Combination Circuit Question School

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Circuit_Jerk

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Hi, I'm doing some electrical homework and ran across this question.
"The hot water for the heating system for a small factory is supplied by a single boiler. The boiler is located in a small building separate from the factory. The power supplied to the boiler is 240V single-phase without a neutral conductor. At the present time, the small building housing the boiler has no inside lighting. The business owner desires that four 100W lamps be installed inside the building. All lamps are to be connected to a single switch. The lamps are to be connected to the present 240V service inside the building. How would you accomplish this task? Explain your answer."

Here's what I got this far, I don't think I'm right but I can't figure out where I went wrong:

240V power supply 4(100W) = 400W lamps wired to single switch

Wire all lamps in parallel with the switch so that the switch turns on all lights but if one light burns out, it won’t affect the other lights. Wire the switch in series with the 240V power supply.

I = 100W/240V = 0.417A per bulb. Total current is summed in parallel circuits.

0.417A * 4 = 1.668A needed through switch into lights.

240V/ 0.417A = 575.54ohms resistance per bulb.

Rtotal= 1/ (1/575.54 + 1/575.54 + 1/575.54 + 1/575.54)

Total resistance of bulbs using reciprocal formula = 143.885ohms

SQRT (P X R) = V

SQRT ( 143.885 * 100 ) = 119.952V

240V – 119.952 = 120.048V drop needed.

P = (1202/36.029) = 399.678 rating for resistor.

R = (120.0482 / 400W ) = 36.029 ohms resistor needed to add with 400W power rating.

So the major issue I've been having is I can't figure out why I'm wrong, I know I am but I can't figure out how to think about this issue differently.

Any advice is helpful. Thanks.
 
1) do you have 120V bulbs or 240V bulbs.

2) you calculate the parallel resistance of 4 bulbs, but then use that resistance to calculate voltage. In that calculation you use the resistance of 4 bulbs but the power of 1. You shouldn't need to recalculate the voltage, but it is a good double check. Done correctly you should get back to your starting voltage.

3) if you actually have 120V bulbs, then the current should be higher and the resistance lower.

4) Zooming out, do you think it is a good idea to use a resistance to drop voltage to the bulbs? Hints: think about the power consumption of the resistor, and think about what happens if a bulb blows. Maybe consider other approaches to getting the desired light.
 
Many ways to skin this beast, but I personally would do a 2+2 setup... two bulbs in series, and two of those strings in parallel. It solves the problems of "Lamps are generally designed for 120v" and "how do you maintain some lighting if one lamp goes out" simultaneously, without having to muck about with extra resistors.
 
To the OP, please give a real profile if you want to continue participating. Thank you
 
winnie said:
1) do you have 120V bulbs or 240V bulbs.

2) you calculate the parallel resistance of 4 bulbs, but then use that resistance to calculate voltage. In that calculation you use the resistance of 4 bulbs but the power of 1. You shouldn't need to recalculate the voltage, but it is a good double check. Done correctly you should get back to your starting voltage.

3) if you actually have 120V bulbs, then the current should be higher and the resistance lower.

4) Zooming out, do you think it is a good idea to use a resistance to drop voltage to the bulbs? Hints: think about the power consumption of the resistor, and think about what happens if a bulb blows. Maybe consider other approaches to getting the desired light.
Click to expand...
  1. The Voltage of the bulbs isn’t given. I'm assuming they are regular bulbs and are 120V? We're doing examples with mostly residential homes. He tells us to assume 120V most of the time.
  2. Good catch, thanks.
  3. Okay. Will recalculate.
  4. I don’t think it’s a good idea, I understand that if a bulb blows, the current will rebalance and maybe burn out the resistor?
 
Only way to do it is two bulbs in series that are in parallel with the other two in series across the 240V line. (If one bulb burns out the other two will still stay lit.) But what answer are they looking for? A theoretical one or a real world one? If I had to do this in an actual installation I would do one of the following:

1) Use 240V bulbs.
2) Use LED lighting that will work with anything from 120 to 277V even 480.
3) Run a neutral.
4) 240 to 120V step down transformer.

-Hal
 
ModbusMan said:
  • Many ways to skin this beast, but I personally would do a 2+2 setup... two bulbs in series, and two of those strings in parallel. It solves the problems of "Lamps are generally designed for 120v" and "how do you maintain some lighting if one lamp goes out" simultaneously, without having to muck about with extra resistors.​
Click to expand...
Thanks, taking your and Winnie's advice about current being higher and resistance lower.

Assuming they are 120V bulbs, I would wire 2 bulbs in series and two parallel circuits.
100W + 100W = 200W per branch.
R = 240^2 / 200W = 288 ohms resistance in each branch.
1/Rtotal = 1/288 + 1/288 = 144ohms resistance total.

I = 100W/120V = 0.833A per bulb.
0.833 * 4 = 3.332A required for 4 bulbs.

SQRT (400 * 144) = 240V

No resistor needed, just wire 2 bulbs in series and 2 branches/parallel circuit?
 
hbiss said:
Only way to do it is two bulbs in series that are in parallel with the other two in series across the 240V line. (If one bulb burns out the other two will still stay lit.) But what answer are they looking for? A theoretical one or a real world one? If I had to do this in an actual installation I would do one of the following:

1) Use 240V bulbs.
2) Use LED lighting that will work with anything from 120 to 277V even 480.
3) Run a neutral.
4) 240 to 120V step down transformer.

-Hal
Click to expand...
I think they're asking for theoretical answers, we're still learning the basics.
 
Circuit_Jerk said:
I don’t think it’s a good idea, I understand that if a bulb blows, the current will rebalance and maybe burn out the resistor?
Click to expand...
There's a much more practical reason why you don't want to use resistors, and it has nothing to do with bulbs burning out (eventually causing the resistor to fail as well).
Circuit_Jerk said:
100W + 100W = 200W per branch.
R = 240^2 / 200W = 288 ohms resistance in each branch.
1/Rtotal = 1/288 + 1/288 = 144ohms resistance total.
Click to expand...
So far, so good, and...
Circuit_Jerk said:
I = 100W/120V = 0.833A per bulb.
Click to expand...
This part is technically wrong but... well... it's homework, so I'll let you figure out why the answer is still 0.833A.
Circuit_Jerk said:
0.833 * 4 = 3.332A required for 4 bulbs.
Click to expand...
Here you go astray... go back to the above and I think you'll see why...
Circuit_Jerk said:
SQRT (400 * 144) = 240V
Click to expand...
This is a good way to double-check your work back in step 1 :)
 
Circuit_Jerk said:
Thanks, taking your and Winnie's advice about current being higher and resistance lower.

Assuming they are 120V bulbs, I would wire 2 bulbs in series and two parallel circuits.
100W + 100W = 200W per branch.
R = 240^2 / 200W = 288 ohms resistance in each branch.
1/Rtotal = 1/288 + 1/288 = 144ohms resistance total.

I = 100W/120V = 0.833A per bulb.
0.833 * 4 = 3.332A required for 4 bulbs.

SQRT (400 * 144) = 240V

No resistor needed, just wire 2 bulbs in series and 2 branches/parallel circuit?
Click to expand...
In an academic thought experiment, this will work. 4 identical resistors in 2 identical parallel paths will cut the voltage in half for each one, and each path gets the full voltage, so each bulb in the first-order level of theory, will experience the voltage drop it's normally built to expect. In reality, there's a lot of other issues to consider, for why I would advise against doing this in practice, for safety reasons.

Examples of why:
1. The bulbs will not be perfectly identical, and real world inconsistencies of incandescent bulbs could exacerbate the likelihood that one of them fails, and that the series group fails. They aren't perfect Ohmic resistors, as you'll notice when you try Ohm-metering a bulb at room temperature: it's a lot less than 144 Ohms.
2. Bulbs/sockets are constructed in a way that "expects" the thread contact to be the grounded neutral, to minimize the risk of accidentally contacting the ungrounded contact.
3. The manufacturer also only anticipated a maximum voltage-to-ground when constructing the equipment, and selecting the insulator strengths. 120V lighting isn't meant to run across 240V and 120V; it's meant to run across 120V and neutral.
4. If it's anything other than incandescent lighting, you can't assume the resistor is a decent model of the device. Other lighting technology like LED's are extremely non-linear.
 
Carultch said:
In an academic thought experiment, this will work. 4 identical resistors in 2 identical parallel paths will cut the voltage in half for each one, and each path gets the full voltage, so each bulb in the first-order level of theory, will experience the voltage drop it's normally built to expect. In reality, there's a lot of other issues to consider, for why I would advise against doing this in practice, for safety reasons.....
Click to expand...
I've done it more than once and lived to tell about it. One time it was to get heat in a well shed to keep the pipes from freezing overnight. I've also done it with apprentices to show how voltage drop works in transformers and motor windings.
 
If the first 2 lights/bulbs are wired in series and one of the bulbs blows, all the lights will go out.
Without know if the building has a neutral, I would use a transformer to get 120V from the 240V and derive a neutral from the transformer.
 
This kind of question really bothers me "... How would you accomplish this task? Explain your answer."
So I've lost count on how many ways that have been mentioned so far. Even though they are all different, they correctly answer the question: "... How would YOU ..."

Now either the question is designed to make the students think about how it might be accomplished (which would be great), or the instructor is really asking "... How would I accomplish this task? Explain you answer." (which kinda stinks)

Now for MY answer. If they want lights now, the next thing they'll want is 120V receptacles. Put in a mini-power zone and wire the place up right, with a light outside at the door, and outdoor receptacles, too.
 
Circuit_Jerk said:
The business owner desires that four 100W lamps be installed inside the building. All lamps are to be connected to a single switch. The lamps are to be connected to the present 240V service inside the building. How would you accomplish this task? Explain your answer."
Click to expand...
TwoBlocked said:
This kind of question really bothers me "... How would you accomplish this task? Explain your answer."
So I've lost count on how many ways that have been mentioned so far. Even though they are all different, they correctly answer the question: "... How would YOU ...
Click to expand...
That's my point also. This is a forum for electricians. If I was an electronics tech designing a voltage divider then I would think along the lines you are. But as an electrician things are different. There are ways to do what this "customer" wants and simply put, never or rarely are lamps put in series. We also have an electrical code to follow which would have something to say about such an installation.

So, how would I accomplish this task? I think I've outlined my options above and depending on what kind of course you are taking I might get an A or an F. :D

-Hal
 
I did once put lamps in series. An industrial oven overheated and burned up the wiring. No drawing, too many electricians got involved, and I was tasked to fixing it over the weekend. 240VAC without a neutral, and neon indicating lights at each end of the oven. I think they had put 240vac to the neon lamps, when 120vac would be more appropriate. I put them in series, but they didn't always work right. Seems that sometimes one would "cascade" before the other and "hog" the voltage. I guess they weren't essential, as they ran it like I left it.
 
Another time I was on a ship that needed drop lights in the cargo holds. This was an oil tanker that was hauling grain of all things. They sent us a bunch of extension cords and lights and said good luck. This being a tanker, the only 120v receptacles where up on the bridge wings. Instead, we came out of the 240V Suez Canal Light receptacle in the bow, ran a 14 ga cable down the centerline and spliced the ends of the extension cords into it. And for bulbs? Well, the ship was a kind of hybrid between US and European. The Edison base fixtures in the berthing areas were 240V, so we used those in the drop lights. And, yeah, sometimes a guy would put a 120V bulb in a 240V fixture with briefly illuminating results!
 
Joethemechanic said:
So does that mean the building is also without a neutral? Or just the boiler circuit is without a neutral?
Click to expand...
It's an academic example. It wouldn't meet code to solve this just with the materials given.

The premise is that the room in question only has the 240V circuit. No information on whether this is center-grounded like we do in the US, or end-grounded like 220V thru 240V circuits are in most of the world.
 
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