAmE]

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?

 

[Tweedle_Dee]

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

 

[TsAmE]

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?

 

[Tweedle_Dee]

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.

 

[DeeNos]

Your calculations for the current and voltage at X and Y are correct. However, in this voltage divider circuit, the voltage at X will always be equal to the voltage at the junction between the 3k and 4k resistors, which is 6V. This is because both sides of the circuit are connected to the same voltage source, so the voltage will be divided evenly between the two resistors. Therefore, there will be no voltage drop across the 4k resistor. Your calculations show that there should be a voltage drop, but this is likely due to rounding errors in your calculations. It is common in circuit analysis to have small discrepancies due to rounding or other factors, but the overall concept and principle of the circuit remains the same.