Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Resistance and Power

Name: Javen
Status: other
Grade: n/a
Location: Singapore
Date: April 4, 2011

Many books says Nichrome is used to make heating element formula P= V*V/R, when V is fixed, the power P dissipated would decreased if R is increased. So using high resistance conductor as heating element would not work as power dissipated is low. What is wrong with this argument?


Here are the relevant formulas V = I*R P = V*I = V*(V/R) = (I*R)*I

There is nothing wrong with your argument. Case 1: V = 10 Volts, Nichrome Resistor = 100 Ohms Resulting current through the Nichrome Resistor = 0.1 Amps Power dissipated in the Nichrome Resistor = 1 Watt

Case 2: V = 10 Volts, Nichrome Resistor = 1000 Ohms Resulting current through the Nichrome Resistor = 0.01 Amps. Power dissipated in the Nichrome Resistor = 0.1 Watts Therefore increasing the resistance in the circuit reduces the dissipated power.

I think you are missing the point in that you are assuming that "high resistivity" means "higher resistivity" which leads you to think that that leads to less power. While "higher resistivity" will result in less power, "high resistivity" is just a set resistance and is a means to install a high value resistor in the circuit.

So how do we get a 60 Watt bulb and a 100 Watt bulb to output their different powers on the same 115 volt house circuit? The 60 Watt bulb incandescent (resistive) element will have a resistance of 0.5217 Ohms The 100 Watt bulb incandescent (resistive) element will have a resistance of 0.8696 Ohms.

The 60 Watt bulb will output 115 V * 0.5217 A = 60 Watts The 100 Watt bulb will output 115 V * 0.8696 A = 100 Watts

Sincere regards, Mike Stewart

It is more a practical issue rather than a theoretical one. Low resistivity materials draw much more current/power than is usually desired, at least at household voltages (100-240 VAC). For example, a nichrome-based heater providing 1000 W of power consumes 8A at 125VAC. If you made the same heater out of copper wire, it would consume almost 500 A of current and fry everything (or blow the circuit breakers). Or you would have to use over 50 meters of copper wire, which is more expensive and difficult to build.

Even so, the resistivity is still often too low, given the power and size requirements. If you look inside a hair dryer, or old aquarium heaters, you may notice the nichrome wire is built as *coiled* wire, in order to increase its total resistivity in a small space.

Paul Bridges


I think the answer relies on the temperature coefficient of Nichrome.

As you have stated, the Power decreases as the Resistance increases. The temperature coefficient of Nichrome is 0.04% per degree C. This means for every degree C increase, the Resistance increases by 0.04%.

In comparison to copper, the temperature coefficient of copper is 0.393% per degree C. So for every increase in C degrees, the resistance in copper increases 0.393%.

So in essence, Nichrome would be a better choice for Power per degree rise in Temperature.

-Alex Viray

Click here to return to the Physics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (, or at Argonne's Educational Programs

Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory