Question

2) In class we considered the circuit shown belovw a) Use the result for the current derived in class to find the voltage Vı across the resistor. b) Find the average power dissipated by the circuit using the equations developed in class and the source voltage Vo(t) and current I(t) c) All the power must be dissipated in the resistor. Check that using just the current and voltage across the resistor gives you the same answer for power dissipation as in part b. d) This circuit is called a high pass circuit. Plot the ratio of the magnitudes of the amplitudes Vil/Vol vs. frequency on a log scale for frequencies between 1Hz and 100kHz with RC 0.001 S. 0 0 l(t) 0

Answer 1

I use the result for the current derived in class to find the voltage v, across the resister capacitance reactance x_c = 1/cw z = squareroot e^2 + xc^2 I(t) = v_n(t)/z v_1 = IR v_1 = v_0/z times R v_1 = v_0 times R/squareroot R^2 + xc^2 b) find the average power dissipated by the circuit using the equation developed in class and the source voltage v_0(t) current I(t) \r\n Pouler dissipated pd = v_0 I cos emptyset = v_0 times v_0/z times R/z pd = v^2_0/z^2 times R others way pd = I^2R partial differential = v^2_0/z^r R S, All the power most be dissipated in the resistor chock that using tact the current and voltage across the resistor gives you the same answer for power dissipation dissipation pd = v_1 I pd = v_0 R/z times v_0/z pd = v_0 R/z times v_0/z pd = v^2_0/z^2 R D) This circuit is called a high pees circuit plot the ratio