Resistor calculation

Resistor calculation

I'm still new to electronics and I am trying to do a calculation. Say I have a 100 Watt bulb that runs at 120V and my source is 240V. I know the bulb will blow but my question is: if I wanted to put a resistor in the circuit how will I calculate the amount of resistance I would need?


Answer 1:

  • “how will I calculate the amount of resistance I would need”

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1) Calculate how much current needs to flow through the bulb at its nominal power rating.

$$I_{nom} = \frac{P}{U} = \frac{100W}{120V} = 0.83A$$

2) Calculate the amount of voltage you need to get rid of. That’s the voltage drop over the reistor R.
$$U_{drop} = 240V – 120V = 120V$$

3) Calculate the resistor value needed to cause such voltage drop
$$R = \frac{U_{drop}}{I} = \frac{120}{0.83} = 145Ω$$

4) Then calculate the power rating of the resistor
$$P = I².R = (0.83)².145 = 100W$$

Such resistor could look like one of these and would get very hot. Because the required voltage drop is exactly 50% in this case, you can use the exact same bulb as well as others have pointed out.

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The efficiency would be:

$$η = \frac{P_{eff}}{P_{tot}} = \frac{100W}{100W + 100W} = 0.5$$ or 50%, which is obviously not good.

2) More efficient alternative: Since you use AC you can replace the resistor with a coil or capacitor. These components can also provoke the required voltage drop but have the advantage that they don’t use real power but only so-called reactive power. In other words they don’t warm up. The apparent resistance they represent is called impedance (X).

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1) Calculate the resistance of the bulb
$$R_{bulb} = \frac{U}{I} = \frac{120}{0.83} = 145Ω$$

2) Calculate the total required impedance
$$X_{tot} = \frac{U}{I} = \frac{240}{0.83} = 290Ω$$

3) Calculate the impedance of the capacitor or coil
$$X_c = \sqrt(X_{tot}² – R_{bulb}²) = \sqrt((290)² – (145)²) = 251Ω$$

4) If the line frequency is 60Hz at the place where you live, this would require a capacitor of:
$$C = \frac{1}{2.π.f.X_c} = \frac{1}{2.π.60.251} = 10μF$$

So if you wire a 10μF/300VAC capacitor like the one below it will also do the trick, with an efficiency of 100%, inefficiencies of the bulb not included of course.

The above only works with incandescent bulbs not LED lighting.

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Answer 2:

If you had to do this for some reason (curiosity, theoretical, etc) you can calculate the needed resistance from knowing the required voltage drop and the nominal current.

The required voltage drop is 120v, and the nominal current of a 100w bulb at 120v is: I=100W/120V , So I = 0.833A.

So a resistor to drop 120v with I=0.833A: V/A = R, 120V/0.833A = 144 ohm.
(As stated this would be the same working resistance as a standard 100w bulb).

But the tough part is that the resistor would also need to be at least a 100W device !

Using two identical 50W bulbs wired up in series would do the trick much easier, but may not be the safest alternative.

Answer 3:

Is this a theoretical question or is this something that you actually want to do?

If this is theoretical, Ignacio Vazquez-Abrams gave you the answer above. However, if this is something that you actually need to do, there are a couple of options.

1) Easiest method: put two identical bulbs in series. Note that there are safety issues in regards to base of the bulb: I’m going to call the bulb that is closest to the live conductor the “high-side” bulb and the bulb that is closest to the neutral connection is the “low-side” bulb. Be sure that the live conductor goes to the center pin of the lamp socket on the high-side bulb, connect the outer base of that bulb to the center pin of the low-side bulb, connect the outer base of the low-side bulb to neutral.

2) Use a 240 to 120V adapter with appropriate rating. There are transformer-based solutions that are simply a 2:1 transformer: 240V in, 120V out. There are also triac-based units that simply act as a fixed dimmer that deliver 120V RMS to the load when fed from a 240V source. Although the triac-based units are intended to be used with things like hair dryers and such, they will operate a light bulb just fine.

3) You might be tempted to run the 120V bulb from the 240V source with a diode in series so as to reduce the power by half. Unfortunately, the bulb will still burn out. Putting a diode in series with the load reduces the power to half but you need to reduce the power to 1/4.