What is the current if the battery is shorted momentarily?

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Homework Help Overview

The discussion revolves around the scenario of a 12V car battery being shorted momentarily, with a focus on understanding the implications of internal resistance and current flow in such a situation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the concept of shorting a battery and its consequences, questioning the current that would flow given the internal resistance. Some participants attempt to apply Ohm's law to calculate the current, while others express confusion about the terminology and implications of shorting a battery.

Discussion Status

There is active engagement with the topic, with participants sharing their understanding and calculations. Some guidance is offered regarding the dangers of shorting a battery and the relationship between voltage, resistance, and current. Multiple interpretations of the term "short" are being explored.

Contextual Notes

Participants note the potential hazards associated with shorting a car battery, including the risk of explosion, and discuss the implications of internal resistance in the context of circuit behavior.

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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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