Question

What is the capacitance when capacitor passes a current of 31.43ma whensupplied with 40v ac with a frequency of 100 hz

Answer 1

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Answer 2

Concept:

The property of an electric conductor, or set of conductors, is measured by the amount of separated electric charge that can be stored on it per unit change in electrical potential.

Given:

Current = 31.43 mA

Voltage = 40 V

Frequency = 100 Hz

Find:

We need to determine the capacitance of a capacitor

Solution:

It has been given to us that the voltage across the capacitor is 40V

Therefore, Voltage × √2πfC = I where, I is the representation of current, f for frequency and C for capacitance

Since the current is given in milliampere, it first needs to be converted into ampere.

Therefore, I = 31.43 mA = 31.43 × 10⁻³ A

Therefore, the equation of voltage, V × √2πfC = I becomes-

√2πfC = I/V

√2πfC = 31.43 × 10⁻³/40

√2πfC = 7.8575 × 10⁻⁴

Taking square on both sides

2πfC = (7.8575 × 10⁻⁴)²

2πfC = 61.74 × 10⁻⁸

C = 61.74 × 10⁻⁸/2πf

C = 61.74 × 10⁻⁸/2 × 3.14 × 100

C = 0.09831 × 10⁻⁸

C = 9.8 × 10⁻¹⁰ F

Thus, the capacitance of a capacitor is  9.8 × 10⁻¹⁰ F.

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