s34n

so if someone put a pair of sylvania ultra bulbs in the foglights it would melt the housing as well? since those are 55w just a lower temperature.

EvansBlue

It shouldn't... HIDs are just hotter than incandescent. They also suck twice the power but at the same wattage. A much simpler way of explaining it is that wattage is relative to brightness, not heat. The amperage is relative to heat.

s34n

gotcha. so in essence the housings shouldnt have melted.

Fast55

Umm, no. In DC circuits, Ohm's law is:

Voltage - Volts (V)

Voltage is the measure of potential energy per unit of charge, and is measured in Volts. A voltage measurement is taken between two points separated by a dielectric or partly conductive material. A voltage can be measured even when no electrons are moving between the points as it is a measure of potential energy and will not be released until the electrons begin to flow. Because the measurement is taken between two points the voltage is also known as a potential difference (p.d.). This is because it is the difference of potential energy between the two points. For example; A measurement with a voltmeter is taken between the two points of a battery and shows a reading of 12V. Now this just means that one terminal is 12V higher than the other. It could be that one terminal is 0V and the other is +12V, or it could mean that one terminal is -6V and the other is +6V. For this situation it does not matter which it is, as either case has a voltage of 12V between the two battery terminals.
This is often useful when designing a circuit where we need a +V and a -V. Two batteries connected in sequence can be considered to have three terminals. One end is +V, the middle where they join is 0V, and the other end is considered -V. You can also use a combination of resistors with a single battery to do the same thing which is known as a voltage divider.
Current - Amps (I)

A current is the flow of charged particles (usually electrons) which is normally produced when a source of e.m.f. is applied to a conductor. The current is akin to the actual number of electrons flowing in the same direction. The amount of current flowing in a conductor is proportional to the applied voltage and the resistance of the material. For example; if a light bulb is connected to a source of e.m.f. such as a battery the current flowing through it would be calculated using I = V / R. Where I is the current, V is the voltage of the battery, and R is the resistance of the light bulb. This relationship is known as Ohms Law. You can see from this example that to double the current flowing in the bulb you would need to double the voltage applied to it.
In a conductor the electrons are flowing from the negative terminal towards the positive, but just to confuse things when talking about electrical currents we say that the current flows from positive to negative
Direct Current (DC)
This is where the current is flowing in one direction at a constant rate. A Battery causes DC to flow in circuits.
Alternating Current
This is where the current is changing with time or oscillating back and forth. A typical mains outlet causes AC to flow in a circuit. In the UK the frequency at which this current oscillates is 50Hz.
Resistance - Ohms (R)

Resistance is a measure of the restriction of the flow of current through a material. All materials except superconductors have a resistance above zero and the value is measured in Ohms (Ω). Metals have lots of 'free' electrons therefore they have a low resistance. In an electrical circuit it is important to use cables with a low enough resistance so that they can adequately carry the necessary current for the application. In high power applications thick wires are used because thicker wires have lower resistance.
Ohms Law and Watt's Law

Ohms law determines the relationship between Voltage (V), Current (I), and resistance (R). The simple formula can be used to determine one unknown variable if the other two variables are known. Related to this is Watt's Law which includes calculations for power (energy per second).
For example; If a 12V battery were connected to a 100 ohm load such as a light bulb, the current flowing in the circuit could be calulated using ohms law.
I = V/R = 12/100 = 0.12 A (120mA)
Another example would be to calculate the power drawn in the circuit.
P = V x I = 12 x 0.12 = 1.44 W
or
P = V<SUP>2</SUP>/R = 12<SUP>2</SUP>/100 = 1.44 W
Power - Watts (P)

Power is a measure of the overall amount of work being done in a system in relation to time (energy used per second). In an electrical system power can be calculated by using the formula P = V I. From this you can see how the voltage and current in a system relate to the overall amount of power used. The unit of a Watt (W) is equivalent to joules per second , therefore one Watt is equal to one joule per second.


In other words, watts are a measure of power, not brightness and directly affect the amount of heat produced. that's why a 35 watt HID bulb in a housing meant for a 27 watt bulb melts it.

Fast55

One thing I did forget to mention is that the actual light produced for a given wattage is higher for HID's. That's why your 35 watt HID kit is in fact, brighter than the 55 watt halogen bulb you typically replace and it is cooler than the halogen bulb (35 watts < 55 watts). It's about 2800-3500 lumens VS 2100 lumens or less from a 55 watt halogen bulb. But it's a lot hotter than the 27 watt bulb in your '07-'10 fog light housing.

73shark

So your tails come on w/ your fogs?

youmeill

Try vvme.com They have 2 different models, but one will melt the lens so I went with the lower watage 6000 & WOW are they bright!

zada230

I've heard on another forum that there are metal housings available but don't remember where. Might try Jeff @ manyadeal.com I contacted them and the correct power connection is coming, but no word on a time frame yet. These should not put the fog housing at risk.