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I am doing a project on the resistance of a wire. I need to know what the p.d is, what you use to measure it, and its unit.

2007-01-29 04:43:31 · 6 answers · asked by jdratbull 2 in Science & Mathematics Physics

6 answers

the p.d. is potential difference and is another name for the voltage.

It is measured in volts with a voltmeter.

2007-01-29 09:12:01 · answer #1 · answered by Ben P 2 · 0 0

p.d between the ends of the wire is equal to the work done in taking a unit charge from one end to the other against the electric field in the wire. The p.d. between the ends of the wire is produced when it is introduced in an electrical circuit having a battery or other source of electrical energy. It is measured by connecting a voltmeter across its ends and reading it.
Its unit is Volt.

2007-01-29 05:14:00 · answer #2 · answered by Anonymous · 0 0

p.d is the poteneial difference btween 2 points in the wire u r measuring.
Voltmeter is used to measure it and p.d has the same unit as voltage.

2007-01-29 11:24:01 · answer #3 · answered by yan 2 · 0 0

Assume the wire is in the form of a resistor.
Let V = voltage across resistor
Let I = current through the resistor
Let R = resistance of resistor.
V = I x R and V is said to be the potential difference across the resistor.
This would be measured in volts by a voltmeter.

2007-01-29 06:15:38 · answer #4 · answered by Como 7 · 1 0

p.d. = potential difference

It's the difference in voltage between different points on the wire. Measured in volts.

Have a guess as to what sort of a meter to use!

2007-01-29 04:55:12 · answer #5 · answered by lunchtime_browser 7 · 1 0

PD = potential difference. Basically it mean the voltage and is measured in volts so use a voltmeter. The resistance of a wire is cause by the electrons having to move through the lattice of metal atoms. It is not helped by faults or impurities in the metal. Remember a solid piece of metal is not one piece. There are many grain boundaries between individual crytstals as well as impurities.

2007-01-29 04:57:44 · answer #6 · answered by Richard T 4 · 0 0

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