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I am currently studying The Art of Electronics, third edition, by Horowitz and Hill. Exercise 1.5 says the following:

C. Power in resistors
The power dissipated by a resistor (or any other device) is \$P = IV\$. Using Ohm’s law, you can get the equivalent forms \$P = I^2 R\$ and \$P = V^2 / R\$.

Exercise 1.5. Show that it is not possible to exceed the power rating of a 1/4 watt resistor of resistance greater than 1k, no matter how you connect it, in a circuit operating from a 15 volt battery.

I don't understand how I am supposed to do this. I've gone back through the earlier pages, but I cannot tell whether I'm just misunderstanding something, or whether the authors have not introduced sufficient information to complete this exercise. Am I supposed to use \$P = IV\$, \$P = I^2 R\$, or \$P = V^2 / R\$ somehow? There's literally 2 sentences of information here, and none of it describes what is being asked in exercise 1.5.

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  • \$\begingroup\$ Given this '15V battery' and a circuit consisting of this 1k resistor, how would you connect them together so cause the highest possible power dissipation (not that there are many options...)? Now calculate the power dissipation (any of those formulas should work and give you the same answer). How does this result compare to the 1/4W rating of the resistor? \$\endgroup\$ – brhans Oct 12 '20 at 14:27
  • \$\begingroup\$ Step 1 : assume the circuit doesn't include a boost converter. \$\endgroup\$ – user_1818839 Oct 12 '20 at 18:49
  • \$\begingroup\$ I am puzzled that the questioner is puzzled ! Surely this is a 'trivial' question ? Where did the confusion arise ? \$\endgroup\$ – Graham Stevenson Oct 13 '20 at 6:35
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A few clues:

If \$ P = \frac {V^2} R \$ then which way will P go with increasing R?

Given that, then what is the worst case for R?

What is the worst case for V (the maximum voltage that you can apply to the resistor)?

You now have both V and R so you can calculate P.

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Voltage Spike Oct 12 '20 at 16:23
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If a circuit has a 15 V battery, what's the highest voltage you could put across your resistor?1

How much power will a 1k resistor dissipate at that voltage?


1 Actually a bit of a trick question.

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  • \$\begingroup\$ Yeah, question is a bit vague in what else is allowed in the circuit or whether it is literally just battery and resistor, in which case there's only 3 ways to connect everything. \$\endgroup\$ – DKNguyen Oct 12 '20 at 14:35
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    \$\begingroup\$ Remeber AoE was written decades before DC-DC switching converters became jelly-bean parts. \$\endgroup\$ – The Photon Oct 12 '20 at 17:09
  • \$\begingroup\$ @ThePhoton: Maybe so, but even back then, high-power DC-to-AC inverters powered by a 12V battery were quite common. So were electronic photoflash units \$\endgroup\$ – Dave Tweed Oct 12 '20 at 20:43

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