Answer:
Its inductance L = 166 mH
Explanation:
Since a current, I = 0.698 A is obtained when a voltage , V = 5.62 V is applied, the resistance of the coil is gotten from V = IR
R = V/I = 5.62/0.698 = 8.052 Ω
Since we have a current of I' = 0.36 A (rms) when a voltage of V' = 35.1 V (rms) is applied, the impedance Z of the coil is gotten from
V₀' = I₀'Z where V₀ = maximum voltage = √2V' and I₀ = maximum current = √2I'
Z = V'/I' = √2 × 35.1 V/√2 × 0.36 V = 97.5 Ω
WE now find the reactance X of the coil from
Z² = X² + R²
X = √(Z² - R²)
= √(97.5² - 8.05²)
= √(9506.25 - 64.8025)
= √9441.4475
= 97.17 Ω
Now, the reactance X = 2πfL where f = frequency of generator = 93.1 Hz and L = inductance of coil.
L = X/2πf
= 97.17/2π(93.1 Hz)
= 97.17 Ω/584.965 rad/s
= 0.166 H
= 166 mH
Its inductance L = 166 mH
