DC electrical question...

[Blacksmith Cycles]

I know this may be slightly off topic for you guys, but it does have to do the basic concept of Ohm's law so I hope you can help me out. I asked someone I know who is a licensed electrician and he referred me to this site.

In a 12VDC circuit with a battery as the power source, a resistor connected in series will eventually drain the battery. Common sense tells me that increasing the resistance will drain the battery faster. In other words a 6 ohm resistor will drain the battery faster than a 3 ohm resistor.

Here's my question...if I use Ohm's law (current = voltage/resistance) then:
12 volts / 3 ohms = 4 amps
12 volts / 6 ohms = 2 amps
Wouldn't a current flow of 4 amps drain the battery faster than a current flow of 2 amps? If so then will increasing the load (resistance) actually drain the battery slower?

Which way of considering this is correct and why?

I know ther is a basic key concept here that i am missing and will probably say "Duh!" when it is pointed out for me.

Thanks in advance for any help!!

 

[iwire]

I Common sense tells me that increasing the resistance will drain the battery faster.

Generally we cannot help DIYs but it does not seem like you are trying to wire a building so I think we can help you out.

At 0 ohms resistance (if that was possible) you would have a direct short and all the current the circuit could delivery would be used.

At say, 100,000 ohms resistance, very little current could flow.

You are just looking at backward.

 

[qmavam]

In a 12VDC circuit with a battery as the power source, a resistor connected in series will eventually drain the battery. Common sense tells me that increasing the resistance will drain the battery faster.

No, increasing the current will drain the battery faster. To increase the current you need to reduce the resistance.

In other words a 6 ohm resistor will drain the battery faster than a 3 ohm resistor.

About now you should be figuring out, that is backwards.

Here's my question...if I use Ohm's law (current = voltage/resistance) then:
12 volts / 3 ohms = 4 amps
12 volts / 6 ohms = 2 amps
Wouldn't a current flow of 4 amps drain the battery faster than a current
flow of 2 amps?

Your math is correct and yes, 4 amps will drain a battery faster than 2 amps.
Mike [/quote]

 

[winnie]

Just to tweak your intuition back into the correct direction, Ohms measure the ability of something to _impede_ current flow. The higher the resistance, the greater the current flow is blocked. The bigger the number, the slower the battery is drained.

There is a different measure, called 'conductance' (the unit of measure used to be called the mho, but some standards group changed the name of the unit to 'Siemen'). The higher the conductance, easier it is for current to flow through the device, and the faster the battery will drain.

-Jon

 

[LarryFine]

BC, welcome to the forum!

Essentially, Ohm's Law states that one volt is capable of pushing one amp through one ohm. Increasing the voltage, or decreasing the resistance, will result in greater current, and vice versa.

 

[ray cyr]

I know this may be slightly off topic for you guys, but it does have to do the basic concept of Ohm's law so I hope you can help me out. I asked someone I know who is a licensed electrician and he referred me to this site.

In a 12VDC circuit with a battery as the power source, a resistor connected in series will eventually drain the battery. Common sense tells me that increasing the resistance will drain the battery faster. In other words a 6 ohm resistor will drain the battery faster than a 3 ohm resistor.

Here's my question...if I use Ohm's law (current = voltage/resistance) then:
12 volts / 3 ohms = 4 amps
12 volts / 6 ohms = 2 amps
Wouldn't a current flow of 4 amps drain the battery faster than a current flow of 2 amps? If so then will increasing the load (resistance) actually drain the battery slower?

Which way of considering this is correct and why?

I know ther is a basic key concept here that i am missing and will probably say "Duh!" when it is pointed out for me.

Thanks in advance for any help!!

Another thing to consider here is terminology. You asked "If so then will increasing the load (resistance) actually drain the battery slower?"

Load and resistance are two different things. Usually the load on the circuit refers to the watts used by the circuit. Since 'watts equals volts times amps' the load increases as the resistance decreases.

 

[charlie b]

What constitutes a "drain" on the battery is the use of energy. A battery has stored energy in some chemical form or other (depending on the type of battery). So the faster you convert its chemical energy to electrical energy, the sooner it will die.

"Power" is a measure of the rate at which energy is being used. One formula for power in a DC circuit is that power equals the square of voltage divided by the resistance. Thus, a smaller resistance means more power, which in turn means a faster use of the energy stored in the battery.