A 150 Omega resistor is connected in series with a 0.250-H inductor. The voltage across the resistor is v_R(3.80 V) *[cos(720*rad/ s)*t]Derive an expression for the circuit current. and Derive an expression for the voltage v_L across the inductor.I got by the equation I v/R.0253A*cos((720*rad/s)*t)and for the other i got by the equation v_LL* (di/dt) -(4.55*V)*sin((720 rad/s)*t)
The voltage across a resistor is in phase with the current and proportional to its resistor, or V_R I*R The voltage drop across an inductor will be value of the inductance, multiplied by the time derivative of the current. In math notation, this becomes V_L L*dI/dt. In a series RL circuit, the current through the inductor will be the same as that through the resistor. We are given the voltage across the resistor. The current through the resistor, and through the inductor as well, will be given by the expression I V_R / R [3.80 * cos (720t)] / 150 0.0253 cos 720t amperes The voltage across the inductor will then be V_L L*dI/dt 0.250 * d[3.80 * cos (720t) / 150] / dt 0.250/150 * 3.80 * d[cos (720t)]/dt 0.250/150*3.80*(-720)sin (720t) -4.56 sin (720t) volts, which is what you got. A note on units, breaking them down into their fundamental MKSA components: volts watts / amps (kg?m?/s?)/s / A kg?m? / (A?s?) Ohms V/I volts/amps kg?m? / (A??s?) Farads Q/V A?s / volts A??s^4 / (kg?m?) Henries V / (dI/dt) volts / (amps/s) [kg?m? / (A?s?)]?s / A kg?m? / (A??s?)
Not quite answering, but just pointing out that your sin and cos expressions could be written as (ex) sin[(720 rad/s) * t] Now when you substitute a value for t, the units inside the brackets become radians after cancellation and then all is well in the ups and downs. I think you have answered the question well, as long as your calculus is done correctly.
