Resistance???

Discussion in 'General Electronics Discussion' started by Young, May 18, 2013.

1. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,501
2,841
Jan 21, 2010
Yeah, I tried. It doesn't work.

2. Young

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0
Feb 6, 2013
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

3. Young

47
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Feb 6, 2013
Type this on google,hyperphics voltage divider

4. komalbarun

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Nov 25, 2011
xxxxx

Last edited: May 19, 2013
5. komalbarun

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Nov 25, 2011
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 !

Last edited: May 19, 2013
6. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
OK, so you changed your post to use the same method that I suggested.

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.

Last edited: May 19, 2013
7. Young

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Feb 6, 2013
Last edited by a moderator: May 19, 2013
8. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
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.

9. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,501
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Jan 21, 2010
And try it yourself with

Vin = 12V
Z1 = 7 ohms
Z2 = 9999999 ohms
ZL = 5 ohms

What is VOut?

Z2 = 999999 ohms is a large number because using that calculator we can't say "I don't actually want a Z2"

10. komalbarun

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Nov 25, 2011
Steve is right.

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

Last edited: May 19, 2013
11. Young

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Feb 6, 2013
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

12. Young

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Feb 6, 2013
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

13. komalbarun

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Nov 25, 2011
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?

Last edited: May 19, 2013
14. Young

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Feb 6, 2013
Power rating 20watts

15. komalbarun

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Nov 25, 2011
and voltage rating is 5V?

16. Young

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Feb 6, 2013
Mistake its 12

17. komalbarun

67
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Nov 25, 2011
Are you using a bulb or an LED? btw the power rating should be mentionned on the bulb...

18. komalbarun

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Nov 25, 2011
So its 5V @ 12W?

19. komalbarun

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Nov 25, 2011
I would still prefer steve to confirm this 20. BobK

7,682
1,688
Jan 5, 2010
A resistor in series with the bulb IS a voltage divider. As Steve said, adding another resistor does nothing but waste power.

Bob  