The discussion centers on calculating the temperature coefficient of a car battery resistor at 0 degrees Celsius, given a voltage drop from 12V to 7.2V while drawing 150 amps. The internal resistance of the battery, denoted as R_i, is calculated as 4.8 V / 150 A, and the outer resistor, R_o, is calculated as 7.2 V / 150 A. To determine the temperature coefficient, a reference resistance at room temperature is necessary, which was not provided in the initial query. The original poster eventually solved the problem independently.
PREREQUISITESElectrical engineers, automotive technicians, and anyone interested in understanding battery performance and resistance calculations in cold temperatures.
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!
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)