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Two parallel plate capacitors are identical, except that one of them is empty and the other contains a material with a dielectric constant of 4.2 in the space between the plates. The empty capacitor is connected between the terminals of an ac generator that has a fixed frequency and rms voltage. The generator delivers a current of 0.29 A. What current does the generator deliver after the other capacitor is connected in parallel with the first one?

Respuesta :

Answer:

current = 1.51 A

Explanation:

Initially the capacitor without any dielectric is connected across AC source

so the capacitive reactance of that capacitor is given as

[tex]x_c = \frac{1}{\omega c}[/tex]

now we have

[tex]i = \frac{V_{rms}}{x_c}[/tex]

here we know that

[tex]i = 0.29 A[/tex]

now other capacitor with dielectric of 4.2 is connected in parallel with the first capacitor

so here net capacitance is given as

[tex]c_{eq} = 4.2c + c = 5.2c[/tex]

now the equivalent capacitive reactance is given as

[tex]x_c' = \frac{1}{\omega(5.2c)}[/tex]

[tex]x_c' = \frac{x_c}{5.2}[/tex]

so here we have new current in that circuit is given as

[tex]i' = \frac{V_{rms}}{x_c'}[/tex]

[tex]i' = 5.2 (i) = 5.2(0.29)[/tex]

[tex]i' = 1.51 A[/tex]

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