Question

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?

Answer 1

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,

$f=\frac{\omega}{2\pi }$

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 $\omega$.

We know, The maximum charge of the generator is,

$Q=CV$

Or,$Q=(35\times 10^{-6}) \times 8.0$

Or, $Q=28\times 10^{-5} C$

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

Now, for the angular frequency we can write that,

$I=\omega\times Q$

Or,$\omega=\frac{I}{Q}$

Now we put the value of $\omega$ in the first equation , we can find that

$f=\frac{\omega}{2\pi }$

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

Or,$f=\frac{1}{2\pi } \times \frac{0.50}{28\times 10^{-5} }$

Or, $f=284.21 Hz$

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

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