Question

connected in series across a standard h rm4 con si era loa a 50 aF capacitor and . 80 mii ductor (6 points each) outlet (a) By what angle is the voltage ahead of the current in this circuit? (b) What is the rms voltage across the capacitor? (c) What is the power dissipated in the inductor (only the inductor), in the capacitor and in the resistor? There are three answers?

Answer 1

In the given problem R = 10 Ohm C = 50 mu F L = 80 mH (a) The phase angle between current & voltage is given by tan phi = X_L - X_C/R Here voltage leads the current with angle & The phasor diagram will be X_L = W L = 2 pi v L = 2 pi (50) 80 mH = 2 pi times 50 times 80 times 10^- 3 H X_L = 25.132 Ohm X_C = 1/2 pi v C = 1/2 pi times 50 times 50 times 10^- 6 F = 63.66 Ohm \r\n R = 10 Ohm Therefore tan phi = 25.132 - 63.33/10 tan phi = -3.8528 phi = -75.45 degree Voltage lags behind the current by 75.45 (b) V_rms = 110 V The resultant impedance of the LCR circuit is Z = squareroot R^2 + (X_L - X_C)^2 = squareroot 100 + (25.132 - 63.66)^2 Z = 39.80 TildeEqual 40 Ohm