Q: How does short circuiting affect the potential difference in a circuit?

[PainterGuy]

"V=IR" and "short circuiting"

hello

Q1:
to find the potential difference across a resistor with resistance R the equation V=IR is used. suppose the resistor is connected to terminals "a" and "b"; and the current is flowing from the point "a" toward point "b". that means point "b" has less potential. but the equation "V=IR" only tells us about the overall potential difference without any regard whether the p.d. is from high to low or from low to high. it's like enclosing the p.d. with modulus "|IR|". but when it is written "V_ab = IR" then attention is given the fact if postential different is from low to high, or from high to low. if current is flowing from a to b, then "IR" would be "-ive" and so on. is what i say correct? any help would be appreciated.

Q2:
when a battery's two terminals a and b are directly connected without any resistor. the current flows from a to b, and this is said to be short circuiting. electrons at point a has maximum potential and when they reach point b they almost have zero potential. i don't get that how we are sure that when electrons reach point b they has lost all their energy or potential. perhaps they still have some of the energy. could you please help me with this? any help is welcome.

 

[mathman]

I don't really understand your questions. However in the equation V=IR, V and R are given and I is the result. In a short circuit it means the current is infinite - that's why fuses blow or circuit breaker break.

 

[diazona]

Q1:
to find the potential difference across a resistor with resistance R the equation V=IR is used. suppose the resistor is connected to terminals "a" and "b"; and the current is flowing from the point "a" toward point "b". that means point "b" has less potential. but the equation "V=IR" only tells us about the overall potential difference without any regard whether the p.d. is from high to low or from low to high. it's like enclosing the p.d. with modulus "|IR|". but when it is written "V_ab = IR" then attention is given the fact if postential different is from low to high, or from high to low. if current is flowing from a to b, then "IR" would be "-ive" and so on. is what i say correct? any help would be appreciated.

I'm not entirely sure I understand you either, but I think you do have a point that usually when applying Ohm's law, you do only calculate the magnitude of the voltage difference, and then you use physical reasoning to determine which end is at the higher potential. It'd probably be more technically correct to write it as ΔV=-IR, since the voltage decreases in the direction of current flow.

 

[jomotron]

Hi. I have a question to do with V=IR.

fairly simple. when R = 0 then current should flow most freely. But by this formula then the voltage should be 0. . . . but don't we need a voltage for current to flow?

what am i missing?

thanks

 

[HallsofIvy]

No, V is fixed. V= I(0) simply means that I will be (almost) infinite. That's why short circuits are bad things!

 

[HallsofIvy]

Q2:
when a battery's two terminals a and b are directly connected without any resistor. the current flows from a to b, and this is said to be short circuiting. electrons at point a has maximum potential and when they reach point b they almost have zero potential. i don't get that how we are sure that when electrons reach point b they has lost all their energy or potential. perhaps they still have some of the energy. could you please help me with this? any help is welcome.

"Potential energy", even electric potential as here, is not an absolute quantity but is always relative to some fixed value. Here, we know that the electrons' potential at point b is 0 because point b is the reference value. Its potential is, by definition, 0.

 

[PainterGuy]

I'm not entirely sure I understand you either, but I think you do have a point that usually when applying Ohm's law, you do only calculate the magnitude of the voltage difference, and then you use physical reasoning to determine which end is at the higher potential. It'd probably be more technically correct to write it as ΔV=-IR, since the voltage decreases in the direction of current flow.

thank you. why delta (v) = -IR, what is reason for including minus sign? could you please tell me? big thanks.

 

[diazona]

As I said, you could justify the minus sign because the voltage decreases in the direction of current flow. So, first you pick a direction to be the positive direction along the wire. If the current is flowing in that direction, you have a positive current, but then moving in the positive direction, the voltage decreases, and it is common to represent a decrease in voltage by a negative number. However, it's conventional to write the equation without the minus sign and to use physical arguments to determine the direction.

A more precise analysis would use the differential form of Ohm's law,
\vec{E} = \rho\vec{J}
Since the quantities involved here are vectors, there would be no ambiguity about the sign.