MHB What Is the Temperature Coefficient of a Car Battery Resistor at 0 Degrees?

AI Thread Summary
The discussion revolves around calculating the temperature coefficient of a car battery resistor at 0 degrees Celsius, with a voltage drop from 12V to 7.2V while pulling 150 amps. Initial calculations provided by a participant indicate the need for both the internal and outer resistances of the battery. However, to determine the temperature coefficient accurately, a reference resistance at room temperature is required. One participant later confirms they solved the problem independently before seeing the responses. The conversation highlights the importance of additional data for precise calculations in electrical engineering contexts.
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Starting a car: the voltage drops from 12V to 7.2 V, it is 0 degrees out, and 150 amps are pulled from the car. What is the temperature coefficient of the resistor?

I have NO idea how to do this. Help please!
 
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Mango12 said:
Starting a car: the voltage drops from 12V to 7.2 V, it is 0 degrees out, and 150 amps are pulled from the car. What is the temperature coefficient of the resistor?

I have NO idea how to do this. Help please!

Hi Mango! ;)

We'll need more information.

From the information you provided, we can only tell that:
$$R_o = \frac{7.2 \text{ V}}{150\text{ A}} \\
R_i = \frac{4.8 \text{ V}}{150\text{ A}}
$$
where $R_i$ is the internal resistance of the battery, and $R_o$ is the outer resistor.

We would need for instance a reference resistance at room temperature of the battery to say anything about temperature coefficient.
Is any such given? (Wondering)
 
I like Serena said:
Hi Mango! ;)

We'll need more information.

From the information you provided, we can only tell that:
$$R_o = \frac{7.2 \text{ V}}{150\text{ A}} \\
R_i = \frac{4.8 \text{ V}}{150\text{ A}}
$$
where $R_i$ is the internal resistance of the battery, and $R_o$ is the outer resistor.

We would need for instance a reference resistance at room temperature of the battery to say anything about temperature coefficient.
Is any such given? (Wondering)

Hey there! Thanks for replying, but I solved this problem on my own before I saw your post.
 
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