Limiting resistor?

[Boognish]

Hello, another homework question here. There are 5 resistors connected in a series circuit.

R1= 70 ohm
P1= 5 w

R2= 20 ohm
P2= 5 w

R3= 100 ohm
P3= 5 w

R4= 50 ohm
P4= 5 w

R5= 50 ohm
P5 = 5 w

The question is, "What is the maximum supply voltage that can be applied to the circuit before the limiting resistor burns open?"

I have determined that the limiting resistor is R3 because it has the lowest current rating. How do I solve this question? Do I take the limit current of R3 and multiply it by R3 to come up with the answer?

Any help is greatly appreciated. I have been stuck on this for some time now.

 

[eric9822]

I have determined that the limiting resistor is R3 because it has the lowest current rating.

They are all 5 watt. How did you come up with that?

 

[Boognish]

I = sq root of P/R. I believe that gives the amount of current that the resistor can take before it blows. Using this formula for each resistor, R3 has the lowest current limit, hence it would blow before the other 4 resistors.

Yes/No? If no, how?

 

[gar]

100127-2244 EST

Boognish:

You picked on the correct answer as to which one will fail first.

Because all the resistors are in series the same current is flowing in each resistor. With that information and knowing that I^2*R = Power tell me what the voltage is across the limiting resistor. and assuming you solve for the limiting resistance operating at 5 watts dissipation. Then what is the voltage across each resistor, and the total voltage across the 5 resistors?

.

 

[Boognish]

oops the total voltage must be calculated. I got 85 V.

Total Resistance = 290 ohm
Total Current = .293
Total Power = 25 w

29.3 is the voltage across the limiting resistor.

 

[Boognish]

100127-2244 EST

and assuming you solve for the limiting resistance operating at 5 watts dissipation. Then what is the voltage across each resistor, and the total voltage across the 5 resistors?

.

I understand everything up until here. Do I take the "limiting current" that I found by SQ root of P3/R3 and multiply that by each resistors resistance value and then add up all of my voltage drops?

 

[gar]

100127-2337 EST

Your total resistance is correct.

Go back and calculate the current thru R3 to dissipate 5 W.

I get 0.22360679775 A. See if you can get this answer or close to it. Four significant digits would be adequate. I always use HP calculators with reverse polish capability.

,

 

[dbuckley]

I have determined that the limiting resistor is R3 because it has the lowest current rating. How do I solve this question? Do I take the limit current of R3 and multiply it by R3 to come up with the answer?

Whereas you've picked the right resistor, for the right reasons, the way you've described it is somewhat backassward. The resistors are in series and thus common sense (or better yet, Kirchoff's law) tells us the same current flows through all resistors. Using a formula wheel, we would note that the power dissapated across a resistance is calculated as RxI^2. Since the term I^2 is the same for each resistor (Kirchoff!), we can ignore it and call it 1. Since that leaves us with P=Rx1, we can conclude that the higher the resistance the more power will be dissapted. The highest value resistance is R3, so that is the resistor that dissapates the most power.

So solve R3 for the voltage that will have it dissapating 5W, back to the formula wheel, and its the square root of PxR.

Now you know the voltage across one resistor you can solve the all resistors problem as a standard potential divider. Remebering Kirchoff, same current in all resistors, the order of the resistors doesn't matter, and the resistors you aren't specifially interested in can be lumped together as one resistor, so simplifying, the problem then becomes the classical potential divider case of a 100R and a 190R resistor in series.

 

[Boognish]

Whereas you've picked the right resistor, for the right reasons, the way you've described it is somewhat backassward. The resistors are in series and thus common sense (or better yet, Kirchoff's law) tells us the same current flows through all resistors.

Well, I can blame my lame teacher or lack thereof for my not properly instructing the class on this. It is not in our text and the teacher doesn't teach. I have been searching out the answer and haven't been able to find it... The teacher just hands out the homework and expects us to know it. After asking him questions, it is apparent that he doesn't know what he is teaching and his expertise is elsewhere.

 

[LarryFine]

Added:Highlight answers to see them:

The resistors will not dissipate the same power, so the 5w muct be the resistors' ratings, and not power seen. The highest-value resistor will see the most power because the current will be the same through all resistors, and it will drop the most voltage.

So, what must be calculated to answer the OP's question is what voltage will create the current necessary to make the 100 ohm resistor dissipate 5w. Ohm's Law tells us that voltage is equal to the square root of (power x resistance), so that gives us 10v.

Now, we have to calculate the total circuit that will make the 100 ohm resistor see 10v, which is easy because we know that each 10 ohms sees 1v. So, with a total resistance of 290 ohms, the total supply must be 29v. We can also figure the current is 0.1a.

 

[Smart $]

Added:Highlight answers to see them:

....

Perhaps they should stay hidden even when highlighted

...and no need to determine voltage across any one or one plus lumped resistors.

P_rating (watts) ? R_3 (ohms) = I_max? (amperes?)
5 ? 100 = 0.05

√I_max? (amperes?)= I_R3 (amperes) = I_circuit (amperes)
√0.05 = 0.223606798 = 0.223606798

I_circuit (amperes) ? R_total (ohms) = E_circuit (volts)
0.223606798 ? 290 = 64.8459713​

Normally I wouldn't post the answer to homework question... yet I don't want to see anyone led astray either.

 

[LarryFine]

Wow. Right you are. :roll:

I worked backwards, using 65v, and came up with a hair over 5w, so your numbers are obviously correct.

I must have missed something between my eyes and my fingers . . . like my brain.

Note to all: Please disregard everything hidden in post #10.

 

[bob]

Using the formula P = I?R the current for 5 watts is as follows

ohms amps
70 -- 0.361
20 -- 0.500
100 -- 0.22
50 -- 0.31
50 -- 0.31
From this list you can see that 0.22 is the max value of current that can flow
through this circuit. Total circuit resistance is 290 ohms.
I = E/R or E = IR E = 0.22 x 290 E = 64 volts. Slightly different from Smarts
answer due to rounding off.

 

[Boognish]

Yep, approximately 64 V is the answer. Thanks for the responses. I understand it now. This helped me get a 95% on my Midterm. Oh, yeah. :grin:

 

[LarryFine]

Yep, approximately 64 V is the answer. Thanks for the responses. I understand it now. This helped me get a 95% on my Midterm. Oh, yeah. :grin:

Drinks on the Boogerman!

 

[Boognish]

Well, I can blame my lame teacher or lack thereof for my not properly instructing the class on this. It is not in our text and the teacher doesn't teach. I have been searching out the answer and haven't been able to find it... The teacher just hands out the homework and expects us to know it. After asking him questions, it is apparent that he doesn't know what he is teaching and his expertise is elsewhere.

Not really relevant to this thread but I will put it out there... I retract the above statement by 100%. My instructor is a great teacher and a very nice person. Too bad that he is retiring, wish I could have him for a teacher in the following semester.

Later,
Ohm

 

[LarryFine]

I retract the above statement by 100%. My instructor is a great teacher and a very nice person. Too bad that he is retiring, wish I could have him for a teacher in the following semester.

Nicely said, and by an obvious gentleman.