Question

EXERCISE 2: USING PHASORS FOR A DRIVEN RC CIRCUIT 10. On the middle set of axes, draw phasors for k and ε using the provided ve phasor as a basis for a circuit in which R- 2Xc. Start by deciding what the peak voltage across the capacitor must be compared to the peak voltage across the resistor and then draw the appropriate Vc phasor We will consider a circuit consisting of a time varying emf in series with a capacitor and resistor. The capactance and resistance are flxed but we can tune the frequency of the source. In an AC circut, each element has a reactance" which can be used to relate the peak current through the element to the peak voltage across the element 9, What does the angular frequency of the emf have to be for the capactor and resstor have the same peak voltage (i.e.for Rand X be the same) 11. On the set of axes on the right, draw phasors for ve and ε using the provided V. phasor as a to basis for a circuit in which R-Xc/2 " Phasors can be used to represent the different parameters for an AC circuit. 12, For each set of phasors, is the capacitor charg ngor discharging at the given instant? How did you determine this? At the angular frequency found in Q9, the phasor diagram (for a general time) is shown in case ( The instantaneous values of the voltages and currents are the horizontaw components of each The lengths of the phasors are the maximum value each parameter can have (the length of the in the figure below phasor current compared to the other phasors is arbitrary The peak emf & the hypotenuse of a triangle formed with the peak voltage for the resistor and capactor Your answers to Q10-13 can be placed directly in the table below. (iN) 0) R=Xc R= Xc/2 13. Which emf has the largest angular frequency? How did you determine this?

Answer 1

R = X_L with frequency f_h as we know X_L = 2 pi f times L R = 2 times pi times f_6 times L implies f_6 = R/2 pi f L as R = X_L implies V_R = V_L Angular frequency w_6 = 2 pi f_6 = R/L - (1) R = 2 times L as X_L = 2 pi f_7 middot L R = 2 times (2 pi f_7) times L = 4 pi f_7 middot L w_7 = R/2L = w_6/2 V_R = R/R + J times L middot E = R/ times L + J times L E_m angle 0 degrees =/+ J E_m angle 0 degrees = 0.894 E_m angle-26.56 as R = 2 times L V_L = J V R/2 = 0.4472 E_m angle -26.56 + 90 = 0.4472 E_m angle 63.44 degrees From the above, if can be that V_R log & V_s by 26.56 degrees and V_L leads V_s by 63.44 degrees