Question

Two coils when connected in series have an equivalent resistance of 20 Ω and when connected in parallel, an equivalent resistance of 5 Ω. Find their resistances ?

Answer 1

Solution :

Let R₁ and R₂ be the resistance of the coils. When joined in series, the end resistance

R =(R₁ + R₂)
∴ R₁ + R₂ = 20 .........(i)

When joined in parallel, the equivalent resistance R is given by

$\frac{1}{ R_{1} } + \frac{1}{ R_{2} } = \frac{1}{R}$

∴ $\frac{1}{ R_{1} } + \frac{1}{ R_{2} } = \frac{1}{5}$

∴ $\frac{ R_{1} + R_{2} }{ R_{1} R_{2} } = \frac{1}{5}$ ...........(ii)

Substituting from (i)

$20R_{1} R_{2} = \frac{1}{5}$

∴ R₁ R₂ = 100 .........(iii)

Or, (R₁ - R₂)² =(R₁ + R₂)² - 4R₁R₂

Or, (R₁ - R₂) = $\sqrt{ (R_{1}+R_{2}) ^{2} - 4 R_{1} R_{2} }$

Now, from (i) & (iii)

R₁ + R₂ = $\sqrt{ (20) ^{2} - 4* 100 } = 0$ .........(iv)

∴ Solving (i) and (ii)

R₁ = 10 Ω
R₂ = 10 Ω

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Answer 2

Given R1 + R2 = 20 Ω
and R1×R2 /(R1+R2) = 5 Ω
So R1 × R2 = 5*20 = Ω^2

(R1 - R2)^2 = (R1 + R2)^2 - 4 R1 R2
= 20^2 - 4 × 100 = 0
So R1 = R2.

So R1 = R2 = 10 Ω.