Question:

A slab of insulating material 130 cm\(^2\) in area and 1 cm thick is to be heated by dielectric heating. The power required is 380 W at 30 MHz. The material has a relative permittivity of 5 and a power factor of 0.05. Determine the necessary voltage.

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For dielectric heating, the required voltage is calculated using: \[ V = \sqrt{\frac{P}{\pi f C \tan \delta}} \] where capacitance is derived from material properties.
Updated On: Feb 10, 2025
  • 837 kV
  • 837 V
  • 652 V
  • 552 V
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The Correct Option is B

Solution and Explanation

Step 1: Dielectric Heating Power Formula The power dissipated in dielectric heating is given by: \[ P = \pi f V^2 C \tan \delta \] where:
- \( P \) = power (380 W),
- \( f \) = frequency (30 MHz = \( 30 \times 10^6 \) Hz),
- \( V \) = applied voltage (to be determined),
- \( C \) = capacitance of the dielectric material,
- \( \tan \delta \) = power factor (0.05). 
Step 2: Capacitance Calculation The capacitance of a parallel plate capacitor is given by: \[ C = \frac{\varepsilon_0 \varepsilon_r A}{d} \] where:
- \( \varepsilon_0 \) = permittivity of free space (\( 8.854 \times 10^{-12} \) F/m),
- \( \varepsilon_r \) = relative permittivity (5),
- \( A \) = area (\( 130 \times 10^{-4} \) m\(^2\)),
- \( d \) = thickness (0.01 m). \[ C = \frac{(8.854 \times 10^{-12} \times 5) \times (130 \times 10^{-4})}{0.01} \] \[ C = 5.75 \times 10^{-12} \text{ F} \] 
Step 3: Voltage Calculation Rearranging the power equation: \[ V = \sqrt{\frac{P}{\pi f C \tan \delta}} \] Substituting the values: \[ V = \sqrt{\frac{380}{\pi \times (30 \times 10^6) \times (5.75 \times 10^{-12}) \times 0.05}} \] \[ V = \sqrt{\frac{380}{\pi \times 8.625 \times 10^{-5}}} \] \[ V = \sqrt{1.4 \times 10^7} \] \[ V \approx 837 V \] 
Step 4: Evaluating options: 
- (A) Incorrect: 837 kV is an overestimation.
- (B) Correct: 837 V matches the computed result.
- (C) Incorrect: 652 V is too low.
- (D) Incorrect: 552 V is incorrect.

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