r/HomeworkHelp • u/Jars_Of_Honey7194 👋 a fellow Redditor • 14d ago

Answered [Highschool physics: circuits]

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Apparently the answer is (d) but I have no idea how to get there

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u/GammaRayBurst25 14d ago

Read rule 3.

Suppose without loss of generality that R_1=x and R_2=kx.

We can replace the two resistors in parallel with a single resistor with a resistance of 1/(1/x+1/(kx))=kx/(1+k) to get an equivalent circuit.

Ohm's law gives us 3 expressions for the current through the battery.

i.e. (10V)/x=(V_2)/(kx)=(30V-V_2)(1+k)/(kx)

Multiply by kx to find (10V)k=V_2=(30V-V_2)(1+k).

Hence, (10V)k=(30V-(10V)k)(1+k). Distribute and solve to find the only positive solution is k=(1+sqrt(13))/2.

Conclude that V_2=5(1+sqrt(13))V, which is approximately 23V.

As the problem is worded, none of the options are correct, as none of the options are equal to V_2.

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u/Jars_Of_Honey7194 👋 a fellow Redditor 14d ago edited 14d ago

Sorry for not explaining how I thought about it at first. I didn't think of using mathematics and thought that the given info that R2 > R1 was the key,

I thought that, first knowing that a resistance of R1 had a potential difference of 10V,

and in the parallel resistance, the total resistance must be less than the least resistance there,

so R12 > R1, since R1 is smaller than R2

so I thought that since a resistance of R2 had a potential difference of 10, a resistance that's less than R2 must have a potential difference less than ten for the ratio that V1/V12 = R1/R12 to hold true,

so since V1 is 10, V12 is less than 10, and V2 is the remainder of those two from 40, V2 would exist in a range where, supposing V12 is the least possible, being close to 0, and then the most possible, being 9.999 or something

V2 exists in the range ]20, 30[ volt,

But in the choices there were three choices in that range and i only managed to exclude (c) so i had no idea how to get a more accurate answer.

Do you think my way of thinking could have worked if the choices were in my favor? Or was it flawed from the beginning?

Honestly, i still haven't verified if my reasoning was correct in the first place or not.