Chapter 23 ---------- 1). The Capacitive Reactance is given by (see formula 23.2) X_C = 1/(2*pi*f*C) where f is the frequency and C is the capacitance. ---> f = 1/(2*pi*X_C*C) = 1/(2*3.14*168 Ohm * 7.5*10^(-6) F) = 126 Hz 4). The capacitive reactance X_C is: X_C = 1/(2*pi*f*C) The capacitance is: ---> C = 1/(2*pi*f*X_C) = 1/(2*3.14*170 Hz * 36 Ohm) = 2.6 10^(-5) F Using now the capacitance we can get capacitive reactance at a new frequency: X_C = 1/(2*pi*f*C = 1/(2*3.14*510 Hz * 2.6 10^(-5) F) = 12 Ohm Another, quicker way to arrive at the solution is to notice that the capacitive reactance is inversely proportional to the frequency. Thus X_C1/X_C2 = f2/f1 ==> X_C2 = X_C1*f1/f2 = 36 Ohm*(170 Hz/510 Hz) = 12 Ohm 10). The Ohm's law for the inductor is V = I * X_L , where I is the current through the inductor, X_L is the inductive reactance. Since X_L = 2*pi*f*L ---> V = I*2*pi*f*L = 2*3.14*750 Hz*0.08 H*0.2 A = 75.3 V 18). Each component of the system has a resistance or reactance. The voltage across each component depends on the current flowing in the circuit. The value of the current depend on the total impedance. First of all let's find the total impedance of the system: Z = sqrt(R^2+(X_L-X_C)^2) where X_L is a reactance of the inductor, and X_C is the reactance of the capacitor. X_L = 2*pi*f*L = 2 * 3.14 * 1350 Hz * 5.3*10^(-3) H = 44.9 Ohm X_C = 1/(2*pi*f*C) = 1/(2*3.14*1350Hz*4.1*10^(-6)F) = 28.8 Ohm Z = sqrt((16 Ohm)^2 + (44.9 Ohm - 28.8 Ohm)^2) = 22.7 Ohm The current in the circuit is equal to I = V/Z = 15 V / 22.7 Ohm = 0.66 A Potential across the resistor is V=I*R = 0.66 A * 16 Ohm = 10.6 V --//-- capacitor is V=I*X_C = 0.66 A * 28.8 Ohm = 19 V --//-- inductor is V=I*X_L = 0.66 A * 44.9 Ohm = 29.6 V Note that the generator voltage 15 V is NOT equal to the sum of the voltages across the elements in the circuit, though they are connected in series.