An unknown circuit element 'X' is connected to an AC source v = V_m ( t) . The instantaneous power P(t) supplied by the source to the circuit is found to be P(t) = -P₀ (2 t) , where P₀ is a positive constant. What is the nature of the circuit element 'X'?

Step-by-step solution

The instantaneous power is given by P(t) = v(t)i(t) . Let the steady-state current be i(t) = I_m ( t + ) . P(t) = V_m ( t) I_m ( t + ) Using the trigonometric identity 2 A B = (A-B) - (A+B) , we get: P(t) = V_m I_m 2 [ ( ) - (2 t + ) ] Comparing this with the given power P(t) = -P₀ (2 t) : The constant term must be zero, so ( ) = 0 = 2 . The time-varying term is - V_m I_m 2 (2 t + ) . For this to equal -P₀ (2 t) , we must have (2 t + ) = (2 t) . Since (2 t - 2 ) = (2 t) , we get = - 2 . A phase angle of = - 2 means