Question

Problem 6.1 PSpice Multisim PartA The current in a 150 H inductor is known to be i seconds. Assume the passive sign convention 25teA for t0, where t is in Find the voltage across the inductor fort>0 O 7.50e (1 500t mV O 3.75e 250t) mV O 7.50e-50"(1-250t) mV O 3.75e SKM (1- 500t) mV O 7.50e-00t) mV 3.75e(+500) mV O 7.50e 250t) mV Submit PartB Find the power at the terminals of the inductor when t 6.6 ms Express your answer to three significant figures and include the appropriate units. pValue Units Submit Part Is the inductor absorbing or delivering power at 6.6 ms? O absorbing power O delivering power Submit

Answer 1

I_h (t) = 25t e^-500t A L = 150 mu H v_h(t) = h di_h (t)/dt = 150 mu d/dt (25 t e^-500 t) = 150 mu times 25 [-500 t e^-500 t + e^-500 t] v_h (t) = 3.75 e^-500 t (1 - 500 t) mv equation (A) i_h (6.6 m sec) = 6.086 mA v_h (6.6 m sec) = -0.318 mv therefore P_h (6.6 m sec) = -1.936 mu Joules equation (B) power is nature implies inductor is delivering power @ E = 6.6 m sec equation (c) E_L (6.6 m sec) = 1/2 times h times i_h (6.6 m)^2 = 1/2 times 150 mu times (6.086 m)^2 E_L (6.6 m sec) = 2.778 n Joules equation (D) \r\n di_h/dt = 0 25 e^-500 t - 25 times 500 te^-500 t = 0 25 e^-500 t = 25 times 500 t e^-500t t = 1/500 = 2 m sec t = 2m sec equation (f) i_h (2 m sec)