Question

In a close circuit of resistance 10 ohm the linked flux varied with time according to relation fi= 6t2-5t+1 at t=0.25 second the current flowing through the circuit is

Answer 1

e = -dfi/dt
differentiate fi wrt to t we get
e = -(12t -5 )
at t = 0.25
e = - ((12*0.25)-5)
hence e = 2V
therfore i = eR
hence i = 2*10 =20A