In a pure inductive circuit, a sinusoidal voltage V(t)=200 250 t is applied to a pure inductance of L =0.02 H . The current through the coil is:

Step-by-step solution

The given voltage is V(t) = 200 (250t) . Comparing this with V(t) = V_m ( t) , we have V_m = 200 V and = 250 rad/s . The inductive reactance X_L is given by X_L = L . Substituting the given values, X_L = 250 0.02 = 5 . The peak current I_m is given by I_m = V_m X_L = 200 5 = 40 A . In a pure inductive circuit, the current lags the voltage by a phase angle of 2 . Therefore, the expression for current I(t) is I(t) = I_m ( t - 2 ) . Substituting the values, I(t) = 40 (250t - 2 ) . Answer: 40 [250 t- 2 ]