When I was an apprentice, I was asked a simple question, not having the initiative to acutally hook up this circuit, I just guessed as to what I thought would be correct and was, of course wrong.

I've asked this question of other electricians just to see if anyone knows off hand and very few people acutally get it right. I thought I'd throw it to you guys and see what the responses are.

If you have a 120 volt circuit with a 40 watt and a 60 watt bulb hooked in series, no other loads on this circuit, when you turn the switch on, which bulb will be brighter? And Why?

Hmmmm. Here is a total shot in the dark, assuming the lights are out

I don't know the answer, but here's my thought process so far:

The bulbs are in series, so the filament becomes a necessary part of the circuit. If one breaks, the circuit opens. So, a 60 watt bulb wants to draw current for 60 watts, which the wire can provide, but the filament of the 40 watt bulb can't provide the current for 60 watts.

OK, how close am I?

[This message has been edited by ThinkGood (edited 10-22-2003).]

If I'd answered that way back when I was an apprentice, I would have done much better.

The 40 watt bulb will be brighter. For discussion, let's assume that light bulbs are strictly resisters. They're not, light bulbs' resistance increases with the temperature of the filament. But for discussion: A 40 W bulb (120V) is 360 ohms, a 60 W bulb is 240 ohms. Now, we have both in series fed by 120V. Total resistance is 600 ohms. The 40 W bulb will see 3/5 of 120V, and the 60W will see 2/5 of 120V (think resistive voltage divider). The current will be the same thru both bulbs, so as the 40 W bulb sees more voltage across it, will dissipate more power. Thus glow brighter.

Now consider that the resistance of a light bulb is lower when cold, the 60W will see even less voltage across it. Which means the 40W will see more voltage. And the 40W bulb's resistance will go even higher as it gets hotter. And thus get a bit brighter. This resistance increase is not linear. But the answer to the question is that the 40w bulb will be brighter.

I've applied this fact on Christmas light strings, some with blinkers. Series/parallel connection. See http://home.netcom.com/~wa2ise/radios/xmassp.html

If you apply 240V on a 120V bulb, its resistance will go to infinity. AKA burned out.

Clyde:
Thanks.

wa2ise:
That's cool!

Thinkgood,
Wa2ise.

Absolutely right. The 40 watt will be brighter. I couldn't have provided a better explanation if I'd thought about it all day.

After I got the question wrong when I was an apprentice, I actually hooked it up and realized that I needed to take the initiative, rather than guess.

I've stumped quite the few electricians with that one though, so it makes me feel a little better knowing that I'm not the only person who was stumped by it.

Again, great answers guys

wa2ise quote....."If you apply 240V on a 120V bulb, its resistance will go to infinity. AKA burned out."

When I was in trade school, I hooked up a regular light bulb to 240V and it worked just fine!

Sandro.

Sandro:

It burned out just fine or it lit up?

There's a similar question relating to capacitors which causes puzzlement sometimes.

You take a 1uF capacitor and a 2uF capacitor, and connect them in series across a battery. Which capacitor would exhibit the higher voltage across its terminals?

Just saw this Thread, and wanted to add some techno-babble to it!

Yes, the 40 Watt Lamp will burn brighter than the 60 Watt Lamp, in fact - almost twice as much!

Here's why:

40 Watt Lamp will draw 14.4 Watts of True Power, and the 60 Watt Lamp will draw only 9.6 Watts - while in this series setup.

Here's the figures:

40 Watt Incandescent Lamp (resistive load):
359.28 Ohms (figured at operating temperature),
60 Watt Incandescent Lamp (also resistive load):
240 Ohms (also figured at operating temperature).

Both Lamps in Series connection results in a total Resistance of 600 Ohms (599.28 actual Ohms). 359.28 Ohms + 240.0 Ohms = 599.28 Ohms (round off to 600 Ohms).
Rt = 600 Ohms.

Total Amperes across this Resistance at 120V = 0.2 Amperes.
It = 0.2 Amps.

Voltage drop across each Element in this Network:

Voltage across 40 Watt Lamp = 72 Volts.
360 Ohms with 0.2 Amps flowing = 72 Volts.

Voltage across 60 Watt Lamp = 48 Volts.
240 Ohms with 0.2 Amps flowing = 48 Volts.

Wattage across Elements:
14.4 Watts across the "40 Watt" Lamp
(72 Volts × 0.2 Amps),
9.6 Watts across the "60 Watt" Lamp
(48 Volts × 0.2 Amps).

Total Wattage in this Node: 24 Watts @ 120 Volts.

***Note: Even without all this math and nonesense, the fact still remains:
The Load With The Higher Impedance / Resistance Will Have A Higher Potential (Voltage) Impressed Across It, When Connected In A Series Fashion.
(not the exact statement, but you get the point!)

So, without even doing the math, one could just simply say "The 40 Watt Lamp Will Be Brighter" and not worry about being wrong!

Throw in some variables, now the odds may change!

Scott35

p.s. Paul;
The smaller Cap has the higher Voltage?
It's being influenced by the fields of both the other Cap and the supply?
Also it holds a lower charge, therefore is a higher Reactance / Impedance (ooops, sorry...opposition... forgot about the DC!!!).

Scott35

Hell-o, New guy here,
just to jump in on the capacitor problem
2 capacitors in series of 1MFD and 2MFD
Scott is correct.
Ct= .666 MFD
Qt = Q1 = Q2
then Qt = Ct x E (assume E=12 VDC)
.667E-6 x 12 = 8E-6

V1 = Qt/C1 = 8E-6/1E-6 = 8 volts
V2 = Qt/C2 = 8E-6/2E-6 = 4 volts

E = V1 + V2 thus 8 + 4 = 12
and thats the way I see it.
Tom

Welcome to ECN Tom.

Yes, that's the answer. The charge, Q, in coulombs is equal to the product of capacitance and voltage.

The capacitors are in series, therefore each will acquire the same charge. Thus the 1uF capacitor must end up with twice the voltage of the 2uF cap.