Answer to Question 2

The mutual inductance between two coils is 240 mH. The current in Coil 1 changes from 15 A to 6.0 A in 12 ms. Calculate: (a) The average EMF induced in Coil 2; (b) The change of flux linked with Coil 2 which consists of 400 turns.

(a) Formula:
Rate of change of current = (6.0 A  15.0 A) ÷ 12 × 10^{3} s =  750 A s^{1}
E = 240 × 10^{3} H ×  750 A s^{1} = 180 V
(b) Formula:
DF = 9.0 A × 0.24 H ÷ 400 = 5.4 × 10^{3} Wb = 5.4 mWb 