Question

The power required for dielectric heating of a slab of 150 centimeter square in area and 2
centimeter thick is 200 W with frequency 30 MHz. The material has relative permittivity of
5 and power factor 0.05. Determine the voltage necessary and current flowing through the
material. If the Voltage is limited to 600 V, What will be value of frequency to obtain the
same heating?​

Answer 1

Answer:

The capacitance of the parallel-plate capacitor formed by the insulating slab is C = ε0εrA/d = (8.854 x 10-12 x 5 x 150 x 10-4)/(2 x10-2) = 33.2 x 10-12F Rsh = 1/(ωC x p.f.) = 1/((2π x 30 x 106) x 33.2 x 10-12 x 0.05) = 3196Ω Now, P = V2/Rsh or V = √(P x Rsh) = √(200 x 3196) = 800V Current I = V/XC = ωCV = (2π × 30 × 106 ) × 33.2 × 10−12 × 800 = 5 A Now, as seen from above P = V2/Rsh = V2/((1/ωC) x p.f.) = V2ωC x p.f. or P ∝ V2f ∴ 8002 x 30 = 6002 x f or f = (800/600)2 x 30 = 53.3