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Batteries and Ohm's Law

  • Thread starter imblum
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  • #1
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Hello, I am just starting to teach myself about electronics, and am already coming up with questions my text is not answering adequately. Unfortunately I don't know anyone who I could ask as I'm not in school.

Consider two 12V batteries, say a car battery and a garage door opener battery.

If you were to connect the terminals of the car battery, you could be killed, though of course if you connected the terminals of your small 12V you would not. Are they leaving a battery rating out somewhere that would tell you the big one is more dangerous? I thought both batteries were just 12V.

Considering Ohm's Law: V=IR, as resistance (R) tends to zero, current (I) will tend toward voltage (V) (right?)
Would a 12V garage door opener battery really be able to put out 12V at 12A with no load?

It also seems that the maximum power: P=IV you could get out of a 12V battery with practically zero resistance (regardless of size) would be just under 144W. Wouldn't this be a a lot for that little battery?

I feel like I'm missing some central stuff. I hope I'm not embarrassing myself by asking silly questions.

thanks,
matt
 
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Answers and Replies

  • #2
krab
Science Advisor
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imblum said:
If you were to connect the terminals of the car battery, you could be killed,...
Huh? Where did you get that notion?

Considering Ohm's Law: V=IR, as resistance (R) tends to zero, current (I) will tend toward voltage (V) (right?)
Also wrong. For given I, if R tends to zero, so does V. Anything multiplied by zero is zero. So V=IR=I times 0 = 0. Generally, though if you have something as large as a car battery, V will remain constant (more or less). Then V=IR can only be satisfied by I going to infinity. So if you drop a crescent wrench on the battery terminals, the current can get large enough that it will burn your hand. It won't kill you though.

Would a 12V garage door opener battery really be able to put out 12V at 12A with no load?
You only get 12A if your resistance is 1 ohm. That's the load. Somehow you seem to be ADDING rather than MULTIPLYING I and R.

It also seems that the maximum power: P=IV you could get out of a 12V battery with practically zero resistance (regardless of size) would be just under 144W. Wouldn't this be a a lot for that little battery?
Again, you get 144W if you are using a 1 ohm load. Indeed, the little battery wouldn't be able to source 12A. What will happen is that the 1 ohm load will fry the battery, bringing its output voltage to near zero. OTOH, a car battery can source 12A with ease. That's typically the amount of current needed to operate 2 headlights. And 2 headlights (in parallel) is just around 1 ohm.
 
  • #3
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thanks

thank you for taking my question. Let me see if I understand.

When you connect the terminals of any battery (R->0), in order to try and maintain whatever voltage, the current becomes unrestricted and tries to go to infinity (A short circuit?). This causes the voltage, and in turn the amperage to fall until the battery dies (like normal wear only faster). It just happens that the car battery can "source" enough amperage to produce the resulting wattage in the form of heat and sparks?
 
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  • #4
mukundpa
Homework Helper
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I think there is a role of internal resistance of the battery as well.
 

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