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I have a 12v adapter and small bulb of about 3.8v. If i give power to bulb with 12 volt adapter and with 100ohm to 10k ohm it doesn't light up and if i directly give it power without the resistor it fuses.

This is the bulb i have

Bulb

When i give power to bulb with 9v battery without resistors it light up perfectly fine.

Adapter gives 1.5amp (mentioned on sticker)

How can i light up this bulb with 12volt adapter without blowing(fuse) it up?. I am not much experienced with electronics.

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  • \$\begingroup\$ Is there any marking on the bulb or base? Voltage or current, or just a type number? Without some specifications on the bulb, it is hard to comment. \$\endgroup\$ Commented Dec 7, 2015 at 23:12
  • \$\begingroup\$ `3.8v' written on the packet it came in \$\endgroup\$
    – Skyyy
    Commented Dec 7, 2015 at 23:14
  • \$\begingroup\$ What is your question? \$\endgroup\$
    – Andy aka
    Commented Dec 7, 2015 at 23:14
  • \$\begingroup\$ How can i light up this bulb with 12volt adapter without blowing(fuse) it up. I am not much experienced with elelctronics \$\endgroup\$
    – Skyyy
    Commented Dec 7, 2015 at 23:15
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    \$\begingroup\$ 9V batteries have very high internal resistance (I've seen as much as 50Ohm). At 0.3A, the battery terminal voltage probably dipped quite significantly (hence not burning out the bulb). \$\endgroup\$ Commented Dec 8, 2015 at 2:10

3 Answers 3

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I found a reference to a #13 radio pilot lamp rated at 3.8 volts, 0.3 amp. To operate that lamp from a 12 volt supply, you need a resistor that will drop 8.2 volts at 0.3 amp. Ohm's Law says that a 27 ohm resistor would work.

If your lamp draws a different current, the you would need a different resistor, but 27 ohms should be a good starting point.

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  • \$\begingroup\$ Thanx i will try and will update you. \$\endgroup\$
    – Skyyy
    Commented Dec 7, 2015 at 23:31
  • \$\begingroup\$ After more calculation, I find you'd need a 2.5 watt resistor, so you really should try to find some 12 volt bulbs (or perhaps use an LED). \$\endgroup\$ Commented Dec 8, 2015 at 0:01
  • \$\begingroup\$ adding 25ohm resistor did work. \$\endgroup\$
    – Skyyy
    Commented Dec 8, 2015 at 0:04
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Many adapters of the type you describe are unregulated giving a higher-than-nominal output when unloaded or even lightly loaded. They tend use the load current of their intended powered device to pull the output voltage down towards the nominal value (12V in your case).

You say that you think your MES lamp is 5V and that the adapter has a capability of 1.5A. Your lamp probably draws between 60 and 300mA, consequently your adaptor is only lightly loaded at perhaps 6% of its capability, causing a high output voltage to appear across your lamp and 'blowing' it!

The value of any series resistance is calculated from Ohm's Law using the formula: R = V/I where V is the difference between the nominal supply voltage (12V) and the lamp's voltage, which in this case gives us the result of 7V.

The value for I is the nominal lamp current, say 100mA (or 0.01A). This gives a resistor value of 7/0.01 = 70 ohms. The nearest E12 resistor value is 68 ohms. If the lamp lights with this value of resistor but is somewhat dim, it means your lamp requires more current (assuming the supply is as described). Try fitting a lower value resistor, say 47 ohms until you reach a satisfactory brilliance.

Your question raises a number of other issues inherent in this situation, such as the required power rating of the resistor and the very fact that whilst this solution is a simple one, it represents a low efficiency solution owing to the energy that is wasted as heat in the resistor.

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The problem is that the resistance of the bulb is probably very low, and you need to use a resistor roughly equal to the resistance value of the bulb, to equalize the voltage drop across both components. The 12v is too high for your 5v bulb, so that's why it burns out when you connect it directly to the 12v. I'm assuming you're connecting it in a parallel?

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  • \$\begingroup\$ When checking with multi meter set on 200M ohm it gives ouput 00.9 \$\endgroup\$
    – Skyyy
    Commented Dec 7, 2015 at 23:22
  • \$\begingroup\$ So probably 1 ohm, which sounds about right, if you connect a 2 ohm resistor across the bulb, you'll drop appx 8volts across the resistor and 4 volts across the bulb, which should be about right. \$\endgroup\$ Commented Dec 7, 2015 at 23:29
  • \$\begingroup\$ Thanx i will try and will update you. \$\endgroup\$
    – Skyyy
    Commented Dec 7, 2015 at 23:31
  • \$\begingroup\$ @philbrooksjazz: he should connect the resistor in series with the bulb, not "across" it (I'd understand "across" as "in parallel", which won't do any good.) Don't forget that the resistance of an incandescent lamp increases as it heat up, so he would want a significantly higher resistance than you suggest. \$\endgroup\$ Commented Dec 7, 2015 at 23:37
  • \$\begingroup\$ So, be careful on one issue - I think Peter's ans. is for a series circuit, and if you're using that, your resistor power rating needs to be close to the bulb, even for a small flashlight bulb it's prob. 1.5 watts, so if you use a standard 1/4 watt resistor is could burn up the resistor. All the current flows through all the components in a series circuit. \$\endgroup\$ Commented Dec 7, 2015 at 23:38

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