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Electricity and Magnetism Level 2

In a circuit we have three resistances, ${ R }_{ 1 }$ , ${ R }_{ 2 }$ and ${ R }_{ 3 }$ . It's known that ${ R }_{ 1 }=a$ , ${ R }_{ 2 }={ a }^{ 2 }$ and ${ R }_{ 3 }={ a }^{ 3 }$ . Also we know that if we create a parallel circuit with ${ R }_{ 1 }$ , ${ R }_{ 2 }$ and ${ R }_{ 3 }$ the total resistance is ${ R }_{ p }$ but if we create a series circuit with them, the total resistance is ${ R }_{ s }$ . If $9{ R }_{ p }={ R }_{ s }$ , calculate the value of ${ R }_{ 1 }$ .

The answer is 1.

5 solutions

Chew-Seong Cheong
Sep 11, 2015

$\color{#D61F06}{\text{Please note that it is actually } R_1 = a \space \Omega, R_1 = a^2 \space \Omega \text{ and } R_3 = a^3 \space \Omega, \text{where } a \text{ is a } \\ \text{dimensionless constant. There is no such thing as } R = R^2=R^3.}$

$\begin{cases} R_p = \dfrac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}} = \dfrac{1}{\frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}} = \dfrac{a^3}{a^2+a+1} \\ R_s = R_1+R_2+R_3 = a + a^2 + a^3 \end{cases}$

Now, we have:

$\begin{aligned} 9R_p & = R_s \\ \Rightarrow \dfrac{9a^3}{a^2+a+1} & = a + a^2 + a^3 \\ 9a^3 & = a(a^2+a+1)^2 \\ 9a^2 & = (a^2+a+1)^2 \\ 3a & = a^2+a+1 \\ a^2-2a+1 & = 0 \\ (a-1)^2 & = 0 \\ \Rightarrow R_1 = a & = \boxed{1} \Omega \end{aligned}$

We note that $R_1 = R_2 = R_3 = 1 \Omega$ .

Wait a second!! Although there isn't any problem mathematically...But how do we explain this dimensionally..

If $R1$ = $a$ , then the units of $a$ must be Ohm. But since $a^2$ is the value of $R2$ , so the units of $a$ must be Ohm^0.5

How does that work out...??

A Former Brilliant Member - 5 years, 9 months ago

No, the wording of the problem is imprecise. $R_1 = a \space \Omega$ , $R_2 = a^2 \space \Omega$ and $R_3 = a^3 \space \Omega$ . $a$ is just a constant and is dimensionless.

Chew-Seong Cheong - 5 years, 9 months ago

@Chew-Seong Cheong no the wording doesn't say so, but i got your point!! Thanks:)

A Former Brilliant Member - 5 years, 9 months ago

@A Former Brilliant Member – What I meant was $\Omega$ should be included to be precise.

Chew-Seong Cheong - 5 years, 9 months ago

This solution does not comply with the laws of Physics . in a parallel section of a circuit current flowing to that section will divide and flow through each resistor based on current squared = voltage/ Resistance. By definition a parallel circuit is a circuit in which the voltage is constant across each branch. Conversely current is constant in a series circuit and voltage is distributed proportionally across each resistor.

Colleen Walsh-Barnes - 5 years, 9 months ago

We note that $R_1=R_2=R_3 = R = 1 \Omega$ . Let the source voltage be $V$ . Then, when the three resistors are in parallel, current flowing through each resistor is $\frac{V}{R}$ so the total current from the voltage source is $I_p = 3\frac{V}{R}$ , therefore the resultant resistance $R_p = \frac{V}{I_p} = \frac{1}{3}R$ . When the resistors are connected in series, then the current $I_s = \frac{V}{3R}$ and the resultant resistance $R_s = \frac{V}{I_s} = 3R$ . This implies that $9R_p = R_s$

Chew-Seong Cheong - 5 years, 9 months ago

If R = 1 Then IP = V/R +V/R + V/R = 3(V/R) = 3V/1 IS = V÷ (R + R + R) = V/3

Regardless of the mathematics the laws of physics is such that for equal resistors RP can never be greater than RS. Also the total Resistance of a parallel circuit is always less than the smallest Resistance. To prove this let 1 resistor in a circuit = 0 then calculate the Resistance in series and parallel

Colleen Walsh-Barnes - 5 years, 9 months ago

@Colleen Walsh-Barnes – Exactly so, here. $R_p = \frac{1}{3} R \color{#D61F06}{<} R_1=R_2=R_3 = R$ and $R_s = 3R \color{#D61F06} {>} R_1 = R_2 = R_3 = R$ and $\color{#D61F06} {R_p < R_s}$ .

Chew-Seong Cheong - 5 years, 9 months ago

Galiève , can you add in the $\Omega$ 's to avoid confusion?

Chew-Seong Cheong - 5 years, 9 months ago

Galiève .
Sep 11, 2015

The resistance of the series circuit is ${ R }_{ s }=a+{ a }^{ 2 }+{ a }^{ 3 }$ , and the resistance of the parallel circuit is ${ R }_{ p }=\frac { 1 }{ \frac { 1 }{ a } +\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { a }^{ 3 } } }$ . Knowing that $9{ R }_{ p }={ R }_{ s }$ , it means that: $\frac { a+{ a }^{ 2 }+{ a }^{ 3 } }{ 3 } =\frac { 3 }{ \frac { 1 }{ a } +\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { a }^{ 3 } } }$ .

Using the arithmetic-geometric-harmonic mean, we know that: $\frac { a+{ a }^{ 2 }+{ a }^{ 3 } }{ 3 } \ge \sqrt [ 3 ]{ a*{ a }^{ 2 }*{ a }^{ 3 } } \ge \frac { 3 }{ \frac { 1 }{ a } +\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { a }^{ 3 } } }$ . But if the harmonic mean is the same as arithmetic mean, it means that is also the same as geometric mean. So mixing what we have, we have this equation: $\frac { a+{ a }^{ 2 }+{ a }^{ 3 } }{ 3 } =\sqrt [ 3 ]{ a*{ a }^{ 2 }*{ a }^{ 3 } } =\frac { 3 }{ \frac { 1 }{ a } +\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { a }^{ 3 } } }$

Let's solve it: $\frac { a+{ a }^{ 2 }+{ a }^{ 3 } }{ 3 } =\sqrt [ 3 ]{ a*{ a }^{ 2 }*{ a }^{ 3 } } ;\quad a+{ a }^{ 2 }+{ a }^{ 3 }=3*{ a }^{ 2 };\quad a(1-2a+{ a }^{ 2 })=0;\quad a{ (a-1) }^{ 2 }=0$ .

Because $a$ is the value of a resistance, it cannot be 0, so $a=1$ .

Thanks for reading and sorry for my english.

Please see my report and tell me where I am wrong.

Department 8 - 5 years, 9 months ago

Tirtha Pratim Jana
Sep 13, 2015

9Rp=Rs After solving we get the equation => (9a^3)/(1+a^2+a^3)=a(1+a^2+a^3) =>. 6a^2 =. 1+2a+2a^3+a^4 Now put the value of a (assume ) =1 See equation is satisfied

Lu Chee Ket
Nov 1, 2015

9 (1/ 3 ohms) = 3 ohms

Answer: 1

Pranav Prakasan
Sep 13, 2015

By trial and error method..select 1 and try

Join the resistance!

The answer is 1.

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