Question

phasor28

Rp 100 Ω Given: At a frequency of 250 rad/sec, the two circuits shown above have the same impedance at terminal-pair a-b Required: Determine the values of Rp and Cp Solution: Rp Cp HF

Answer 1

C_p = X_c = 1/j w c_p Z_ab = R_p(1/j w c_p)/R-p + (1/j w c_p) = R_p/R_p j w c_p + 1 Z_ab = 100 + 1/j w(40 mu F) = 100 - j(1/w (40 mu F) (w = 250 rad/sec) z_ab(series) = 100 - j 100 Given z_ab(series) = z_ab(parallel) \r\n z_ab(parallel) = R_p/(1q + R_p j w c_p) times 1 - R_p j w c_p/(1 - R_p j w c_p) implies (a + b) (a - b) = a^2 - b^2 = R_p (1 - R_p j w c_p)/1^2 - R^2_p j^2 w^2 c^2_p = R_p - j R^2_p w C-p/1 + R^2_p w^2 c^2_p (Because j^2 = -1) separating into Real & imaginary terms) z_ab (parallel) = R_p/1 + R^2_p w^2 c^2_p - j R^2_p w c_p/1 + R^2_p w^2 c^2_p z_ab (series) = 100 - j 100 comparing Real terms, R_p/1 + R^2_p w^2 c^2_p = 100 rightarrow 1 implies R_p = 100(1 + R^2_p