Question

When 2 resistors of resistances r1 and r2 are connected in parallel the net resistances 3 ohm when connected in series its value 16 ohm what is the value of r1 and r2?

Answer 1

Answer:

$\large{\underline{\boxed{\sf 12 \Omega \: and \: 4 \Omega}}}$

Explanation:

Given that;

When the two resistors R₁ and R₂ are connected in series, the equivalent resistance is 16 $\Omega$

⇒ R₁ + R₂ = 16

⇒ R₂ = 16 - R₁

Also, it is given that,

When the two resistors R₁ and R₂ are connected in parallel, the equivalent resistance is 3 $\Omega$

⇒ $\sf \dfrac{1}{R_1}+ \dfrac{1}{R_2}= \dfrac{1}{3}$

⇒ $\sf \dfrac{R_1 + R_2}{R_1 R_2}= \dfrac{1}{3}$

Since, R₁ + R₂ = 16,

⇒ $\sf \dfrac{16}{R_1 R_2}= \dfrac{1}{3}$ ..... (i)

On cross-multiplying :

⇒ $\sf R_1 R_2 = 48$

Now, Substituting R₂ = 16 - R₁ in (i)

⇒ $\sf \dfrac{16}{R_1 R_2}= \dfrac{1}{3}$

⇒ $\sf \dfrac{16}{R_1 (16 - R_1)}= \dfrac{1}{3}$

On cross multiplying ;

⇒ R₁ ( 16 - R₁) = 48

⇒ 16R₁ - (R₁)² = 48

⇒ R₁² - 16R₁ + 48 = 0

On splitting the middle term ;

⇒ R₁² - 16R₁ + 48 = 0

⇒ R₁² - 12R₁ - 4R₁ + 48 = 0

⇒ R₁(R₁ - 12) - 4(R₁ - 12) = 0

⇒ (R₁ - 12)(R₁ - 4) = 0

⇒ R₁ = 12 or R₁ = 4

When, R₁ = 12 :

R₂ = 16 - 12

R₂ = 4 $\Omega$

When, R₁ = 4 :

R₂ = 16 - 4

R₂ = 12 $\Omega$

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Hence, the resistance of two resistors R₁ and R₂ are 12 $\Omega$ and 4 $\Omega$

Answer 2

Heya!

Here's your answer!!

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See the picture..

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Thanks for the question!!