Question

The peak amplitude of the sinusoidal voltage source in the circuit shown in the figure is 210 V , and its frequency is 5000 rad/s. The load resistor can be varied from 0 to 4000 Ω, and the load capacitor can be varied from 0.1 μF to 0.5 μF. (Figure 1)

Ro 0.6 H 1 12 ks

Part A. Calculate the average power delivered to the load when Ro=2000 Ω and Co=0.2 μF.

Part B. Determine the settings of Ro and Co that will result in the most average power being transferred to Ro. Find Ro.

Part C. Find Co.

Part D. What is the most average power in Parts B and C?

Part E.If there are no constraints on Ro and Co, what is the maximum average power that can be delivered to a load?

Part F. What is the value of Ro for the condition of Part E?

Part G. What is the value of Co for the condition of Part E?

Please answer all parts with numbers not just variables.

Answer 1

Given the circuit Z_L = j omega L = j 2pifL = j 2pi times 5000 times 0.6 = j 18.85 j Ohm Find V_th: - V_th = V_g times 12 times 10^3/6 times 10^3 + 12 times 10^3 = 210 times 2/3 = 140V Find Z_th: - Z_th = 61/12 + j 18.85 = 6 times 12/6 + 12 + j 18.85 = (4 + j 18.85) k Ohm \r\n Part A: - Z_C_0 = -j/omega c = -j/2 pi fc = -j/2pi times 5000 times 0.2 times 10^-6 = j 159.16 Ohm I = V_th/-Z_th + R_0 + Z_th = 140/(4 + j 18.85)k + 2000 + (-j 159.16) = (7.13 angle -72.2) mA Average power = Vertical-bar I/squareroot 2 Vertical-bar ^2 times R_0 = (7.13 times 10^-3/squareroot 2)^2 times 2000 = 50.84 m omega Part (B): - max average power transfer from source to load when Z_L = Z*_th Z_L = (4 + j 18.85)* = (4 - j18.85) k Ohm I = V_th/Z_th + Z_L