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Try reading this and see where you made your mistake.

http://sound.westhost.com/articles/meters.htm

Roy

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Hi ace,This I did with 5 wtt resistors and hooked the meter to the pack ( my DVM read 152v) and to my delight the 12v meter hung at about 16…… for about 5 seconds than poof!!! Up went the resistors. Fried to a crisp. What gives! Any ideas?

My ohmslawmath says about 2.9 watts in the resistor. Maybe a bad resistor??? Was it rated for that high a voltage?

major

Actually about 3.7W (and about 0.66W in the meter resistance) since there is about 102V across the 2800 Ohm resistor due to resistive voltage division with the 500 Ohm in the meter. Why did your DVM read 152V if it is a 120V pack? A voltage of 152V would give about 5.9W dissipation in the resistor explaining why it blew up. Sounds like you need larger resistance if the pack will be at 152V.0.036Amp * 120V = 4.32 Watts.

Voltage across external R = external R*pack voltage/(external R + meter R)

Power dissipated in external R = voltage across external R squared/external R

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O.K. Maybe we can get this straight. He has a 500Ω meter, 18V full scale. Meaning movement needs 18V / 500Ω = 0.036A at full scale deflection. He then put a 2800Ω resistor in series with the meter, connects to a battery and saw 16V on the meter. So the meter current was 16/18 * 0.036A = 0.032A. The power in the resistor is I²R = 0.032A² * 2800Ω = 2.8672W.Actually about 3.7W.....

Is that not correct

edit: see post #11

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+-1.0 kohms--1.80 kohms-----meter---500 ohms

_ ------------------------------------------

I think i have left over 5 wtt resistors and I'll try a parallel/series arangement like this.

+1.0 k----1.8k--------meter/500 ohm.

+1.0 k----1.8k

- -----------------------------------

hopefully this will split the load? I'll let you know. thanks again.

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the system was being charger at the timeActually about 3.7W (and about 0.66W in the meter resistance) since there is about 102V across the 2800 Ohm resistor due to resistive voltage division with the 500 Ohm in the meter. Why did your DVM read 152V if it is a 120V pack? A voltage of 152V would give about 5.9W dissipation in the resistor explaining why it blew up. Sounds like you need larger resistance if the pack will be at 152V.

Voltage across external R = external R*pack voltage/(external R + meter R)

Power dissipated in external R = voltage across external R squared/external R

---[-----2.8K----2.8K-----]-------500ohm meter-------

[-----2.8K----2.8K-----]

this would allow you to use 5 watt resistors.

Or go to Radio Shack and get 3 packs of 33K ohm ½ watt resistors and connect 12 of them in parallel this gives you

----2.75K-----------------------500ohm meter-----

It is 10% off but pretty close for being able to buy the parts locally.

Put a 500mA fuse on that to save your meter.

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O.K. This bugs me. Something isn't right here. If he had 152V with a load of 3300Ω (2800Ω in series with a 500Ω meter) then the current would be 0.046A. And 5.94W in the resistor. But that would mean that the meter was seeing 23V, not about 16V as he said.O.K. Maybe we can get this straight. He has a 500Ω meter, 18V full scale. Meaning movement needs 18V / 500Ω = 0.036A at full scale deflection. He then put a 2800Ω resistor in series with the meter, connects to a battery and saw 16V on the meter. So the meter current was 16/18 * 0.036A = 0.032A. The power in the resistor is I²R = 0.032A² * 2800Ω = 2.8672W.

Is that not correct

I'm guessing this is the case and the resistor failed because 5.94W is above the 5W rating. Maybe the meter was pegged and bounced back to 16

It looks like you are going to have to just build the meter and see if it blows up. It is confusing because the meter goes up to 18V, and the meter apparently has a resistance of 500 ohms. This gives 18V/ 500 ohms = 0.036 Amps for full scale deflection. To achieve full scale deflection with 120V you would need a 2.8K resistance in series with the meters 500ohms this gives 120V/ 3300 ohms = 0.036Amp flowing through the meter circuit. If you want to measure 180V you would need 5K of resistance to achieve 180V/ 5000ohms = 0.036Amps, so 4.5k in series with the 500 ohms of the meter. It is a voltage divider 90% of the voltage is dropped a crossed the resistor 10% a crossed the meter. Sorry for the confusion, but even us monkeys figure it out if you get enough of us together.

Maybe through a picture of your meter up?

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