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### Network # Resistance???

#### Young

Feb 6, 2013
47
How can i calculate the resistance required to get a specific voltage from another voltage,...for example how much resistance do i need to get 5v from a 12v.....,

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Resistors can't do that (well, they can, but possibly not in a useful way).

What are you actually trying to do?

#### Young

Feb 6, 2013
47
I want to have a power supply of 12v(dc) and i want to break it down to 5v(dc) so that i can use it to power a 5v(dc) bulb

#### komalbarun

Nov 25, 2011
67
google for "resistor as potential/voltage divider"

What is your bulb's power rating?

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#### Young

Feb 6, 2013
47
Komalbarun thank you very much God bless you,you are blessed...imediately i typed it on google i got the exact thing i was looking for ..an online calculator,all i need to do is to type an input voltage,an output voltage and 1 of the resistor value it immediately gives me the second resistor required,i dont beleive i have been using this same voltage divider calculator before but i didnt know i could use it to calculate the resistance i thought all i could do is to type the input voltage,R1,R2 and then get the output voltage all i was doing before was guesing R1 and R2 with a known input voltage to get an output voltage not until today when God used you..thank komalbarun God bless.,am so happy

#### komalbarun

Nov 25, 2011
67
Thx btw you can also use ratio:

2 resistors of ratio 7:5 will divide the voltage 7V : 5V.

Vdc = 12 V

Resistors are of ratio 7: 5.

R 1 = 7 ohms
R 2 = 5 ohms

Voltage drop on R2 = 5 ohms/ (7+5) ohms * 12V = 5 volts

Now if you need a specific current flow (say 5mA?).

Since V = IR
R = 12/0.005
R = 2400 ohms

Applying ratio:

R2 / 2400 * 12 v = 5 v (or any required V-drop)
R2 * 12 = 12000
R2 = 12000 / 12
R2 = 1000 ohms

Hence R1 = 1400 and R2 = 1000

Check it out in the online calculator But ! normally if you are connecting the bulb in parallel to R2 then since R2 an Rbulb will be in parallel, V-drop will be much less here - That's what I think but pls heed the advice of steve and bobK).

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#### BobK

Jan 5, 2010
7,682
Komalbarun thank you very much God bless you,you are blessed...imediately i typed it on google i got the exact thing i was looking for ..an online calculator,all i need to do is to type an input voltage,an output voltage and 1 of the resistor value it immediately gives me the second resistor required,i dont beleive i have been using this same voltage divider calculator before but i didnt know i could use it to calculate the resistance i thought all i could do is to type the input voltage,R1,R2 and then get the output voltage all i was doing before was guesing R1 and R2 with a known input voltage to get an output voltage not until today when God used you..thank komalbarun God bless.,am so happy
Not really. As soon as you connect a bulb to the circuit, it no longer holds the same voltage because you have changed one of the resistors.

What you need to make a 5V bulb run of 12V is, first to know the current the bulb takes at 5V. After you know that, you use a single resistor in series that will drop 7V at the current the bulb takes. The bulb itself is one of the resistors in the voltage divider.

For example, if the bulb takes 100mA:

V = I R
(12-5) = 0.1 R
7 = 0.1 R
70 = R

You also need to size the resistor for power. The power is:

P = V I

In the example:

P = 7 * .01
P = 0.7

So a 1 Watt resistor will work. And it will get hot.

Bob

#### Young

Feb 6, 2013
47
Thank you very much bobk and komalbarun,bobk i will take note of what you said,before i forget komalbarun where did you learn how to use ohms law like that..please tell i can see you are a talented ohmslawer

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
edit: I really should refresh the page before I spend so much time repeating almost exactly what BobK had typed several hours previously...
edit2: Actually, I did add a bit more.

Komalbarun thank you very much God bless you,you are blessed..

All true..

Well it would have been if the advice were good (and it's not).

For powering a bulb you could use a simple series resistor. Beware that for a 5V globe running from 12V, the resistor will dissipate more power than the bulb. Assuming the bulb is a low power one this may not be a problem.

If you know the bulb's power or current requirements you can calculate the resistor you'd need.

Le's assume it's 1W.

The current is determined by P = V * I, so I = P/V = 1/5 = 0.2A

If you already know the current, you could skip that step.

Given a current of 0.2A, a bulb voltage of 5V and a power supply voltage of 12V, you want a resistor which will drop (12-5) = 7V at 0.2A.

Using V = I * R and rearranging to give us R = V/I, we calculate the resistor as 7/0.2 = 35 ohms.

You can't get a 35 ohm resistor, but you could use the next highest easily available value, 39 ohms.

Now we need to calculate the power dissipated in this resistor. This is given by P = I^2 * R (you can derive that from the two formulae given above). This si 0.2 * 0.2 * 39 = 1.56W

Because you want to leave some margin for safety (and so you don't burn your finger if you touch the resistor), you would select one with a power rating of at least 2W, but probably higher would be better, even a 5W one would not be too far over the top.

So, get a 39 ohm 5W resistor.

But remember, you'll need to repeat these calculations for the actual power or current requirement of your bulb.

Another option is to use a 5V voltage regulator. This is a little bit more expensive, but will allow you to run a load that varies from nothing to about 1A with no problems with changing voltage. This will require a 7805 regulator, preferably some small capacitors and almost certainly a heatsink.

If you really only need to run a single bulb, a resistor may be a viable alternative.

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#### komalbarun

Nov 25, 2011
67
Thank you very much bobk and komalbarun,bobk i will take note of what you said,before i forget komalbarun where did you learn how to use ohms law like that..please tell i can see you are a talented ohmslawer

I listened to my teacher in class - and did some research on my own. Seriously ( I still do need to learn a bit more on it though).

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#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
I listened to my teacher in class - and did some research on my own. Seriously ( I still do need to learn a bit more it though).

Essentially, your answer would have been correct if you had considered the bulb as one of the resistors in the voltage divider.

This works because the load (the bulb) can be considered a resistor at its operating point.

To use a voltage divider is to place a resistor in parallel with the bulb. This is rarely a good idea because the best it's going to do is waste power.

#### Young

Feb 6, 2013
47
Steve please can you tell me the effect of a low power

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
I don't understand your question.

#### Young

Feb 6, 2013
47
I mean if a load requires 2v for example,and a 2amp that is 4watts of power and you give it 2v at 0.5amp making a power of 1watt what will be the effect,because when ever i see a load rated for example 6v and 0.5amp,all i consider is the 6v i dont even care about the current what is the effect of this?

#### Young

Feb 6, 2013
47
Steave am going to try it on my circuit simulator now them am going to give you the result

Last edited:

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
I just got a glimpse of your post before you edited it. All I can say is, good luck. (Because you're going to need it)

If you have a load that requires 2A at 2V, you simply *can't* give it 2V at 0.5A. At 2V it will draw 2A, and at 0.5A it will drop a lower voltage (0.5V -- assuming it is a resistive load).

My method *specifically* uses ohms law. The voltage divider approach mis-uses it.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Edit, oh, and by the way 0.07k is not 7 ohms, it is 70 ohms.

Make your life easier by doing all calculations in Volts, Ohms, Amps and Watts. Do the conversion later.

#### Young

Feb 6, 2013
47
Steave your calculation was wrong i tried it on my circuit simulator and i didnt get anything near 5v using 39ohm what i got was more than 5v i got 8v or 12 becos i wasnt sure of the conversion of ohm to kilo ohm so i used 0.039kohm and 39000kohm..komalbarun was right i use his and i got 5.01v using a voltage divider the online voltage divider i used already has a place to fill the load resistance and it calculates all of it together,thank you once again komalbarun God bless you,you are a saviour,this is the link of the voltjage divider "hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html" thanks

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
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.

#### Young

Feb 6, 2013
47
Check the link of the voltage divider please

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