Newbie needs help with basics
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Posted by John Fields on March 18, 2006, 11:12 am
Please log in for more thread options wrote: --- LOL, yup, you're right! I also missed the bit on the bottom of the page where the bulb makes reference to being burned out. Thanks, -- John Fields Professional Circuit Designer | |||||||||||||||||||||||||||||||||||||||||||||||||||||||
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Posted by Dorian McIntire on March 16, 2006, 7:25 pm
Please log in for more thread options "hdjim69" Wrote: > Hi, I'm trying to learn electronics, I'm right at the beginning,
> just learned Ohms law and Watts law. I understand both. I understand > how to calculate both. However, I went to this tutorial on > Electronicsworkbench.com and I'm confused about how they come up with > these numbers. > > In this example there is a series circuit with a 15v battery, a 10w/12v > light bulb with a voltmeter reading the voltage across the bulb and you > have to choose between 3 different resistors that will enable the bulb > to light - 1ohm, 100ohm and 5ohm. If you connect the 1ohm, the > voltmeter reads 15volts. How can that be ? In a series circuit each > load has some resistance and the sum of all the loads must equal the > source. So, what about the resistor ? It has to drop some voltage but > according to this, it's dropping zero. And if you connect the 100 > ohm resistor, the voltmeter reads 1.88v. How did they come up with > that number ? And the 5ohm reads 12v across the voltmeter. But > according to Ohms law, E = IxR, if the resistor is 5ohms and the source > is 15v, then 15/5 = 3amps, that part I get but they are saying it's > dropping 3 volts ?? huh? Again, how did they get that value ? > Consider: 10 W (Watts) / 12 V (Volts) = 0.833 A (Amps) required by the bulb to light without burning out or burning to dimly. The applied voltage is 15 V so 3 V must be dropped by the current limiting resistor to light the lamp properly. 3 V / 0.833 A = 3.6 Ohms required to drop 3 V at the required 0.833 A. The closet resistor to this value is the 5 Ohm resistor OR 3 V / 1 Ohm = 3 A - too high 3 V/ 100 Ohms = 30 MA - too low 3 V / 5 Ohms = .6 A - about right | |||||||||||||||||||||||||||||||||||||||||||||||||||||||
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Posted by redbelly on March 17, 2006, 8:12 am
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Dorian McIntire wrote: > 3 V / 0.833 A = 3.6 Ohms required to drop 3 V at the required 0.833 A.
> > > > The closet resistor to this value is the 5 Ohm resistor The real question is, why don't they just provide the correct 3.6 ohms as one of the choices? Mark | |||||||||||||||||||||||||||||||||||||||||||||||||||||||
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Posted by hdjim69 on March 17, 2006, 11:15 am
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So to summarize what I have learned: 1) Find total resistance in curcuit 1a) use formula R=E^2/P if R is unknown but P & E are known 2) Find (total) current in circuit by using RT and VT 3) Find voltage drop of each load and ensure it's adequate for each load Excellent ! I learned a lot from this post !! One small question...why do we square the voltage in the formula R=E^2/P ? Why can't we just use the actual voltage ? TIA | |||||||||||||||||||||||||||||||||||||||||||||||||||||||
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Posted by redbelly on March 17, 2006, 11:53 am
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hdjim69 wrote: > One small question...why do we square the voltage in the formula
> R=E^2/P ? Why can't we just use the actual voltage ? > > TIA Think of these two equations: P = E*I and I = E/R Let's assume you believe those. Combine them and you get P = E * E/R or E^2 / R So we have to use E^2, not just E. Mark | |||||||||||||||||||||||||||||||||||||||||||||||||||||||
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>They show it burned out. Ed