Yes. And here's why.

# Rheostat dimmers

Old dimmers, used a variable resister to dim the light. Lets look at a simple example.

We can find total resistance (RT), by adding up all the resistance.

RT = R1 + R2 = 0 Ohms + 144 Ohms = **144 Ohms**

Then we can find the total current (IT).

IT = ET / RT = 120V / 144 Ohms = **.83A**

We'll then calculate the voltage across each resistive load.

E1 = IT * R1 = .83A * 0 Ohms = **0V**

E2 = IT * R2 = .83A * 144 Ohms = **120V**

Finally, we'll calculate the total wattage (WT)

WT = V^2/R = 120V ^2 / 144 Ohms = **100 Watts**

Lets see what happens when we increase the resistance of R1

RT = 200 Ohms + 144 Ohms = **344 Ohms**

IT = 120V / 344 Ohms = **.349A**

E1 = .349A * 200 Ohms = **69.77V**

E2 = .349A * 144 Ohms = **50.23V**

WT = 120V ^2 / 344 = **41.86 Watts**

As you can see, we've increased the resistance of R1 and effectively reduced the voltage across R2. And now we have a dim light.

# Thyristor dimmer

Modern dimmers use a TRIAC, to reduce the amount of time the light is on. However, because of the circuitry in the dimmer, there is not a direct 1:1 energy savings. Dimming the light to 50%, will not equate to a 50% savings in electricity.

A typical waveform in an AC system would look like this.

A TRIAC prevents electricity from flowing every time voltage reaches 0, something like this.

So you end up with a waveform that looks like this.

With the TRIAC, the light is actually turning off and on 120 times per second. With every cycle, you're saving a small amount of power. Is it enough to actually see on your electric bill? I guess it would depend on how long the lights are on, and what percentage they are dimmed.

2If you prefer dimmed lights and would never want the lights to be brighter you could always install lower output bulbs instead. – ChrisF – 2012-01-26T10:17:30.420

It is better to have less light on, then all the lights dimmed. – Walker – 2012-01-26T12:57:27.757

4For both answers: It's not true that a simple resistor does not save power. As Power = Voltage² / Resistance, and Voltage is always 230V (or 110V depending on country), the consumed power actually drops. – Nikodemus – 2012-01-26T13:45:28.377

I guess @Nikodemus comment comes from eliminating

`I`

(current) in`P = V*I`

and`V = I*R`

. But to understand this better it helps me to think about how, as the resistance is increased, current (and hence power) must drop, because the job of mains is to keep`V`

from sagging at all under load. – wim – 2012-01-26T14:02:46.867