An a. c. voltage is applied to pure inductor. The current in the inductor

Step-by-step solution

For a pure inductor with inductance L connected to an AC voltage source V = V₀ ( t) , the inductor voltage is V_L = L dI dt . Equating the applied voltage and inductor voltage gives: V₀ ( t) = L dI dt Solving for the current by integration: dI = V₀ L ( t) dt I = V₀ L ( t) dt = V₀ L ( - ( t) ) = - V₀ L ( t) Using the identity - = ( - 2 ) , the current becomes: I = V₀ L ( t - 2 ) Letting I₀ = V₀ L denote the peak current: I = I₀ ( t - 2 ) Comparing with the voltage V = V₀ ( t) , the current lags by a phase angle of 2