Problem

(a) Given a 48. 0 -V battery and $24.0-\Omega$ an…

02:16

Problem 5 Medium Difficulty

Your car’s 30.0-W headlight and 2.40-kW starter are ordinarily connected in parallel in a 12.0-V system. What power would one headlight and the starter consume if connected in series to a 12.0-V battery? (Neglect any other resistance in the circuit and any change in resistance in the two devices.)

Answer

Headlights would consume:
$$P _ { 1 } = 29.26 \mathrm { W }$$
Starter would consume:
$$P _ { 2 } = 0.3658 \mathrm { W }$$

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Physics 102 Electricity and Magnetism

College Physics for AP® Courses

Chapter 21

Circuits, Bioelectricity, and DC Instruments

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Video Transcript

to solve this question, we must begin by finding resistance is off headlamp or headlight, and the starter So 30 What? So let's go for 30 words first. So 30 worst means power is equal to be I from this equation we can use 30 is equal to 12 times I. So from this we can find out that I, in this case, is so two divided by 12 which is equal to 2.5 amperes. So from this we can find the resistance off headlight which is equal to read about it by eye. Obvious 12 divided by 2.5. This is full point eight. Now, this is resistance off headlight. Similar on a similar lines. We confined resistance off starter So started taste 2400 words, we and I. Yeah, Um, he's 12 I So I is equal to 200 MPs. From this, you can find the resistance is equal to 12. The voltage applied divided by 200. This is equal to 0.6 Or one could simplify this calculation by immediately substituting we as i r i r. Then it will be I square are. And from this we confined I again or so. This can also be simplified as we squared about it by our. So this will be much more easier, much better to use. No, now that we have both, resistance is so this is for starter in Siri's net. Resistance will be addition off these two resistance is this will be zero for 4.867 So the potential job across this headlight on DDE the starter will depend upon the resistance is so let's find the fraction of resistance that headlight carries. This is equal to 4.8, divided by four point 86 4.8 divided by 4.6, this is equal to 0.9877 So this much off fraction off 12 world will be across the headlamp headlight. And for starter that is equal to this is 0.6 divided by 4.86 equal to zero point 012 See now let's find the potential difference across them. It's a potential difference. Across the headlight will be. The fraction off. Resistance is that is R H. Divided by our end times. The potential applied that is 12 so 12 times zero point 98 77 is equal to 11 point 85 and potential difference across the starter is again. The same procedure we we use instead of 0.9877 were used 0.0 123 This gives me 0.14 81 for one could also do 12 minus this amount. So this is also a cultural minus 11.85 depending on what accuracy one uses. So to find the power across these two objects, we must also know the current current is equal to potential Applied, divided by the resistance, the potential is 12 net pretension and metre distance is, as we know, 4.86 So the current here will be equal to two point 47 and piers now we can use equation for power across headlight will be able to be etch times I, which is equal to we itches 11.85 current is this value. Let's multiply them. This is equal to 29 point 26 What? And p starter again The same question But the S I on again the same multiplication. Give me a zero point 37 all 3658 Since the answer can be kept in four significant digits. What

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