Can someone explain reducer 'amps' to me?

Dela

Threw a new reducer on the new cart - went with a cheap 48 to 12v 10 amp just to get started - instantly got a buzzing sound on my speakers with this reducer - assuming it's the reducer as it didn't happen with my 30 amp reducer from old cart.

Just ordered a new 30 amp reducer to try to solve the buzz but am curious - what exactly are amps and what is the difference between 10 and 30 amps?

First to answer '20 amps' gets detention

BGW

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hyper1

amps is like pressure in a water hose, The water is the voltage (12v) and the amps is the pressure at which it is delivered. Hopes this helps.

hyper1

your noise is coming from the current. You may need to use a filter if you don't want to change reducer's.

bubba4dad

The difference in the 10 amp vs the 30 amps is just the amount of load it can supply.

Example: Power (Watts) is equal to Current (Amps) x Voltage

So if you have a 12 volt, 10 amp supply you can only power up 120 watts.

If you have a 12 volt, 30 amp supply you can power up 360 watts.

As for the buzzing sound, I am sure its probably because it is just a cheap converter.

bubba4dad

This is actually an incorrect comparison.

If you want to compare to water in a hose:

The "pressure" would be the voltage and the Flow rate (GPM) would be the Amps.

E36Racer

Try grounding the amp to the main pack negative and see if the noise goes away. The negative output of the reducer should be in the same circuit as the main pack.

I had a similar issue when I first installed my lithium pack because I just hooked everything up at first so I could take it for a spin and then went back and cleaned up the wiring. My cart was down for a couple weeks building the pack so I was itching to use it again... When I cleaned up the wiring I added a buss bar to connect all of the grounds together (pack, charger, reducer and fuse panel ground block). Wala, turned the radio on and noise was gone.


Sent from my iPhone using Tapatalk

hyper1

Yea I was trying to explain it as simple to enable someone to get the idea without all the technical stuff. You are correct so here is how it works.

A neat analogy to help understand these terms is a system of plumbing pipes. The voltage is equivalent to the water pressure, the current is equivalent to the flow rate, and the resistance is like the pipe size.
There is a basic equation in electrical engineering that states how the three terms relate. It says that the current is equal to the voltage divided by the resistance or I = V/R. This is known as Ohm's law.
Let's see how this relation applies to the plumbing system. Let's say you have a tank of pressurized water connected to a hose that you are using to water the garden.
What happens if you increase the pressure in the tank? You probably can guess that this makes more water come out of the hose. The same is true of an electrical system: Increasing the voltage will make more current flow.
Let's say you increase the diameter of the hose and all of the fittings to the tank. You probably guessed that this also makes more water come out of the hose. This is like decreasing the resistance in an electrical system, which increases the current flow.
Electrical power is measured in watts. In an electrical system power (P) is equal to the voltage multiplied by the current.
The water analogy still applies. Take a hose and point it at a waterwheel like the ones that were used to turn grinding stones in watermills. You can increase the power generated by the waterwheel in two ways. If you increase the pressure of the water coming out of the hose, it hits the waterwheel with a lot more force and the wheel turns faster, generating more power. If you increase the flow rate, the waterwheel turns faster because of the weight of the extra water hitting it.
A neat analogy to help understand these terms is a system of plumbing pipes. The voltage is equivalent to the water pressure, the current is equivalent to the flow rate, and the resistance is like the pipe size.
There is a basic equation in electrical engineering that states how the three terms relate. It says that the current is equal to the voltage divided by the resistance or I = V/R. This is known as Ohm's law.
Let's see how this relation applies to the plumbing system. Let's say you have a tank of pressurized water connected to a hose that you are using to water the garden.
What happens if you increase the pressure in the tank? You probably can guess that this makes more water come out of the hose. The same is true of an electrical system: Increasing the voltage will make more current flow.
Let's say you increase the diameter of the hose and all of the fittings to the tank. You probably guessed that this also makes more water come out of the hose. This is like decreasing the resistance in an electrical system, which increases the current flow.
Electrical power is measured in watts. In an electrical system power (P) is equal to the voltage multiplied by the current.
The water analogy still applies. Take a hose and point it at a waterwheel like the ones that were used to turn grinding stones in watermills. You can increase the power generated by the waterwheel in two ways. If you increase the pressure of the water coming out of the hose, it hits the waterwheel with a lot more force and the wheel turns faster, generating more power. If you increase the flow rate, the waterwheel turns faster because of the weight of the extra water hitting it.
In an electrical system, increasing either the current or the voltage will result in higher power. Let's say you have a system with a 6-volt light bulb hooked up to a 6-volt battery. The power output of the light bulb is 100 watts. Using the equation I = P/V, we can calculate how much current in amps would be required to get 100 watts out of this 6-volt bulb.
You know that P = 100 W, and V = 6 V. So, you can rearrange the equation to solve for I and substitute in the numbers.
In an electrical system, increasing either the current or the voltage will result in higher power. Let's say you have a system with a 6-volt light bulb hooked up to a 6-volt battery. The power output of the light bulb is 100 watts. Using the equation I = P/V, we can calculate how much current in amps would be required to get 100 watts out of this 6-volt bulb.
You know that P = 100 W, and V = 6 V. So, you can rearrange the equation to solve for I and substitute in the numbers.
I = 100 W/6 V = 16.67 amps
What would happen if you use a 12-volt battery and a 12-volt light bulb to get 100 watts of power?
I = 100 W/12 V = 8.33 amps
So, this latter system produces the same power, but with half the current. There is an advantage that comes from using less current to make the same amount of power. The resistance in electrical wires consumes power, and the power consumed increases as the current going through the wires increases. You can see how this happens by doing a little rearranging of the two equations. What you need is an equation for power in terms of resistance and current. Let's rearrange the first equation:
I = V/R can be restated as V = I*R
Now you can substitute the equation for V into the other equation:
P = V*I substituting for V we get P = I*R*I, or P = I2*R
What this equation tells you is that the power consumed by the wires increases if the resistance of the wires increases (for instance, if the wires get smaller or are made of a less conductive material). But it increases dramatically if the current going through the wires increases. So, using a higher voltage to reduce the current can make electrical systems more efficient. The efficiency of electric motors also improves at higher voltages.
This improvement in efficiency is what drove the automobile industry to consider switching from 12-volt electrical systems to 42-volt systems in the 1990s. As more cars shipped with electric-powered amenities — video displays, seat heaters, "smart" climate control — they required thick bundles of wiring to supply enough current. Switching to a higher-voltage system would provide more power with thinner-gauge wiring.
The switch never happened, because carmakers were able to boost efficiencies with digital technology and more efficient electric pumps at 12 volts. But some new models employ hybrid systems with a separate 48-volt generator to power advanced features like idle shut-off while increasing overall system efficiency.

Dela

Thanks guys - helpful info here.

I have a new reducer on the way - hope it solves the buzzing noise.

Hyper1 - my brain just exploded reading that

bubba4dad

Funny how googling Amps can make you and electrical engineer....lol

https://science.howstuffworks.com/en...uestion501.htm

radioman

I think you got this amps voltage analogy bass ackwards.