What is the current if the battery is shorted momentarily?

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    Battery Current
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SUMMARY

When a 12V car battery with an internal resistance of 0.4 ohms is shorted momentarily, the resulting current can be calculated using Ohm's Law. The formula I = V/R yields a current of 30 A (12V / 0.4 ohms). Shorting a car battery poses significant risks, including potential explosion due to the high current flow through minimal resistance. Understanding the implications of short circuits is crucial for safety and proper handling of batteries.

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  • Ohm's Law
  • Understanding of electrical resistance
  • Basic knowledge of battery specifications
  • Familiarity with short circuits
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  • Research the safety protocols for handling car batteries
  • Learn about the effects of internal resistance in batteries
  • Study the principles of short circuits and their consequences
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Electrical engineers, automotive technicians, and anyone interested in battery safety and electrical circuit behavior will benefit from this discussion.

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If a 12V car battery has an internal resistance of .4 ohms, what is the current if the battery is shorted momentarily?

Maybe this is stupid, but if the battery is shorted, it shouldn't be delivering any current, right?
 
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I hear that one should not try to short a car battery (connect the + and - terminals directly with a wire of negligible resistance) because they have a tendency to explode. This is because in circuit loop of such paltry resistance, the current drawn will be HUGE. Think about it...the battery supplies an EMF of 12V. There has to be a corresponding (and equal) potential drop around the circuit. The only resistance available to provide such a voltage drop is the internal resistance of the battery, because the potential drop is given by V = RI. So 12V = RI ==> I = 12V/.4Ohm = 30 A. Did I get this right?
 
cepheid said:
I hear that one should not try to short a car battery (connect the + and - terminals directly with a wire of negligible resistance) because they have a tendency to explode. This is because in circuit loop of such paltry resistance, the current drawn will be HUGE. Think about it...the battery supplies an EMF of 12V. There has to be a corresponding (and equal) potential drop around the circuit. The only resistance available to provide such a voltage drop is the internal resistance of the battery, because the potential drop is given by V = RI. So 12V = RI ==> I = 12V/.4Ohm = 30 A. Did I get this right?

That makes sense... I guess I didn't know what it meant to "short" a battery. I thought it meant it went dead.
 
Yeah...same thing it means as in a "short circuit". The current takes the path of least resistance, and if the that path is not the one you intended (i.e., there is a 'short' or unintended conducting pathway somewhere), then the increased current means bad news.
 

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