Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
What resistor is required to dim a 150 watt light bulb to half power of 12 volt ciruit?
What resistor would be required to dim a 150 watt light bulbto half power on a 12 volt circuit?
Anybody?
Thanks
7 Answers
- billrussell42Lv 71 decade agoFavorite Answer
Resistance of bulb is
P = E²/R
150 = 12²/R
R = 0.96 ohms
assuming the resistance is linear with temperature, which it is not,
half power is 75w
75 = E²/0.96
E = 8.49 volts (note that this is not half the voltage)
I = E/R = 8.49/0.96 = 8.84 amps
resistor has to drop 12–8.49 = 3.51 volts
R = E/I = 3.51/8.84 = 0.40 ohms
rating of resistor
P = EI = 3.51 x 8.84 = 31 watts, use a 50 watt resistor.
edit, fixed dumb arith mistake.
.
- Peter HLv 71 decade ago
newly-wed's calculation gives you the effective resistance of the 150 watt bulb, but not the resistance required to give half the power. If the power is reduced to 75 watts, then the current flowing is 75/12 = 6.25 amps and the total resistance is 12/6.25 = 1.9 ohms. In theory you subtract 0.96 ohms for the bulb to get the value for the external resistance, but in practice the resistance of the bulb at half power is slightly less than at full power, so the figure is only approximate. The external resistor will still be dissipating 30-40 watts and needs to be appropriately specified (e.g. 60 W). This is a very inefficient way of reducing the lamp power. It is better to use some electronics to drop the voltage.
- 5 years ago
Other posters gave great answers. The one that thought you had a two strip light must have thought you had a 3 foot tank, which would have equated to that wattage. The coral referenced by others is very valid, and still applicable. You can add things like branching montipora to the SPS. Your leathers are moderate light leathers, and should do fine. Basically, stick with any coral that is low, low-moderate, and moderate light requirement classification and you will be golden! On the leathers you have and the feather dusters, be sure to feed a filter feeder food to keep them healthy. The other posters are correct, the yellow tail damsel is a small monster and the domino grows into a large monster! The blue damsel would be expected to be the lesser aggressor, but still not exactly peaceful. If doing the anemone, know they move! They can be difficult with coral, especially in a small tank, as they tend to sting or outright move on top of the coral!!! So, if you chose to attempt a bubble, wait on coral, if you want coral, ditch the anemone plan.
- LarryLv 41 decade ago
P = IE
IE = P
I = P/E = 150/12 =12.5 Amps
R = E / I = 12/12.5 = 0.96 ohms
BULB RESISTANCE 0.96 ohms
check: P = V^2 /R = 144 / .96 = 150 W
Current required for 75 watts thru .96 ohm bulb.
(assumes bulb of constant resistance)
P=I^2 R
I^2 = P / R
I^2 = 75/.96 = 78.125
I = 8.8388 Amps
Total resistance to limit current to 8.8388 amps when 12 V supplied:
Rtotal = 12V / 8.8388 A = 1.3577 ohms
Rneeded = Rtotal - Rbulb = 1.3577 - .96 =
0.3977 ohms
check: 12v / Rtotal = 12/1.3577 = 8.838 amps
I^2 R = 8.838^2 times .96 = 74.99 W
- How do you think about the answers? You can sign in to vote the answer.
- 1 decade ago
power is P=V.I you don't have I but you want to solve the problem using resistors, so using V=I.R, replace I by V/R
Now you have P=V.(V/R)=V^2/R
rearrange this equation so you have R on the left side, R=V^2/P
now put in the numbers, R=(12)^2/150=0.96 ohm
- 1 decade ago
You can use a light dimmer regulator(electronic) for reducing the light discharge from the bulb for practical purposes,and (or)use I=E/R.formula.