Question

Variation of 31.56: The L-R-c Parallel Circuit. A resistor, r, and capacitor are connected in parallel to an ac source with voltage amplitude V and angular frequency co. a) Use Kirchoff s Laws, V IR, Q C/V. V Ldi/dt, and i dq/dt to determine the voltage across each element. Now, let the source voltage be given by v Vcos (ot. b) Show that the instantaneous voltages vR, Vi, and v at any instant are equal to v and that i igti tic, where i is the current through the source and iR, iL, and ic are currents through the resistor, inductor, and the capacitor. c) What are the phases of iR, iL and ic with respect to v? Use current phasors to represent i, iR, iu, and i In a phasor diagram, show the phases of these four currents with respect to v d) Use the phasor diagram of part b) to show that the current amplitude i for the current l through the 2 0.5 source is given by I R+( lc -IL) e) Show that the result of part c) can be written as EVZ with 1/2 1/R3 +(oC-1/ole)

In this question I just wondering about question (a)... Actually I soved other questoons but I don't undertand question (a).. .. help me..

Answer 1

As given the components are in series parallel so voltage across each of component in same as that of source voltage But current is different for all so by applying Kirchoff's law I_S^2 = I_R^2 + (I_L - I_C)^2 by the help of phasor diagram I_S = squareroot I_L^2 + (I_L - I_c)^2 And Z = squareroot R^2 + (X_L - X_c)^2 squareroot R^2 + (wL - 1/Wc)^2 Now voltage across each element is V = IR squareroot I_R^2 + (I_L - I_c)^2 times squareroot R^2 + (wL - 1/Wc)^2 \r\n Now by applying kirchoff's law in loop ABGHA I_RR - V = 0 V = I_R R 1 In loop BCFGB X_c I_c - I_R R = 0 X_cI_c = I_RR 2 in loop CDEFC X_LI_L - X_cI_c = 0 X_LI_L = X_CI_C 3 Now from 1 , 2 & 3 I_R R = X_LI_L = X_CI_C = V I_R R = wLI_L = 1/W_c I_C = V