How Does Internal Resistance Affect Battery Performance in a Car?

[mutineer]

Homework Statement

"A car battery has an e.m.f. of 12V and an internal resistance of 0.04 Ω. The starter motor draws a current of 100A. a.) What is the terminal p.d. of the battery when the starter motor is in operation? b.) If the headlamps are rated at 12V, 36W, what is their resistance? c.) To what value will their power output decrease when the starter motor is in operation?"

Homework Equations

E=I(R+r)

The Attempt at a Solution

The first two answers are easy to get. 8V and 4ohms. I can't get the last one( the lamps I am assuming are in parallel). The book says the answer is 16W.

 

[Simon Bridge]

with motor, terminal voltage drops to 8V

power through lamps is $V^2/R$

 

[gneill]

The question as stated has a few problems with being vague. It states that the starter motor draws 100A, but is that its rating with an ideal 12V source or is it with the given "imperfect" battery with a 0.04Ω series resistance? Does it thus draw 100A at 8V?

When the headlamps are switched on they will draw additional current through the battery's series resistance, so that voltage drop will be slightly larger; The starter motor will not be getting the 8V it did before. Will it still draw 100A?

These considerations can make a small but noticeable difference in the result. The book result seems to ignore these details and assumes that when the lamps are switched on, no additional current is drawn from the battery over the 100A that were being drawn previously.

 

[Simon Bridge]

@Gneill: it is a set question. We are told that the starter motor draws 100A ... we are not told how that is determined, true, but determined it has been[*]. It is not given as a "rating", but as one of the facts of the problem.

It is drawing 100A - what is the terminal voltage? See? Not vague.
The terminal voltage must be 8V for it to draw 100A ... modeled by working out the effective resistance of the motor. If it were drawing 0A, the terminal voltage would be 12V.

The headlamps, according to the book answer, draw an additional 2A. (16W, 8V, => 2A)

However - the lamps are an additional load on the battery so the terminal voltage should be further reduced. - I get 7.8957V instead of 8V... but we can only keep 1 significant figure - thus 8V it is. (@mutineer: working this out would probably be worth marks.)

So turning on the lamps does not make any noticeable (within margin of error) difference.

The model is a bit dodgy - real lamps' resistance depends on applied voltage (and temperature), and the motor is unlikely to be so well behaved either.

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[*] this is not unusual - physics problems very often provide figures without mention of how they were obtained. Quite a lot can hinge on a single word, or absence of one (in this case).

 

[asteriatic]

I would like to clarify and explain the concepts involved in this problem. The internal resistance of a battery refers to the resistance within the battery itself, which can cause a decrease in the battery's voltage when a current is drawn from it. This internal resistance, denoted as "r" in the equation E=I(R+r), is usually small but can have a significant effect on the battery's performance.

In this problem, the car battery has an e.m.f. (electromotive force) of 12V, which is the maximum voltage it can provide when no current is being drawn from it. However, when the starter motor draws a current of 100A, the voltage drops due to the internal resistance of 0.04 Ω. Using the equation E=I(R+r), we can calculate the terminal p.d. (potential difference) of the battery when the starter motor is in operation as 8V.

Moving on to part b of the question, we are asked to find the resistance of the headlamps. We know that the headlamps are rated at 12V and 36W, which means they draw a current of 3A (using the equation P=IV). Using Ohm's Law (V=IR), we can find the resistance of the headlamps to be 4Ω.

Finally, in part c, we are asked to determine the decrease in power output of the headlamps when the starter motor is in operation. This decrease occurs because the starter motor is drawing a significant amount of current from the battery, causing a decrease in the battery's voltage. Using the same equation P=IV, we can calculate that the power output of the headlamps decreases from 36W to 16W (since the voltage has decreased from 12V to 8V).

In summary, understanding the concepts of internal resistance and e.m.f. is crucial in solving this problem. By applying relevant equations and using basic principles of electricity, we can accurately solve for the terminal p.d. of the battery, the resistance of the headlamps, and the decrease in their power output when the starter motor is in operation. I hope this explanation helps in understanding the problem and its solution.