|Apr15-05, 02:16 PM||#1|
Here's the question:
"A car battery has an e.m.f. of 12V and an internal resistance of 0.04 ohms. 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?"a.) I get the terminal p.d. for the first part as follows:
V = E - IR = 12 - 0.04 x 100 = 8 volts
b.) Now, as I understand the next part, the starter motor has not been engaged yet and the headlamps are presumably in parallel. But if the battery has an e.m.f of 12V, how can it power them? Moreover, how can their resistance be calculated as asked? Presumably the E.M.F. of 12 is equal to the lost volts (I x 0.04) plus the voltage across the lamps. But since the voltage across the lamps isn't going to be less than 12 (is it?), isn't one of the terms in the equation E = Ir + V going to have to be zero (?!)
[ c.) If I can get part b.) right, I don't think there will be any problems with c.) ]
I'm obviously misunderstanding something here. Can someone explain?
|Apr15-05, 03:03 PM||#2|
The headlamps are simultaneously rated for voltage and power. This assumes that both maxima are attained at the same current. This is independent of the rest of the circuit.
Is there a relation that involves voltage, power and resistance ?
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