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A 35 μfμf capacitor is connected across an ac generator that produces a peak voltage of 8. 0 vv. the peak current is 0. 50 aa. part a what is the oscillation frequency?

Sagot :

The frequency of the oscillation is (f)=  284.21Hz

How do we calculate the frequency of the oscillation?

To calculate the frequency we used the formula,

[tex]f=\frac{\omega}{2\pi }[/tex]

Here , we are given,

I= The peak current across an ac generator = 0.50 A.

V= The peak Voltage across an ac generator = 8.0 V.

C= The capacitance of the capacitor = 35 μf.

We have to find the angular frequency of  an ac generator [tex]\omega[/tex].

We know, The maximum charge of the generator is,

[tex]Q=CV[/tex]

Or,[tex]Q=(35\times 10^{-6}) \times 8.0[/tex]

Or, [tex]Q=28\times 10^{-5} C[/tex]

So, The maximum charge of the generator is (Q)= [tex]28\times 10^{-5} C[/tex]

Now, for the angular frequency we can write that,

[tex]I=\omega\times Q[/tex]

Or,[tex]\omega=\frac{I}{Q}[/tex]

Now we put the value of [tex]\omega[/tex] in the first equation , we can find that

[tex]f=\frac{\omega}{2\pi }[/tex]

Or,[tex]f=\frac{1}{2\pi } \times \frac{I}{Q}[/tex]          [Note: from above[tex]\omega=\frac{I}{Q}[/tex]]

Or,[tex]f=\frac{1}{2\pi } \times \frac{0.50}{28\times 10^{-5} }[/tex]

Or, [tex]f=284.21 Hz[/tex]

Therefore, from the above calculation we can conclude that the frequency of the oscillation is 284.21 Hz.

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