48 ohms
To find the resistance in ohms of a 194 12-volt bulb, you can use Ohm's Law (R = V/I). The 194 bulb typically has a current rating of about 0.25 amps. Using this information, the resistance would be approximately 48 ohms (R = 12V / 0.25A).
The formula you are looking for is W = I x E.
yes
No, you cannot use a 12v halogen bulb with a 6v AC adapter. The voltage of the bulb must match the voltage of the adapter for them to work properly. Using a higher voltage bulb with a lower voltage adapter can cause the bulb to burn out or potentially damage the adapter.
To calculate the minimum fuse rating needed for a 36W bulb on a 12V circuit, divide the wattage by the voltage (36W / 12V = 3A). Therefore, a minimum 3A fuse would be sufficient for a 36W bulb on a 12V circuit.
The resistance can be calculated using Ohm's Law, which states that resistance (R) equals voltage (V) divided by current (I). In this case, R = 12V / 5A = 2.4 ohms.
To find the resistance in ohms of a 194 12-volt bulb, you can use Ohm's Law (R = V/I). The 194 bulb typically has a current rating of about 0.25 amps. Using this information, the resistance would be approximately 48 ohms (R = 12V / 0.25A).
Resistance = voltage / current = 12V / 2.5mA = 12V / (2.5 x 10-3 A) = 4.8 x 103 ohm
no
ful circut diagram ac to dc adapter 12v dc
Yes, generally it is 12v 21w.
The formula you are looking for is W = I x E.
yes
yes
No, you cannot use a 12v halogen bulb with a 6v AC adapter. The voltage of the bulb must match the voltage of the adapter for them to work properly. Using a higher voltage bulb with a lower voltage adapter can cause the bulb to burn out or potentially damage the adapter.
Yes,they will. But the bulbs won't be very bright: Current will flow through all the filaments. However, with 6 times the impedance the power in each bulb will be tiny, almost invisible. For example, a 12V 6W bulb has a resistance of 24 ohms. 6 of them will have a total resistance of 144 ohms (nominally, filament resistance changes with temperature). 12V divided by 144 ohms gives 83 mA current. Using P = I2R, each bulb will have (.083)2x24 W, or 1/6 W. In a very dark room you might see a faint orange glow, but not much more.
To calculate the minimum fuse rating needed for a 36W bulb on a 12V circuit, divide the wattage by the voltage (36W / 12V = 3A). Therefore, a minimum 3A fuse would be sufficient for a 36W bulb on a 12V circuit.