Would some one please explain this circuit , & how to calculate the o/p Voltage?

 

Homework ? Did you try it ?

 

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.

 

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

 

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.

 

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

Does that mean you figured out how to solve that problem?

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

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?

 

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.

thank you , but actually my problem was with the Voltage at (B) while connecting Load 2

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 completely 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?

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.

 

I just wanna know what happens at the Point "B" when we connect a Parallel load at point "A".

 

I just wanna know what happens at the Point "B" when we connect a Parallel load at point "A".

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?