How to Design a Circuit Diagram to Independently Control Multiple Light Bulbs?

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Discussion Overview

The discussion revolves around designing a circuit diagram to independently control five light bulbs with specific voltage requirements. The focus is on using a 12 V power supply, two resistors, and five switches to achieve the desired functionality while ensuring that the voltage remains within specified limits for each bulb.

Discussion Character

  • Homework-related
  • Technical explanation
  • Exploratory
  • Debate/contested

Main Points Raised

  • Participants discuss the need for each light bulb to be independently controlled and the voltage requirements for each (12 V for bulbs 1 and 2, 9 V for bulbs 3, 4, and 5).
  • Some participants propose using resistors to maintain the required voltages for the light bulbs.
  • One participant mentions that they have only two resistors available for the design, which raises questions about how to effectively use them.
  • Suggestions include using a voltage divider to reduce the voltage for bulbs 3, 4, and 5, with calculations provided for determining the appropriate resistor values.
  • There is a clarification that the number of bulbs needing 9 V is three, not two, prompting a reevaluation of the design strategy.
  • Participants express uncertainty about the limitations of using only two resistors for all three 9 V bulbs and the implications for the circuit design.

Areas of Agreement / Disagreement

Participants generally agree on the need for a voltage reduction method, such as a voltage divider, but there is no consensus on how to effectively implement the design with the constraints of using only two resistors for multiple bulbs. The discussion remains unresolved regarding the best approach to meet all criteria.

Contextual Notes

Participants note the importance of ensuring that the voltage remains within a 5% tolerance for the light bulbs, which adds complexity to the calculations and design. There are also unresolved questions about the feasibility of using only two resistors for three bulbs.

Cisneros778
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Homework Statement


Given =
5 - light bulbs
Light bulbs 1 and 2 each dissipate 180 mW at 12 V
Light bulbs 3, 4, and 5 each dissipate 45 mW at 9V.

1 - 12 V supply

2 - Resistors (Range 10-1000 Ω)

5- switches

The design must meet the criteria:
Each light bulb can be independently turned on and off
Light bulbs 1 and 2 must get 12 V from the power supply when on.
The other light bulbs must get 9V (+/- 5%) from the power supply when on.

Homework Equations


V = IR
P = I2R

The Attempt at a Solution



Light bulbs 1 and 2 have 800 Ω each.
Light bulbs 3, 4, and 5 have 1800 Ω each.

I can solve the diagram easily when I have only one light bulb on each time. It is when I have multiple light bulbs on that the voltages change and no longer fit the criteria.
 
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Cisneros778, what could be the idea behind allowing the use of resistors? http://imageshack.us/a/img706/5864/hidden14.gif
 
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The 2 resistors will keep all light bulbs within their allowed voltages.
 
Cisneros778 said:
The 2 resistors will keep all light bulbs within their allowed voltages.
"2 resistors"http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon5.gif There is no limit of 2; you may use as many as you like. :wink:

Sketch a diagram to illustrate your idea.
 
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I actually have only two resistors to work with. I mean, the light bulbs are resistors as well, but apart from that I only get to use two resistors for my diagram. :-/
 

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Okay, I figured out the first two light bulbs, having them parallel will give them 12 V. For light bulbs 3,4 and 5 I need to have 9V running through each one of them. So, I'm guessing to use the same strategy by having them parallel but I only have the 12 V source. So I believe here is where the 2 resistors come into play. How to include them, is what puzzles me.
 
any suggestions?
 
Are the five switches all SPST? :smile:
 
Yes, just on or off
 
  • #10


If I understand you correctly, you just need to know how to reduce the voltage using two resistors? The simplest solution would be a voltage divider. I'll leave it to you to look that up and figure out exactly how to implement it in this situation, but it shouldn't be too difficult. (Maybe steer clear of the Wikipedia page and look for a simpler explanation if you're new to circuit analysis.)
 
  • #11
For bulbs 3 and 4. Use..

Power = Voltage * current

to work out the current going through the bulb.

Then use one resistor in series with each bulb to drop the 12V down to 9V.

V=I*R
so
R=V/I

where the drop across the resistor V = 12-9 and I is the current you calculated above.

However the answer is not a standard resistor value in the 5% range.

http://ecee.colorado.edu/~mcclurel/resistorsandcaps.pdf

Try nearest value up. Recalculate the voltage to check it's within 5%. It should be according to my calc.
 
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  • #12


Humm Ok I missed that there are THREE 9V bulbs not TWO. Will have a rethink.

Does the problem really say that ONLY two resistors can be used for all three bulbs?
 
  • #13


thegreenlaser is correct. Look at a voltage divider. Bit of maths to do to ensure the voltage does not go out of tollerance with one or three bulbs connected.
 
  • #14


thegreenlaser said:
If I understand you correctly, you just need to know how to reduce the voltage using two resistors? The simplest solution would be a voltage divider.
That is certainly the simplest solution! This makes for a good challenge question, too.
 

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