Question

a student has three resistors of resistances 2Ω, 4Ω and 6Ω. How can he connect all these resistors to get equivalent resistance of 1.09Ω?

Answer 1

Given :

A student has three resistors of resistances 2Ω, 4Ω and 6Ω.

To Find :

How should we connect them to get equivalent resistance of 1.09Ω.

Solution :

First of all we need to know following formulae.

★ Equivalent of series :

• R = R₁ + R₂ + ... + Rₙ

★ Equivalent of parallel :

• 1/R = 1/R₁ + 1/R₂ + ... + 1/Rₙ

In order to get equivalent resistance of 1.09Ω, we have to connect them in parallel to each other as shown in attachment.

Let's find equivalent resistance;

$:\implies\sf\:\dfrac{1}{R}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}$

$:\implies\sf\:\dfrac{1}{R}=\dfrac{6}{12}+\dfrac{3}{12}+\dfrac{2}{12}$

$:\implies\sf\:\dfrac{1}{R}=\dfrac{11}{12}$

$:\implies\sf\:R=\dfrac{12}{11}$

$:\implies\:\underline{\boxed{\bf{\purple{R=1.09\Omega}}}}$

Answer 2

Given :

A student has three resistors of resistances 2Ω, 4Ω and 6Ω.

To Find :

How should we connect them to get equivalent resistance of 1.09Ω.

 

Solution :

First of all we need to know following formulae.

★ Equivalent of series :

R = R₁ + R₂ + ... + Rₙ

★ Equivalent of parallel :

1/R = 1/R₁ + 1/R₂ + ... + 1/Rₙ

In order to get equivalent resistance of 1.09Ω, we have to connect them in parallel to each other as shown in attachment.

Let's find equivalent resistance;

= 1/R = 1/2+1/4+1/6

= 1/R = 6/12+3/12+2/12

= 1/R = 11/12

= R = 12/11

= R = 1.09Ω