I cant figure out what the potential difference between this two points is.

[aznking1]

i know that the p.d measured across XY is the p.d. across resistor B in this picture:

http://img812.imageshack.us/img812/7665/68689848.png

But what if resistor B is removed like in this picture:

http://img707.imageshack.us/img707/7798/41743055.png

What is the p.d across XY now? Is it 0? Anyone care to explain? Would greatly appreciate your help!

 

[ehild]

i know that the p.d measured across XY is the p.d. across resistor B in this picture:

http://img812.imageshack.us/img812/7665/68689848.png

But what if resistor B is removed like in this picture:

http://img707.imageshack.us/img707/7798/41743055.png

What is the p.d across XY now? Is it 0? Anyone care to explain

Yes, it is zero.
The points X and Y are connected with a piece of wire supposed to have zero resistance.
Ohm's law says that the voltage across a resistor is RI. I is defined by the emf of the battery and R, and multiplying it with zero resistance, you get zero voltage.

ehild

 

[aznking1]

ooo thanks then does that mean there will be no reading on the voltmeter?

 

[ehild]

There will be reading on the voltmeter, it will be "0".

ehild

 

[aznking1]

oh one last question. If that's the case, then why do voltmeter measure the pd across the battery in this circuit?

http://img405.imageshack.us/img405/1755/84586970.png

Is the above circuit the same as this circuit?:

http://imageshack.us/photo/my-images/542/24112736.png/

If so then shouldn't the potential difference across XY be 0 and voltmeter reading be zero as well?

 

[kushan]

in both the above cases you are taking out potential difference across an ideal wire , which will always be zero .
Thing to note V=RI , and R= 0 .

 

[ehild]

It does not. It will measure the pd across the battery if you remove the wire at the bottom of the figure.

Usually the batteries have some internal resistance. Connecting the terminals with a wire, a very big current would flow trough that wire and the battery, including the internal resistance. The potential would fall across the internal resistance, and the pd between the terminals becomes zero.

Connecting the terminals with a wire, to "short" them, is dangerous. The current destroys the battery and makes the wire and battery so hot that it can make fire. ehild

 

[aznking1]

ok! I get it now! Thanks for your help guys! Really appreciate it.

 

[kushan]

ehild , I have a confusion .
The thing is potential difference across ideal wire is zero (always )
and pd across battery is the EMF of the battery
Arent these two statements contradicting each other in the above figure ?