Question

Show the work out but try to make simple as much as u can please

Question 2. (10+10+15-35 marks) For the circuit shown in Figure 2, the frequency of the voltage source is 500 Hz. Find: (10 marks) a) the reactance of the capacitor and inductor. b) the impedance presented to the voltage source expressed in the form zo'o. c) the RMS and peak value of the current i in the circuit. (10 marks) (15 marks) Ft 50Ω 2452 10 sin 1mH Figure 2: Circuit for Question 2

Answer 1

Given circuit a) Reactance of the capacitor X_c = 1/j w c X_c = 1/j 2 pi f_c Therefore w = 2 pi f X_c = -j/2 pi times (600) times 2 times 10^- 6 X_c = -159.15 j which is in Ohm. Reactance of the Inductor X_L = j w L X_L = j times 2 pi f times L = j times 2 pi times 500 times 10^- 3 = j pi X_L = 3.14 j Ohm \r\n (b) The impedance presented the voltage source. Node analysis at V_in - i + V_in/50 - j 159.15 + V_in/2 + pi j i = V_in (1/50 - 159.15 j + 1/2 + pi j) V_in/i = 1/(1/50 - 159.15 j + 1/2 + pi j) V_in/I = 2.083 + 3.15 j = Z Z (Impedance) = V/I = 2.083 + 3.15 j Z = 3.776 56.53 degree (c) RMS and peak value of current i