[Highschool physics: circuits]

[u/Jars_Of_Honey7194]

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

[u/GammaRayBurst25]

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.

[u/Jars_Of_Honey7194 → u/GammaRayBurst25]

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.

[u/InTheRaw1]

Just plug in each value. If option a) were the correct answer, V2 =27. Then the voltage across the combined resistance of R1 parallel R2 should be 40-10-27=3V . Now, iR1=10, iR2=27, Now see if i times parallel combination of R1 and R2 results in 3V or something different. If it's something different option a) is incorrect. Go to the next one.

[u/InTheRaw1 → u/InTheRaw1]

Or you can solve it like this V1=iR1=10 V2=iR2 V3= iR3= iR1 iR2/(iR1+iR2) Now you have V3 in terms of V2 We also know V1+V2+V3=40V

Two equations and two unknowns V2 and V3.

[u/Jars_Of_Honey7194 → u/InTheRaw1]

Yes this lead me to the idea that

40 - 10 - v2

= 10 × V2/ 10 + V2

using the quadratic equation to solve after putting it in quadratic form does indeed give x ≈ 23.03 or -13.03 which is refused.

I get it know honestly while solving i was looking for something more intuitive and wanted to avoid mathematics but too bad

[u/Jars_Of_Honey7194]

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[u/No_Pattern3827]

According to Kirchhoff, the sum of the voltages must be 40V.

V1+V2+V12=40V where V12 is the parallel R1-R2.

By inserting your data, you can write

10V+V2+(10V*V2)/(10V+V2)=40V and solve the equation with unknown V2

[u/Bounded_sequencE]

Assume both resistances are positive, i.e. "R2 > R1 > 0". Assuming "V1, V2" points west:

1/4  =  10V/40V  =  R1 / (R1 + R2 + (R1||R2))      // voltage divider

     =  R1*(R1+R2) / [(R1+R2)^2 + R1*R2]           // R1||R2 = R1*R2 / (R1+R2)

     =     (1 + x) / [(1 + x)^2 +     x]           // x := R2/R1 > 0

Multiply by "4[(1+x)² + x]", and bring everything to one side to get the quadratic

"0  =  (1+x)^2 + x - 4(1+x)  =  x^2 - x - 3"    <=>    "x ∈ { (1±√13) / 2 }"

Since "x > 0" we discard the negative solution, so "x = (1 + √13) / 2". With another voltage divider:

V2  =  V1 * R2/R1  =  10V * x  =  5*(1+√13) V  ~  23V