# Can I connect 500K resistor directly to the mains?

I would like to test what will happen if I put a high resistor between 220V AC terminals.

My calculations say that it is possible, but I asked this question to know the practical point of view before testing it.

I = 220 / 500k = 0.44 mA

P = I * V = 0.44 * 220 = 96.8 mW

If the maximum power that a resistor can handle is 0.125 watt, the previous calculations allows me to do this experiment. Is that right?

If the maximum power of my resistor is 0.125 watt and I need to apply 0.25 watt to it, can I use two series resistors?

One more thing: Can I calculate the input impedance of a circuit and then connect a suitable resistor in series to the circuit so that I no longer need a transformer? I know that it is too dangerous but is it possible?

simulate this circuit – Schematic created using CircuitLab

• It depends mostly on if you hold it with your bare hands... Commented Oct 22, 2015 at 8:12
• @PlasmaHH Of course I won't. I will not touch anything. I have a bread board and wood bearings. Commented Oct 22, 2015 at 8:30
• You may exceed the specified max voltage for the breadboard. Please check that before connecting mains to it. Or better, use a screw terminal block instead.
– user16324
Commented Oct 22, 2015 at 8:44
• Just plug it in and see what happens, don't be surprised if it gets hot. Commented Oct 22, 2015 at 12:17

The answer is 'it depends'. Resistors also have a maximum voltage rating. If yours is rated for >350v (240v AC peak = 240x1.41), it should be ok.
Example datasheet: http://www.vishay.com/docs/28722/sfr16s25.pdf
You will want to keep a 50% 'margin' on both wattage and voltage rating to be on the safe side and also keep an eye on the enclosure - if the temperature around the resistor(s) gets too high, you need to de-rate the allowable power dissipation.
On your second question, you are correct. Using two resistors of half the value in series will halve the voltage across them. Since the current is the same, you will halve the power dissipation. This also allows you to use a lower voltage rating.

• 250 V is not a sufficient rating for 220 VAC. The peaks will be over 310 V. You need a resistor rated for at least 350 V, preferably 400 V. Commented Oct 22, 2015 at 11:32
• @OlinLathrop the important part is the datasheet actually says that too, "Operating voltage, Umax. AC/DC" Commented Oct 22, 2015 at 15:37
• so yes, I agree the full voltage rating should deal with the max possible instantaneous voltage including the fact that AC voltage goes ~1.4x higher than what is written, and then from that, do the x1.5/x2 safety factor. Unless of course you like living on the edge and enjoy a flash-over/spark across the resistor at 50/60Hz Commented Oct 22, 2015 at 15:39
• Thanks all, I've added a correction for AC to be on the safe side.
– RJR
Commented Oct 24, 2015 at 4:20
• I do not recommend using one resistor. In life it happens that the purchasing department will get cheaper resistor, incorrectly missing its inappropriate lower voltage rating, especially with SMD resistors. the solution is be safe and always put 2 resistors in series (of course with half of the desired resistance each). This is what I use in my designs for 240Vrms AC. Or go crazy, for industrial high reliability devices I have seen 4 resistors in series in 240Vrms AC. Commented Jan 21, 2019 at 14:17

Keeping the power dissipation within ratings is necessary, but not always sufficient. Your power calculations are correct.

In practice if the resistor is rated for 500V or more (peaks on 220VAC are well over 300V nominal) it will likely be fine, at least for brief connection or if reliability is not important. If the resistor is connected in parallel with some inductance (a motor or transformer) across the mains, and the power to the pair is switched, it may well not be fine. The energy store in the inductance can cause the voltage to rise to more than 1kV, dspending on exactly where in the AC cycle the pair is switched off.

For high reliability and safety it is best to use a part that is designed for this kind of service such as Vishay's VR25 series. They can withstand high transient voltages (many kV).

If you want to use regular resistors, putting two in series helps with he voltage rating issue as each will see half the voltage if the resistances are equal.

Yes you can derive a lower voltage power supply by using a dropping resistor, but it's horribly inefficient, as well as being potentially dangerous. For example if you powered two 2V red LEDs connected in inverse parallel through the 500k resistor the power supply would be less than 1% efficient. Still possibly useful- but keep in mind that the LEDS would have to be surrounded by insulation since contact could prove fatal. The epoxy molding would not be sufficient.

Can I calculate the input impedance of a circuit and then connect a suitable resistor in series to the circuit so that I no longer need a transformer?

Yes, you can do it under three conditions:

• if the current is negligibly small (e.g. a neon test lamp), so that the dissipated power is small

• if the current (load resistance) is relatively constant, so that the load voltage is relatively constant

• if the load has extremely low resistance (in this case, the combination of the mains voltage source and the resistor in series acts as an AC current source supplying a current load)

You can connect a capacitor in series to a resistor to reduce the dissipated power; however, the load must pass the two half-waves of the current.

This arrangement can be danger if the load is disconnected. Then the mains voltage can be applied to you through the resistor (like in the case of the neon test lamp).

A clever trick is to utilize the dissipated power by replacing the ballast resistor with a useful load. Travel hair dryers exploit this idea where the resistor is replaced by a heater and the load is a DC motor (of course, a full wave diode rectifier is connected before the motor).