A Simple Parallel Resistance Problem

Valdaquende

Joined Aug 1, 2011
7
Friends - I am, fairly late in life, trying to learn electronics in my spare time using a high-school/college textbook ('Electronics Fundamentals' by Floyd). I'm about 200 pages in, at this point. I'm working on parallel-resistance networks and ran into this problem:

A certain parallel network consists of five 1/2-Watt resistors with the following values: 1.8kΩ, 2.2kΩ, 3.3kΩ, 3.9kΩ and 4.7kΩ. As you slowly increase the voltage across the parallel circuit the current slowly increases. Suddenly the current drops to a lower value.

a) Excluding power supply failure, what happened?
b) What is the maximum voltage you should have supplied?
c) Specifically, what should be done to repair the circuit?

Of course, a resistor has failed. In deriving the maximum voltage, I reasoned that the maximum power for this circuit would be 2.5W as power is additive in parallel circuits.
I calculated a total resistance for the network of 561Ω.
I used what I could recall of my college algebra to manipulate the formula:

P=IV into V=√PR

and then used that to calculate the maximum voltage as 37.45v. The book's answer, though, is 30v. They do not show the steps; only the result.

I am the sort of person who finds it difficult to let go of an unsolved problem. In this case, I don't know if my approach is wrong, if my algebra is faulty or if the book is incorrect. If someone could set me on the right track, I'd be grateful. Thank you for the time and energy spent on such an elementary problem.

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ericgibbs

Joined Jan 29, 2010
12,231
hi,
Which resistor do you think will fail first due to excessive heating.?
Why have you included the value of the failed resistor.?

E

Jony130

Joined Feb 17, 2009
5,221
Try to calculate the power dissipated in each individual resistor for V = 37.45V

RBR1317

Joined Nov 13, 2010
608
Why calculate the total power dissipation in all resistors? Only one resistor failed, and it failed when it's power dissipation exceeded ½W. You need to look at power dissipation in the resistors individually.

LesJones

Joined Jan 8, 2017
3,085
I think eric's question is a good clue to take you in the right direction. Remember each resistor will have the same voltage across it.

Les.

WBahn

Joined Mar 31, 2012
26,398
Of course, a resistor has failed. In deriving the maximum voltage, I reasoned that the maximum power for this circuit would be 2.5W as power is additive in parallel circuits.
Bad reasoning -- and exactly the bad reasoning that the author anticipated you would make because it is so commonly made. So don't feel bad.

Think about way you think that power is additive in parallel circuits?

Consider a physical example. I want to drive a car across a ditch. I find a steel rail and a wooden timber that are long enough to span the ditch. I know that the car weighs 3000 lb and that the rail can support 2500 lb while the timber can support 1000 lb meaning that, together, they can support up to 3500 lb. Is it safe to drive across the ditch with the left tires on the rail and the right tires on the timber?

Do you see the fault in the reasoning? Now see if you can apply it to the resistor problem. Keep in mind that what is evenly applied to the resistors in parallel is NOT the current and is NOT the power, it is the voltage.

Valdaquende

Joined Aug 1, 2011
7
ericgibbs and Jony130 - Thanks so much for considering this and answering. I think, in retrospect, that I was too linear in my thinking; I was hung up enough on the discrepancy between my result and the book's that I didn't go further - I should have.

The power dissipated at 30v is: R1 = .5W; R2 = .409W; R3 = .27W; R4 = .23W and R5 = .19W

None of these would, in theory, have failed at this voltage, as they are all at or below their ratings but clearly exceeding 30v would cause failure at some point. The first resistor to fail would be R1, as it would be dissipating more than its rated amount of heat

The power dissipated at 37.45v is: R1 = .779W; R2 = .63W; R3 = .43W; R4 = .36W and R5 = .3W
Clearly wrong; in this case, both R1 and R2 would fail.

I guess my next question would be 'What's wrong with my math?' Algebraically, it seemed like a piece of cake but clearly the result is wrong. Can you set me straight?

bertus

Joined Apr 5, 2008
21,247
Hello,

Have a look at the resistor values.
You have seen that R1 will fail first.
At what voltage will have R1 0.5 Watts to dissipate?

Bertus

WBahn

Joined Mar 31, 2012
26,398
The problem isn't the math, it is the reasoning. You reasoned that resistors in parallel will share power equally. But this is only true if the resistances are equal. Resistors in parallel share the voltage equally -- but the power dissipated in a given resistor is not a function only of the voltage across it, but also the resistance itself. The smaller the resistor, the more power will be dissipated at the same voltage. So the power limitation of the parallel network is imposed by the smallest resistor (if all of the resistors have equal power ratings).

Which resistor would set the power limitation if they were all in series?

Valdaquende

Joined Aug 1, 2011
7
WBahn - Thank you. In considering power across parallel resistances as 'additive', I forgot that they are proportional.

Bertus - Thank you, thank you. I not only made the mistaken assumption that the power usage was constant across the resistors but also failed to see that I could have figured this out one resistor at a time. I didn't need to calculate the network's equivalent resistance at all. Sometimes I get so hung up in the concepts that I fail to see the most straightforward approach. Thanks!

Valdaquende

Joined Aug 1, 2011
7
WBahn - In a series circuit, the largest resistor would fail first, as its dissipated power is directly proportional to its resistance. Conversely, in a parallel circuit, the smallest resistor would fail first as its dissipated power is inversely proportional to its resistance.

Valdaquende

Joined Aug 1, 2011
7
I see now. Or at least I see another foot or two further through the fog of the things I don't know.

I want to thank you all for your help. The word 'educate' comes from the Greek word 'educos', meaning 'to draw from within'. By giving me hints instead of an answer, you have helped me do that. A truly satisfying introduction to AAC forums.

Many Thanks!

dl324

Joined Mar 30, 2015
12,678
I not only made the mistaken assumption that the power usage was constant across the resistors but also failed to see that I could have figured this out one resistor at a time. I didn't need to calculate the network's equivalent resistance at all.
Since you're new to electronics and AAC, you can save those who try to help you time by posting a schematic that's easy to access; like this:

From the problem description, something happened when voltage was increased to a certain point; current went down so something increased in resistance.

The power equation that deals with R and V is
$$\small P = \frac{V^2}{R}$$
rearranging, you get
$$\small V= \sqrt{RP} = \sqrt{1800 ohms * 500mW} = 30V$$

The text is assuming that the resistor will fail at the rated wattage, which isn't necessarily the case. For reliability, most designers would derate from a component's max rating.

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WBahn

Joined Mar 31, 2012
26,398
For a bit of real world engineering (doesn't apply to this problem as far as the expected answer is concerned), textbook resistors have this interesting property that they can either dissipate an unlimited amount of power (if we don't have a power rating for them) or they can dissipate up to their power rating and then fail instantly as soon as we go above it. This is fine for early-on educational purposes where we are trying to learn basic concepts, but real world resistors don't act that way. As they get close to their rated power limits several bad things happen -- the value changes more significantly and you get closer to causing permanent changes in the resistance even after it cools down, but total failure of the resistor usually doesn't happen until well above the rated power level. So in practice we leave ourselves plenty of margin. For many things power related we live a factor of two, so if you need a resistor to dissipate 0.25 W you use a resistor rated for a minimum of 0.5 W.

WBahn

Joined Mar 31, 2012
26,398
I see now. Or at least I see another foot or two further through the fog of the things I don't know.

I want to thank you all for your help. The word 'educate' comes from the Greek word 'educos', meaning 'to draw from within'. By giving me hints instead of an answer, you have helped me do that. A truly satisfying introduction to AAC forums.

Many Thanks!
Ah, if only most of the folks coming here for help understand this!!!

Thank you for saying so and I look forward to seeing you around the boards in the future.