I have connected an Incandescent light bulb (R1) in series with a capacitor (C1) to the power line like so:

enter image description here

With the capacitors I have, only for capacitor values C1 = 17uF, 20uF and 22uF does the bulb glow.

For capacitor values C1 = 1uF, 2uF and 2.2uF the bulb does not glow.

The cold resistance of the bulb is 19 Ohms.

When the bulb glows, I got the following readings from my Kill-a-watt:

  • Current draw: 0.45A
  • Voltage: 117.4vac
  • PF: 0.91
  • Wattage: 48W (52VA)

From wikipedia, I read the cold resistance of tungsten-filament lamps is about 1/15 the hot-filament resistance when the lamp is operating

This agrees with the readings on the Kill-a-watt as the current through the circuit would be 0.45A if the resistance of the bulb was indeed 15 times the cold resistance (19 * 15 = 285 Ohms. Hence, 117.4/285 = 0.41A).

What I don't understand is why the bulb lights for capacitor values 17uF, 20uF and 22uF but not for 1uF, 2uF and 2.2uF.

With resistor value of 19 Ohms, capacitor values 17uF, 20uF and 22uF should have a cut off freq of more than a few Khz, hence not letting the 60Hz AC through?

Why I say a cut off freq of more than a few Khz:

Because T = RC, and here if R = 100Ohm, C = 20uF, then RC = 100 * (20 / 10^6) => f = 1/T = 500Hz which is much more than 60Hz AC and hence no signal should be out.

(Of course, my understanding wrong as I can see light on, but I want understand where I am wrong.)

Does it imply the higher value caps I am using are damaged (although they read the correct values)?

  • \$\begingroup\$ Are you using regular electrolytic capacitors or the bipolar type? What voltage are the capacitors rated for? \$\endgroup\$
    – W5VO
    Commented Aug 11, 2012 at 23:49
  • 1
    \$\begingroup\$ @W5V0: bipolar type, rated for 240VAC \$\endgroup\$
    – sekharan
    Commented Aug 12, 2012 at 0:46

2 Answers 2


The smaller capacitors will not pass as much current at 60Hz, so the bulb will not light (or glow as brightly)

We can calculate the expected wattage:

For 1uF:
1 / (2 * pi * 60 * 1e-6) = 2652 Ohms.

If we add this to the bulb resistance (say 50 ohms as it will be somewhere in between cold and hot) we get 2700 ohms (we can be rough here as the bulb resistance makes little difference)

For a 115V line, that's 115 / 2700 = ~43mA.
So the apparent wattage will be around 115V * 43mA = ~5W
The real wattage dissipated by the bulb will be much less, around 100mW.
We calculate this by I^2 * R. So with a 50 ohm filament resistance we get 0.043^2 * 50 = 92mW.

Not enough to light the bulb up. For the 2.2uF the result would be around 900mW (may glow a bit?)

If we do the same for the 22uF capacitor, we get:
1 / (2 * pi * 60 * 22e-6) = 120 Ohms.
sqrt(120^2 + 285^2) = 309 Ohms.
115V / 309 = ~372mA
115V * 372mA = ~43W (~39W dissipated by filament)

So you see the larger cap makes a big difference.

  • \$\begingroup\$ Your answers amazing. My brother posting original questions as he think I not being able to ask proper but you also answer my older questions. So if I put voltmeter across bulb, I should be see 117VAC for all cases but bulb not light up always? \$\endgroup\$
    – sekharan
    Commented Aug 12, 2012 at 0:48
  • 2
    \$\begingroup\$ If you put the meter across the bulb, the voltage you read will depend on the capacitor you are using (the voltage will be the current through the filament times it's resistance - so it will only be a few volts in the case of the 1uF - 2.2uF capacitors) \$\endgroup\$
    – Oli Glaser
    Commented Aug 12, 2012 at 0:54
  • 1
    \$\begingroup\$ You don't show the actual calculation - if you update we can have a look. I'm not sure about the CFL, it depends on how it's driven, if you just want to measure the power with the Kill-a-watt whilst running it in it's mounting then that should be okay, but I'd probably avoid trying to feed it through the capacitor without knowing more about it. \$\endgroup\$
    – Oli Glaser
    Commented Aug 12, 2012 at 1:02
  • 1
    \$\begingroup\$ Because it (the voltage across the bulb) of a simple 1 pole RC filter doesn't drop to 0 after the cutoff, the cuttoff is only -3dB = 0.707 of the input value. At 60 Hz the example in your question will be only ~4.5dB down (0.6 of the input) \$\endgroup\$
    – Oli Glaser
    Commented Aug 12, 2012 at 3:46
  • 1
    \$\begingroup\$ Not quite sure what you mean here. By 0.6, I mean 0.6 * the input voltage. So if the input voltage is 115V, the voltage across the bulb will be 115 * 0.6 = 69V. I suggest reading up on passive (RLC) network theory. Search for "impedance triangle" also, that should bring up some useful links. \$\endgroup\$
    – Oli Glaser
    Commented Aug 12, 2012 at 6:44

You have a high pass filter design. So you want the cutoff frequency closer to 60Hz for a brighter bulb. Increasing the capacitance decreases the cutoff frequency. That is why it is glowing for the higher values and not the lower.

  • \$\begingroup\$ Can you please be show the calculations? I try one example, but I arrive at much higher values than 60Hz and by that logic, blub should not glow at all in any case! \$\endgroup\$
    – sekharan
    Commented Aug 12, 2012 at 1:09

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