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Engineering

Physics

Physical Quantities

Question

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550Ω

910Ω

990Ω

505Ω

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$\frac{R_{1}}{R_{2}} = \frac{x}{100 - x}$

$\frac{R_{2}}{R_{1}} = \frac{x - 10}{100 - (x - 10)} = \frac{x - 10}{110 - x}$

$(\frac{x}{100 - x}) (\frac{x - 10}{110 - x}) = 1$

x² – 10x = 11000 – 210x + x²

200x = 11000

x = 55

R₁ = 550