Question

A parallel-plate capacitor having square plates 4.50 cm on each side and 8.00 mm apart is placed in series with an ac source of angular frequency 650 rad/s and voltage amplitude 22.5 V, a 75.0 ? resistor, and an ideal solenoid that is 9.00 cm long, has a circular cross section 0.500 cm in diameter, and carries 125 coils per centimeter. What is the resonance angular frequency of this circuit?

Answer 1

Given data, Side of square plate, s 4.50 cm Distance between plates. D 8.00 mm Length of ideal solenoid, 1-9.00 cm Diameter of cross section of solenoid,d Number of coils per centimeter, n 125 0.500 cm Calculate the inductance of an induced solenoid Substitute the nl for M Substitute the known values 0.5 L = (4TX 10-)(125x9)' = (34.66×10-3) H 34.66 mH Calculate the resonance angular frequency. LC Substitute | for Substitute the known values 34.66x10)* 8.85x10-12x4.5x 4.5 8.00x10 = 1.2688×10° rad/s Therefore, the resonance angular frequency is 1.2688x10 rad/s