1 Answer

Jeremy Herzog · Answer 1 · 2020-03-11T12:05:42+0000

Answer:

The frequency is 65.25 Hz.

Step-by-step explanation:

Given that,

Capacitor = 9.38 μF

Reactant = 260.0 ohm

We need to calculate the frequency

Using formula of reactant

$X_(c)=(1)/(2\pi fC)$

Where, f = frequency

C = capacitor

X_(c) =reactant

Put the value in to the formula

$260.0=(1)/(2*\pi* f*9.38*10^(6))$

$f=(1)/(2*\pi*9.38*10^(-6)*260.0)$

$f=65.25\ Hz$

Hence, The frequency is 65.25 Hz

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