In a particular R-L series circuit a voltage of 10V at 50 Hz produces

1.	In a particular R-L series circuit a voltage of 10V at 50 Hz produces a current of 700mA while the samevoltage at 75 Hz produces 500 mA. What are the values of circuit constants?
Answer» (i) Z = √(R² + (2π x 50L)²) = √(R² + 98696L²); V = IZ or 10 = 700 x 10^-3√(R² + 98696L²) √(R²+ 98696L²) = 10/700 × 10⁻³ = 100/7 or R² + 98696 L² = 10000/49 ...(i) (ii) In the second case Z = √(R² + (2π x 75L)²) = √(R² + 222066L²) ∴ 10 = 500 x 10^-3√(R² + 222066L²) i.e √(R² + 222066L²) = 20 or R² + 222066L² = 400 (ii) Subtracting Eq. (i) from (ii), we get 222066 L²− 98696 L² = 400 − (10000/49) or 123370 L² = 196 or L = 0.0398 H = 40 mH. Substituting this value of L in Eq. (ii), we get, R² + 222066 (0.398)² = 400 ∴ R = 6.9 Ω

In a particular R-L series circuit a voltage of 10V at 50 Hz produces a current of 700mA while the samevoltage at 75 Hz produces 500 mA. What are the values of circuit constants?

Answer»

(i) Z = √(R² + (2π x 50L)²) = √(R² + 98696L²);

V = IZ or 10 = 700 x 10^-3√(R² + 98696L²)

√(R²+ 98696L²) = 10/700 × 10⁻³ = 100/7

or R² + 98696 L² = 10000/49 ...(i)

(ii) In the second case Z = √(R² + (2π x 75L)²) = √(R² + 222066L²)

∴ 10 = 500 x 10^-3√(R² + 222066L²) i.e √(R² + 222066L²) = 20 or R² + 222066L² = 400 (ii)

Subtracting Eq. (i) from (ii), we get

222066 L²− 98696 L² = 400 − (10000/49) or 123370 L² = 196

or L = 0.0398 H = 40 mH.

Substituting this value of L in Eq. (ii), we get, R² + 222066 (0.398)² = 400

∴ R = 6.9 Ω

Discussion

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