# How to calculate a resistor

Discussion in 'Homework Help' started by mikelynch, Aug 7, 2009.

1. ### mikelynch Thread Starter Member

Jun 30, 2007
23
0
I have a 5V battery powering a lamp that requires 60mA , but when cold the lamp will try and draw 600mA. The lamp uses 3V+

I wanted to vfind out the resistance of the lamp -> (5 - 3)/.06 A = 33 Ohms

or when lamp is cold (5- 3)/.6 A = 3 Ohms

When I want to calulate the current flowing through the circuit I also did this
when the lamp was cold. 5V/3A = 1.6 Amps

Or (5 - 3)/3A = 0.6 Amps.

I dont know which way to get the current. Then I have to work out what resistor to use.

Sorry about the poor png, how do i save a picture so that it comes up clear. I have PAINT and Photoshop.

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Last edited: Aug 7, 2009

Jul 7, 2009
1,585
141
It's not clear exactly what your question is (actually, you didn't ask a question about the circuit). It sounds like you have a 3 V lamp that draws 60 mA. That means it has a resistance of 3/0.06 = 50 Ω. To run this lamp from a 5 V supply, you need to drop 2 V across a resistor in series with the lamp. Thus, the resistance needed is 2/0.06 = 33.3 Ω. The total resistance is 83.3 Ω and we'd better get 60 mA of current with that: 5/83.3 = 0.06 A. Thus, things check. A resistor with a power rating greater than 3*3/33.3 = 0.27 W is needed -- probably a half watt resistor.

If the lamp draws 0.6 A when cold, that means its cold resistance is 3/0.6 (you didn't specify the voltage across the lamp when this current is drawn, so I assumed it was the same as in the first paragraph). That means the resistance is 5 Ω, or 10 times less than the heated resistance (reasonable if the current was 10 times higher). If the start-up current is too large, you'll have to find a way other than a plain resistor to limit the current.

3. ### mikelynch Thread Starter Member

Jun 30, 2007
23
0

1. Why did you multiply by three to get Resistor Watts 3* 3 / 33.3 = 0.27 W

2. How do i calculate 1/2 Watt resistor into Ohms ? Or can resistors with the same Ohmns have totally different Wattage thresholds ? In either case can this be simply calculated ?

I guess the answer will be that a resistor of around 33 Ohms and 1/2 Watt is available ?

Thx again.

Mike.

Last edited: Aug 8, 2009
4. ### Wendy Moderator

Mar 24, 2008
20,766
2,536
I think it may be a typo, the equation is Watts = V²/R, which is another way of saying Watts = V*I (ohms law is in play here, the textbook equation is P=E*I). In this case the V would be 2V, the voltage drop of the resistor, so 2V²/33.3Ω = .12W. As a check figure V*I, or 2V*0.06A = 0.12W. A 33 Ω¼W resistor would suffice.

The logic was good, but brain farts can happen anywhere and anywhen to anyone, voice of experience.