(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