Ohms Law and Batteries Got me Confused

Thread Starter

kring123

Joined Feb 6, 2010
15
This may be a newbie question but I'm having a difficult time getting my brain around how much current batteries produce.

For example, Someone not skilled in electronics (like myself) might assume that a standard 9 volt battery would produce 9 amps of current. I'm assuming that there is no resistance to hold back the 9 volts so I = V/R would be 9. Is there some built in resistance that I need to account for?

Another example is when you connect 2 wires directly from a 9 volt battery you get a small spark and the battery slowly gets hot. However, when you connect 2 wires from a 7.2V battery pack you get a large spark and a VERY hot battery pack. Logic tells me that 9 V should produce more current than 7.2 V.

Also, why can you get a shock from a car battery but when you touch 2 terminals on a 12V drill battery you can't feel anything?

I'm obviously missing something here. Is current more a function of the size of the battery cell and not so much the voltage?

If someone can indulge me and help me understand some of these simple questions it would be appreciated!
 

t_n_k

Joined Mar 6, 2009
5,455
I would be surprised if anyone really received a shock from a 12V car battery. Perhaps you are thinking of the shock you get from an open auto spark plug or spark plug lead. That can hurt! It is possible under extremely low skin resistance conditions to get a shock from a 32V AC supply.

It is electrical current that causes sensation, pain or death in humans. The higher the voltage the easier it is to break down the initial skin resistance and thereby permit a higher current flow.

Check this link in Wikipedia

http://en.wikipedia.org/wiki/Electric_shock

With respect to the batteries shorted out and getting hot. It is the values of a battery's internal resistance and EMF that will determine the current flow via an externally applied short circuit. This will vary over the duration of the short being applied. The EMF will diminish as the electro-chemical capacity is depleted (discharged) and the internal resistance will increase in like manner. The heat generated within the battery under shorted condition will be the result of all these things coming into play. The power generated at any instant in the battery's internal resistance will give you some idea of the effective heat dissipation over time. The heat will be related to the Joule resistance heat loss which is proportional to the square of the current flow and the effective internal resistance. This heat will also effect the electro-chemical reaction.
 

Wendy

Joined Mar 24, 2008
23,415
Everything has resistance, even batteries. The internal resistance in batteries is what limits how much current they can provide. A small 9V battery has much greater resistance than a large 12V car battery, so the small battery is good for 500ma max while the car battery is good for 500A. As a battery drains the internal resistance goes up.

I've gotten mild shocks from 12V, when I was hot and sweaty. The sweat dramatically lowered my skin resistance.
 

Markd77

Joined Sep 7, 2009
2,806
For V=IR you are using R=1 ohm. If you connect the terminals directly R is much less than 1 ohm in the connection but as the previous replies have said the battery has internal resistance. It's very low in NiCad batteries which can produce several amps when shorted even though the voltage is only 1.2
 

mbohuntr

Joined Apr 6, 2009
446
A battery is like a bottle of water. The max current is more a factor of the opening size (internal resistance)and the reserve capacity is like the water inside.The max current is based on the internal resistance, not (1 ohm ) as the others have said. If you discharge them with a direct short, you usually damage the battery. (and maybe yourself.) Compare a "9v" with a "C" for mAH of reserve capacity,and internal resistance, and you get the picture. Search the e-book here as well as the older posts, and you will learn a lot.
 
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