An ac generator has a frequency of 1110 Hz and a constant rms voltage. When a 4560 resistor is connected between the terminals of the generator, an average power of 0.270 W is consumed by the resistor. Then, a 0.0850H inductor is connected in series with the resistor, and the combination is connected between the generator terminals. What is the average power consumed in the inductorresistor series circuit? Number n Units v When only the resistor is present, the average power is P = Imms R, and the current is given by Irms = Vrms / R, according to rms = Irms R. Therefore, we find in this case that Resistor only P = 12 R = rms R = rms R R When the inductor is also present, the average power is still P = Im R, but the current is now given by Irms = Vims/ Z, according to Vrms = Irms Z, where Z is the impedance. In this case, then, we have Vrms 2 V2 Resistor and inductor P = 17 R = Z R = Z2 Dividing this equation by the analogous result for the resistor-only case, we obtain P Resistor and inductor V2 RIZZ R- P Resistor only V2 IR Z2 In this expression, Z is the impedance and is given by equation Z = \\ R2 + (X1 - Xc)as Z = \\ R2 + X}, where XL is the inductive reactance and is given by X1, = 2x f L. With these substitutions, the ratio of the powers becomes P Resistor and inductor R2 P Resistor only R2 + ( 2 n f 1 ) 2 The power consumed in the series combination is P Resistor and inductor = P Resistor only R2 R' +( 2x / 1)?] (456 2)2 = (0.270 W) 0.100 W (456 2)-+[2x(1 1 10 Hz)(0.0850)12 As expected, more power is consumed in the resistor-only case