Question

Use node analysis to determine the node voltages at node 1,2 and 3 and also the voltage across the 10mH inductance. Given V(t)-10c°s(30000t+π/4) volts. 10mH R1 C0.2AF R2 300Ω

Answer 1

V (t) = 10 Cos (3000 at + pi/4) = 10 angle pi/4. w = 30,000. X_L = jwL = j 30 times 10^3 times 10 times 10^-3 = j 300 Ohm. C_C = -j/wC = -j/30 times 10^3 times 0.2 times 10^-6 = -j 166.67 Ohm. By nodal analysis at node 2, KCL. I_1 + I_2 + I_3 = 0. V - 10 angle 45 degree/1000 + V/-j 166.67 + V/300 + j 300 = 0. V/1000 + V/-j 166.67 + V/300 squareroot 2 angle 45 degree = 10 angle 45 degree/1000 Voltage at node (2) rightarrow V = 1.965 angle -13.39 V. Voltage at node (1) rightarrow V (t) = 10 angle 45 degree V. Voltage at node (3) rightarrow V_3 = V times 300/300 + 300 j (Voltage division rule). V_3 = 1.309 angle -58.39 V. Voltage across inductor rightarrow V - V_3 = V_L = 1.389 angle 31.62 V.