# Help With Loaded potentiometer voltage divider Quiz

Discussion in 'Homework Help' started by amirengineer, Apr 12, 2013.

1. ### amirengineer Thread Starter New Member

Aug 13, 2008
4
0
Would some one please explain this circuit , & how to calculate the o/p Voltage?

2. ### mrmount Active Member

Dec 5, 2007
59
7
Homework ? Did you try it ?

amirengineer likes this.
3. ### WBahn Moderator

Mar 31, 2012
17,716
4,788
You need to ask more specific questions and show YOUR efforts to solve YOUR homework. That will give us a starting point for a discussion. As it is, we have absolutely no idea what about this circuit might be causing you problems. Perhaps its because you have no idea what a potentiometer is, or perhaps it's something else entirely. We are not mind readers.

amirengineer likes this.
4. ### amirengineer Thread Starter New Member

Aug 13, 2008
4
0
thanks to all of you , i was very confused about what actually happens when we put a load in parallel with the pot..

now i have another specific question also on V.Divider ,here it is

5. ### screen1988 Member

Mar 7, 2013
310
3
When only load 2 is connected, the equivalent resistance of two resistors 3k and 700Ω will be 3kΩ + 700Ω = 3.7kΩ
Now you have the circuit with resistor 1k2 is in series with 3.7k || 5k.
From this you can find the voltage at A.
When voltage at A has been solved, you can again find the voltage at B because it is a voltage divider with two resistors 3k and 700Ω.

amirengineer likes this.
6. ### WBahn Moderator

Mar 31, 2012
17,716
4,788
Does that mean you figured out how to solve that problem?

The question is reasonably specific, now you just need to show us YOUR efforts to arrive at a solution. They don't have to be correct, but it gives as a look into how you are thinking and approaching things and where you are having problems.

If only Load 2 is connected, then you can forget Load 1. So redraw the circuit with Load 1 compltely removed. Now, can the top two resistors be combined into a single effective resistance? If so, then doesn't the circuit look just like your original circuit?

amirengineer likes this.
7. ### amirengineer Thread Starter New Member

Aug 13, 2008
4
0
thank you , but actually my problem was with the Voltage at (B) while connecting Load 2

I'd solved it but i think i answer it the hard way, here is what i did

i simplified the Circuit into just a DC source & a Resistor then get the total current after that i used the current divider to get the current that path through the 700 ohm and then multiply both the branch current & the 700 ohm Resistor...it gives me a v.close answer but I'm suspected that there is an easier way to do it(this suppose to be a Voltage Divider question not current Divider) beside i want to know what happens physically at Point (B) when we connect the 5K load , not just how to solve it.

8. ### amirengineer Thread Starter New Member

Aug 13, 2008
4
0
I just wanna know what happens at the Point "B" when we connect a Parallel load at point "A".

9. ### WBahn Moderator

Mar 31, 2012
17,716
4,788
So, for this case, we are only talking about a single load (namely Load 2), correct? In other words, we can ignore Load 1 -- that switch is going to remain open.

Q1) What is the voltage at B in terms of the voltage at A?

Q2) Does the answer to Q1 depend on whether Load 2 is connected or not?

Q3) What happens to the voltage at A when Load 2 is connected? Does it go up or go down?

Q4) Given the answers to the above, does the voltage at B go up or go down when Load 2 is connected?