# Voltage divider question

• TsAmE

## Homework Statement

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

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?

#### Attachments

• Voltage divider 1.png
3.1 KB · Views: 388

## Answers and Replies

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

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?

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.