Design a circuit diagram

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  • Thread starter Cisneros778
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  • #1
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.
 

Answers and Replies

  • #2
NascentOxygen
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Cisneros778, what could be the idea behind allowing the use of resistors? http://imageshack.us/a/img706/5864/hidden14.gif [Broken]
 
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  • #3
Cisneros778
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The 2 resistors will keep all light bulbs within their allowed voltages.
 
  • #4
NascentOxygen
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The 2 resistors will keep all light bulbs within their allowed voltages.
"2 resistors"http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon5.gif [Broken] 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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  • #5
Cisneros778
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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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Cisneros778
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https://mail-attachment.googleusercontent.com/attachment/u/2/?ui=2&ik=c0da1462ed&view=att&th=139d7c80e1de131f&attid=0.1&disp=inline&realattid=f_h78jmhci0&safe=1&zw&saduie=AG9B_P_MKhT_3deMhgRG8lpbZ3yh&sadet=1347945000944&sads=cscxVT049sxfN8g2eBCSKNZUd3k&sadssc=1

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.
 
  • #7
Cisneros778
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any suggestions?
 
  • #8
NascentOxygen
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Are the five switches all SPST? :smile:
 
  • #9
Cisneros778
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Yes, just on or off
 
  • #10
thegreenlaser
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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
CWatters
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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
CWatters
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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
CWatters
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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
NascentOxygen
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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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