Assuming the voltage of the negative terminal is 0, how calculate the voltage of point A and B? And can anyone draw a simpler equivalent circuit for me. God bless the people helping me.

 

picture

 

beside, why am I not allowed to use the image extension on this board?

Edited August 29, 201114 yr by little boy

Is this homework? What are your thoughts about this circuit so far? What have you tried?

 

It requires a couple of assumptions and working through some algebra using some laws of electricity that you probably know.

Edited August 29, 201114 yr by ewmon

What you have here is known as a loaded, probably unbalanced, bridge circuit. You will find a lot of information in this link . http://www.hallikainen.org/rw/theory/theory6.html

Edited August 29, 201114 yr by TonyMcC

I even didn't know it is called bridge circuit, therefore I had not searched the internet. Now I found a possible solution.

 

voltage A=V[R2/(R1+R2)], voltage B=V[R4/(R3+R4)]

 

It is deduced by My link, please check whether I am correct.

Edited August 29, 201114 yr by little boy

It's not really a bridge circuit because of R5. You may want to use Kirchhoff's current law that deals with the currents flowing into and out of nodes.

Edited August 29, 201114 yr by ewmon

It's not really a bridge circuit because of R5. You may want to use Kirchhoff's current law that deals with the currents flowing into and out of nodes.

 

R5 makes it a loaded bridge. Ratios of the other resistors will determine whether the loaded bridge is balanced. If unbalanced, you need to use quite a complicated bit of circuit theory. There are quite a few different ways to tackle the unbalanced loaded bridge. e.g Thevenin, superposition or a set of Kirchoff equations.

Try looking up Wheatstone bridge.

Try looking up Wheatstone bridge.

Unfortunately this will only take you so far. The Wheatstone Bridge is usually considered to have a galvanometer in place of R5. A Galvanometer is an instrument which can detect very small currents. Circuit resistor values are adjusted with the sole aim of getting zero current through the galvanometer. This indicates the circuit is balanced. Under balanced conditions voltages VA and VB could be calculated from basic principles as the galvanometer could be completely ignored. As soon as the galvanometer is replaced by a resistor and the circuit examined under unbalanced conditions things get much more complicated. http://www.hallikainen.org/rw/theory/theory6.html

Edited August 30, 201114 yr by TonyMcC