Why V = IR Can't Be Used to Define p.d.

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V = IR cannot be used to define potential difference (p.d.) because it only applies to ohmic materials and does not account for non-linear devices where the relationship between voltage and current is not constant. Potential difference refers to the energy per unit charge used between two points in a circuit, while electromotive force (EMF) represents the energy provided to unit charge. Even when current is zero, a potential difference can still exist, as seen in devices like electric motors that generate back EMF. Understanding the distinction between p.d. and EMF is crucial, as both are measured in volts but represent different concepts in electrical circuits. Thus, V = IR is limited in its applicability when defining p.d. in various scenarios.
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Homework Statement


V = IR is used to define resistance. why can't it be used to define p.d.?


The Attempt at a Solution



i suggested that sometimes the current can be zero, but there is a p.d. so the equation does not hold.

but my teacher says that the answer is more towards the concept of variables, independent and dependent. i can't understand how?

thanks
 
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Can you please tell us what p.d is ? It is not obvious what you are referring to.
 
errm p.d refers to the potential difference between two points.
 
V=IR applies only to ohmic materials. There are a lot of devices (known as non-linear devices) for which V=IR doesn't hold ie. the potential difference across the device is not some unknown constant multiplied to the current through it. I don't get what your teacher is trying to say.
 
PD is the amount of energy per unit charge used between two points in a circuit. compare this to EMF which is the amount of energy provided to unit charge between two points. PD is across a resistor or any component that uses energy - bulb, wire, motor.

Currrent is zero when there is no PD across a component, however there is emf. consider electric motor. EMF provides current. motor moves and generates back EMF which provides back current. sum of current in the wire is reduced, so accordingly(ohm's law), PD is reduced - but the EMF of motor is there along with back EMF as this is a motor that is turning and simultaneously generating electricity as well.

make sure you know PD is joules per coulomb taken by a component, EMF is joules per coulomb given to a component - both are VOLTS.
 
thnks:)
 
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