Question

Item 15 < 15 of 15 A 1.15-k22 resistor and a 600-mH inductor are connected in series to a 1350-Hz generator with an rms voltage of 122 V. Part A What is the rms current in the circuit? TO AED PE IF ? Irms = 2.33 mA Submit Request Answer Part B What capacitance must be inserted in series with the resistor and inductor to reduce the rms current to half the value found in part A? EVO AED F ? C = nF Submit Request Answer

A 1.15-kΩ resistor and a 600-mH inductor are connected in series to a 1350-Hz generator with an rms voltage of 12.2 V .

Part A: What is the rms current in the circuit?

Part B:What capacitance must be inserted in series with the resistor and inductor to reduce the rms current to half the value found in part A?

Please show with steps, thank you!

Answer 1

12.2 2,199 x 10-3 (1.15X103)2 +1508608 8978C 2 12.2 → 1.15x103 + so 86-0- 84 78C 7:1996/03 taking square of both sides 1•3225 X 1067 5086.8- 178.89 T-4376 x 10-6 84780 5086.84 8478 C |(508 = 103.5337 X 106–1:3225 X 106 taking square root. of both sides - 5086.8 = 102.21x106 84786

य | 50866 - 1010-9.9 84780 50 23.। 84180 f 844825पर c- 23.5x10 110-9 F = 23.5 nF Ko_capacitance insented capaciton_is_23.5 nF . है