# Thread: Electrical engineering question

1. ## Electrical engineering question

The circuit of FIGURE 2 shows a 10 kΩpotentiometer with a 5 kΩ load.Determine the position of the slider on the‘pot’ when the voltage across points ‘XX'is 3 V.

Really struggling with this one. I understand that it'll end with a quadratic, but that's about it. Could really use some direction!!

2. ## Re: Electrical engineering question

equivalent circuit ...

total resistance in the parallel section is 1/2 the resistance of the section in series ...

$\dfrac{5R}{5+R} = \dfrac{10-R}{2}$

... where $R$ is in $k\Omega$

3. ## Re: Electrical engineering question

Hi, thanks for replying.

So, if we know the voltage across points 'XX' is 3V, then of course the voltage across the first resistor is 6V.

So R=5kohm ??

So how can I determine the position of the slider? What other math do I have to do?

4. ## Re: Electrical engineering question

Okay, so I've got this.

As the question says, 'the voltage across points ‘XX' is 3 V.' Meaning 'R' must be 5kohm as it has the same voltage across it as its parallel 5kohm resistor does, right?

And as you said, 'total resistance in the parallel section is 1/2 the resistance of the section in series ...'

So because the voltage drops from 6V to 3V (50%) between '10kohm-R' and the 'R' & '5kohm' parallel resistors.

Can we conclude that the slider position on the 'pot' will be 50% of the way down (0.5) ?

Seems a little wordy to me, like there should be more maths involved??

Am I close or way off?

5. ## Re: Electrical engineering question

Most sliding potentiometers are linear. Set halfway should give the desired result.