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Resistance???

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Yeah, I tried. It doesn't work.
 

Young

Feb 6, 2013
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The resistance of the load is 5ohm,R1 is 3.7ohm,R2 is 5ohm,try it an you will still get 5ohm,this is the link "hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html" thanks
 

komalbarun

Nov 25, 2011
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Young -- tell me, what resistor values did you use for your voltage divider and how did you model the lamp in the simulator?

I think it's clear that you simply don't realise that once you connect any load to the voltage divider the voltage will fall (and that was essentially the point of the first response I made in this thread).

komalbarun is a false prophet.

I try to help where I can .. and dude..am also here to learn...and I also edited my post where I have used the ratio approach, pls check it before insulting people..it's because of people like you that people like me don't get discouraged at learning electronics !
 
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(*steve*)

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I try to help where I can .. and dude..am also here to learn...and I also edited my post where I have used the ratio approach, pls check it before insulting people..it's because of s**ts like you that people like me don't get good at electronics!

OK, so you changed your post to use the same method that I suggested.

Unfortunately, nobody gets notified of your changes. I certainly don't re-read every post before I reply.

Far better to post another post with corrections if there are posts following the one you want to retract.

edit: Actually, I'll retract that. Your edited post is still wrong if the bulb draws any appreciable power. Either that or you need to use very low values resistors which will eat up an unnecessary amount of power.
 
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(*steve*)

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Let me explain again, for the benefit of Young and komalbarun...

If you have a voltage divider let's say the 7 ohms and 5 ohms suggested by komakbarun.

If you put 12 volts across it, you will get 5 volts across the 5 ohm resistor.

Now let's put a load on it. Let's assume our load is 5 ohms (that would be similar to a 5V 5W lamp).

The voltage the load sees is now 12* ((5||5)/((5||5) + 7)

5||5 is 2.5 ohms so

V = 12*(2.5/(2.5+7)) = 12* (2.5/9.5) = 12* (0.2632) = 3.158 volts.

Clearly it's not 5V.

My solution would be to put a 7 ohm resistor in series with the bulb, and the voltage across the bulb would be...

5V

But apparently it's far too complex to get this right...

Now, my solution depends on Young answering the question of the lamp rating in Watts, or the current in Amps, which he has not, despite being asked more than once.

Now, it's also possible to allow for the resistor in parallel with the bulb.

You could have a resistor of 3.5 ohms and a resistor of 5 ohms.

Unloaded, this would produce a voltage of 7.06V, but with the bulb in place you would again get 5V.

However this is simply wasting the power flowing through the 5 ohm resistor.
 

(*steve*)

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komalbarun

Nov 25, 2011
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Steve is right.

A resistor in series to the bulb is much better than a V-divider.
 
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Young

Feb 6, 2013
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Steave calm down when ever you input a number into the calculator it loads number into other blank boxes dont care about that all i want you to do is make sure z1,z2,zl,and v input is typed before you conclude.the only box you should leave alone are the v unounted and vout and you will get 4.8v i think thats enought note z1=3.7ohm(not 7ohm),z2=5ohm,zl(load resistance)=5ohm the vout(voltage ouput even after the load resistance)=4.8v,v unmounted is vout put before appling a load resistance
 

Young

Feb 6, 2013
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Steve is right.

A resistor in series to the bulb is much better than a V-divider.

i didnt say its oot a better way but his calculation was wrong i used 39ohm as he calculate with a voltage of 12v i got 8v this may even damage my load thats why i went with your cos i got 4.8-5v even with the load resistance,all i did was asume the load resistance was another resistance and i inputed it to d voltage divider calculator and found out i needed r1=3.7,r2=5,load resistance=5 and i got 4.8-5v try it and see
 

komalbarun

Nov 25, 2011
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ok got it so its like this :

Rp = total R of R2 and RL in parallel...
Rp = r2xRL / r2+rL = 25 / 10 = 2.5

Rtotal= R1+Rp = 3.7 + 2.5 = 6.2

VRp = (2.5/6.2) * 12
= (approximately) 0.4 x 12
=(approximately) 4.8V

But Since VRp = I x Rp;

I = VRp / Rp
I = 4.8 / 2.5 = 1.92 A

Power Dissip. in load will be(I think) :

PL = IV = 1.92 X 4.8 = 9.2W !

Does your bulb withstand a power dissip. of more than 9.2Watts?
 
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komalbarun

Nov 25, 2011
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ok got it so its like this :

Rp = total R of R2 and RL in parallel...
Rp = r2xRL / r2+rL = 25 / 10 = 2.5

Rtotal= R1+Rp = 3.7 + 2.5 = 6.2

VRp = (2.5/6.2) * 12
= (approximately) 0.4 x 12
=(approximately) 4.8V

But Since VRp = I x Rp;

I = VRp / Rp
I = 4.8 / 2.5 = 1.92 A

Power Dissip. in load will be(I think) :

PL = IV = 1.92 X 4.8 = 9.2W !

Does your bulb withstand a power dissip. of more than 9.2Watts?

I would still prefer steve to confirm this ;)
 

BobK

Jan 5, 2010
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Steve is right.

A resistor in series to the bulb is much better than a V-divider.
A resistor in series with the bulb IS a voltage divider. As Steve said, adding another resistor does nothing but waste power.

Bob
 
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