ANSWER TO CHAPTER 31 HOMEWORK

1.  (a)  We find the impedance from

                           Z = X_C = 1/2pfC = 1/2p(600Hz)(0.036 x 10^-6F) = 7.37 x 10^-6W = 7.4kW.

         (b)  We find the peak value of the current from

                           I_peak = Ö2I_rms =  Ö2(V_rms/Z) =  Ö2(22kV)(7.37kW) = 4.2A

                The frequency of the current will be the frequency of the line:  600Hz.

2.  The reactance and impedance in the circuit are

                          X_L = 2pfL = 2p(60Hz)(35 x 10^-3H) = 13.2W.

                          X_C = 1/2pfC = 1/2p(60Hz)(20 x 10^-6F) = 133W.

                          Z =[R² + (X_L - X_C)²]^1/2 = [(2.0W)² + (13.2W - 133W)²]^1/2 = 119W.

          (a)  The rms current is

                          I_rms = V_rms/Z = (45V)/119W) =  0.38A.

          (b)  We find the phase angle from

                          cosf = R/Z = (2.0W)/(119W) = 0.0168.

                 Because X_C > X_L, the current leads the voltage, so f =  -89°.

          (c)  The power dissipated is

                         P = I_rms²R = (0.38A)²(2.0W) = 0.29W.

3.   Because the circuit is in resonance, we find the inductance from

                        f_o = (1/2p)(1/LC)^1/2;

                       18.0 x 10³Hz = (1/2p)[1/L(220 x 10^-6F)], which gives L = 3.55 x 10^-7H.

          If r is the radius of the coil, the number of turns is

                       N = l_wire/2pr.

          If d is the diameter of the wire, for closely-wound turns, the length of the coil is

                       l = Nd.

         Thus the inductance of the coil is

                      L = m_oAN²/l = m_opr²(l_wire/2pr)²/Nd = m_ol_wire²/4pNd;

                     3.55 x 10^-7H = (4p x 10^-7T.m/A)(12.0m)²/4pN(1.1 x 10^-3m), which gives N =  3.68 x 10⁴ turns

4.  (a) From the expression for V, we see that V_o = 0.95V, and 2pf = 754s^-1.

                      X_C = 1/2pfC = 1/(754s^-1)(0.30 x 10^-6F) = 4.42 x 10³W = 4.42kW.

                      X_L = 2pfL = (754s^-1)(0.0220H) = 16.6W = 0.0166kW.

              The impedance of the circuit is

                       Z = [R² + (X_L - X_C)²]^1/2 = [(23.2kW)² + (0.0166kW - 4.42kW)²]^1/2 =  23.6kW.

              We find the phase angle from

                       tanf = (X_L - X_C)/R = (0.0166kW - 4.42kW)/(233.2kW) = -0.191, which gives f = -10.8°.

          (b)  The power dissipated is

                       P = I_rms²R = (V_rms/Z)²R = (V_o/ZÖ2)²R = 1/2(V_o/Z)²R

                           = 1/2(0.95V/2.36 x 10³W)²(23.2 x 10³W) =  1.88 x 10⁵W.

          (c)  The rms current is

                      I_rms = I_oÖ2 = V_o/ZÖ2 = (0.95V)/(23.6 x 10³W)Ö2 = 2.846 x 10^-5A =  2.8 x 10^-5A.

                 The rms reading across the elements are

                      V_R = I_rmsR = (2.85 x 10^-5A)(23.2 x 10³W) =  0.66V.

                     V_L = I_rmsX_L  = (2.85 x 10^-5A)(16.6W) =  4.7 x 10^-4V.

                      V_C = I_rmsX_C = (2.85 x 10^-5A)(4.42 x 10³W) = 0.126V.