Least Power rule of thumb

[jeepnjeff]

In this resistor problem is it alway the rule of thumb that the furthest resistor from voltage sorce will dissipate the least power.

 

[birdman0_o]

NO do the math. I just did it and got

R1= 1.69W
r2 = 0.4418W
r3 = 1.344 W
r4 = 0.1152 W <---- lowest in this circuit
r5 = 1.0092 W
r6 = 0.6728 W

If you do not know how to do the math ask, don't make a general guess rule as you will get many questions wrong.

Mike

 

[jeepnjeff]

I must be doing my series parallel math wrong. Do figure R6 series With R5 then R5 and R6 parallel with R4 to get Req. Then I get lost what to do next.

 

[Pommie]

I can't see how a 200Ω (R4) resistor in parallel with a 500Ω (R5+R6) can dissipate less power. My guess is R6 - maybe I should do the maths.

Mike.
Edit, I make it,
R1=1.757
R2=1.402
R3=0.477
R4=0.243
R5=0.058
R6=0.039

 

[ericgibbs]

hi,
I tried it LTSpice just for the heck of it.

Rounded up some values, got R6 is least.

I'm going to recheck the sim.

EDIT: Im not satisfied that some of these values are correct.???

 

[Pommie]

Something wrong there Eric,

I for R1 is Sqrt(4/100) = 200mA
V R1 = 0.2*100 = 20V
V R2 = 30-20 = 10V
W R2 = 10^2/200 = 0.5W

Weird thing is the values for R4-6 match mine.

Mike.

 

[MrAl]

Hi Jeff,

That rule does not apply here although it may be true by pure coincidence.
For example, consider a 10 megohm resistor connected directly across the
30v supply battery. It is right at the dang source and yet it will dissipate
very very little power. It is clearly not the furthest from the source.

To find out the power in each resistor you could use Thev/Norton equivalents
to reduce the network one element at a time, or you could collapse the
network one element at a timie and work backwards.
I suspect that you already used the 'collapse' method so maybe i can explain that...
Starting with R5 and R6, calculate the total resistance and then replace both
of them with one resistor that has the calculated resistance. Then, calculate
that resistance in parallel with the next resistor back which is R4. Once you have
the parallel calculation, replace those two with that one resistor. Continue in
this way until you have only two resistors in series, then calculate the voltage.
That will be the voltage for the junction of R1 and R2 and R3. From there
you can repeat the process to get the next node voltage.
Once you have all the node voltages, you can calculate the power in each
resistor and compare them.

BTW, i get 38.9 milliwatts for the power in R6 also, which is the lowest of all.

 

[SPDCHK]

I can't see how a 200Ω (R4) resistor in parallel with a 500Ω (R5+R6) can dissipate less power. My guess is R6 - maybe I should do the maths.

Mike.
Edit, I make it,
R1=1.757
R2=1.402
R3=0.477
R4=0.243
R5=0.058
R6=0.039

I agree with Pommie,
I get slightly different values but that's due to the number of decimals used.

R1 = 1.7569W
R2 = 1.3947W
R3 = 0.4811W
R4 = 0.2376W
R5 = 0.057W
R6 = 0.038W

 

[ericgibbs]

Something wrong there Eric,

I for R1 is Sqrt(4/100) = 200mA
V R1 = 0.2*100 = 20V
V R2 = 30-20 = 10V
W R2 = 10^2/200 = 0.5W

Weird thing is the values for R4-6 match mine.

Mike.

hi Mike,
I agree, something amiss.?

Added more results to sim diagram.

As I am still studying LTSpice, I just try some posters diagrams in order to learn and find the 'funnies' in the sim.

 

[SPDCHK]

hi Mike,
I agree, something amiss.?

Added more results to sim diagram.

As I am still studying LTSpice, I just try some posters diagrams in order to learn and find the 'funnies' in the sim.

Eric, the voltages don’t tie up.

Your sim shows 16.74V, 6.97V and 2.79V. That only totals to 26.5V

My calcs give me 13.255V, 9.808V and 6.893V. Totals 29.956V

 

[ericgibbs]

hi SPD,
I'm curious too, going to relook at it.

Its looks like I am misreading the results, I'll let you know where I have gone wrong.

EDIT:
Rechecked and tidied, similar results.
SPD: I suspect you are adding the wrong set of voltages.

Would appreciate any feedback on this.

 

[Pommie]

Looks like we're all now singing from the same hymn sheet. Hopefully, the OP isn't discouraged by our internal debates.

From debate does understanding come. Can't remember who said that first, so it may have been me.

Actually, between all our debating we missed the actual question. Luckily, Mr Al answered it.

Mike.

 

[jeepnjeff]

MrAL
I apploigize for my elementy electronic skills. I now know how to solve for total resistance and I am trying to follow your advice on finding voltage for each node to calculate power, but I cant figure how you guys are comming up with the R voltage for each.

 

[Pommie]

You should end up with a combined resistance of 226.3Ω which gives you a total current of 30/226.3 = 0.1325A. Use this value to find the voltage across R1 (100*0.1325) and hence find the voltage across R2 (30-13.25). Knowing the voltage across R2 allows you to calculate the current in R3 etc.

Mike.

 

[jeepnjeff]

Ok I've been able to get to that point on my own. when you subtract 30-13.3 you V2 is 16.7. which is .0835amps. Ok I think I follow so now you Subtract (I1-I2) to get I3. Then use ohms to figure R3 voltage?

 

[Pommie]

You can do it that way or you can pretend that R2 is now a 16.7V battery and repeat the calculation to get the voltage across R3 and R4.

Mike.

 

[Leftyretro]

These kind of problems can be a pain when first learning but they really help you nail down basic DC theory and ohms law. They also help you with how to best approach a problem, as something you have to calculate something you don't need in order to be able to calculate what you do need.

So it's something we all had to go through and is the best proven way to learn and understand the subject. I suspect someone just using a simulator to find the answer would miss a large learning opprutinity. Once you understand the basics then the simulator is a great tool for saving you time.

Lefty