[Physics] Why does a higher wattage incandescent light bulb have a lower resistance value than a lower wattage incandescent light bulb

electric-circuitselectric-currentelectrical-resistancepowervoltage

I am an electrician and know through experience that resistance in an electrical circuit causes heat. An incandescent light bulb's light is a by-product of heat, so why does a 100w bulb have a lower resistance than a 25w bulb? It seems counter intuitive to me. Please help me understand.

Best Answer

You probably know that $P=IV$ (power is current times voltage), right? And Ohm's Law, $I=\frac{V}{R}$.

If you substitute $\frac{V}{R}$ for $I$ in the power equation, you will find your answer.

What confuses me, is that apparently for an incandescent light bulb, less resistance=more heat?

That's what the math says: $P=\frac{V^2}{R}$. Power (in Watts) is equal to the square of the Voltage divided by the resistance.

What's happening is, the national power grid does a very good impersonation of a constant voltage source. No matter what you connect to your power line, if the voltage is 120V when you're not drawing any current, it still will be 120V when you're drawing tens or hundreds of amperes.

Ohm's law says that if the voltage is constant, then there will be more current for a smaller resistance. The power law says that if the voltage is constant, then more current means more power.

Therefore, if the voltage is constant, there will be more power consumed by a smaller resistor.

...does this mean that a heating element, such as in a dryer or oven, must have less resistance than the wires supplying the power?

Good question. No. It doesn't mean that.

The water heater and the wires form a circuit called a voltage divider. Physically, there are three resistors; One conductor from the panel to the heating element, the heating element itself, and the other wire back to the panel.

The math is easier though if we just consider two resistors:

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The sum of the voltages dropped by each resistor in the loop, $V_1+V_2$, must add up to the supply voltage. (That means I was lying a little bit when I said that the voltage supplied to your water heater was constant. It's not. When current flows in the circuit, the resistance of the wires steals some of the voltage.) Meanwhile, the current flowing around the loop, $I$, must be the same everywhere because there's no place else for it to go.

If we say that $R_1$ represents the resistance of the wires, and $R_2$ represents the resistance of the heating element, you can see that $V_1$ will proportionately less that $V_2$ if $R_1$ is less than $R_2$.

That's important because Ohm's Law and the power law apply separately to each resistor: $P_1=\frac{V_1^2}{R_1}$, and $P_2=\frac{V_2^2}{R_2}$.

Another law you can derive is $P=I^2R$, or in this case, $P_1=I^2R_1$ and $P_2=I^2R_2$. So, we actually want the resistance of the wires to be much less than the resistance of the heating element because when they are wired in series, the smaller resistor will dissipate proportionately less power.

I don't have time to go into more detail, but here's a link to a tutorial/experiment that you can work through if you want to get a better feel for it:

http://www.ibiblio.org/kuphaldt/electricCircuits/Exper/EXP_3.html#xtocid113033

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