Question

Two resistances are in the ratio 1:2. If these are connected in parallel,their equivalent resistance becomes 8 ohm.Calculate the value of each resistance.

Answer 1

Given that, two resistances are in the ratio 1:2 and both of them are connected in parallel.

Let us assume that, first resistance i.e. R1 is 1x and second resistance i.e. R2 is 2x.

If R1 and R2 are connected in parallel then, their equivalent resistance becomes 8Ω.

1/Rp = 1/R1 + 1/R2

1/8 = 1/1x + 1/2x

1/8 = (2x + 1x)/2x²

1/8 = 3x/2x²

Cross-multiply them

1(2x²) = 8(3x)

2x² = 24x

x² = 12x

Divide by x on both sides

x²/x = 12x/x

x = 12

Therefore,

R1 = 1x = 1(12) = 12Ω

R2 = 2x = 2(12) = 24Ω

Answer 2

Given :-

• 2 resistances are in the ratio 1 : 2.

• The resistances are connected in parallel, due to which, the equivalent resistance becomes 8 Ω.

To find :-

The value of each resistance.

Solution :-

Let the first Resistance, R₁ = 1x and the second resistance, R₂ = 2x.

As they are connected in parallel, so the equivalent resistance becomes :

$\displaystyle{\sf{\frac{1}{1x}+\frac{1}{2x}=R_{Equivalent} }}$

$\text{According to the question,}\\\\\displaystyle{\sf{\implies \frac{1}{x}+\frac{1}{2x}= }\frac{1}{8} }$

$\displaystyle{\sf{\implies \frac{2+1}{2x}=\frac{1}{8} }}\\\\\displaystyle{\sf{\implies \frac{3}{2x}=\frac{1}{8} }}$

$\displaystyle{\sf{\implies 2x=24}}\\\\\displaystyle{\sf{\implies x=12}}$

∴ So, the value of R₁ is 12Ω and the value of R₂ is 2*12 = 24Ω.