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An A.C. generator generating an e.m.f of...

An A.C. generator generating an e.m.f of `epsilon =300 sin (100pi)` t is connected to a series combination of `16muF` capacitor, 1 H inductor and 100 `Omega` resistor. What is the frequency of A.C.?

A

100 Hz

B

50 Hz

C

300 Hz

D

25 Hz

Text Solution

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The correct Answer is:
To find the frequency of the alternating current (A.C.) generated by the A.C. generator, we can follow these steps: ### Step 1: Identify the given equation The equation for the electromotive force (e.m.f) generated by the A.C. generator is given as: \[ \epsilon = 300 \sin(100\pi t) \] ### Step 2: Compare with the standard form The standard form of the A.C. voltage equation is: \[ \epsilon = E_0 \sin(\omega t) \] where \(E_0\) is the peak voltage and \(\omega\) is the angular frequency. From the given equation, we can identify: - \(E_0 = 300\) volts - \(\omega = 100\pi\) rad/s ### Step 3: Calculate the frequency The relationship between angular frequency (\(\omega\)) and frequency (\(f\)) is given by: \[ \omega = 2\pi f \] To find the frequency \(f\), we can rearrange the equation: \[ f = \frac{\omega}{2\pi} \] ### Step 4: Substitute the value of \(\omega\) Now, substituting the value of \(\omega\): \[ f = \frac{100\pi}{2\pi} \] ### Step 5: Simplify the expression The \(\pi\) terms cancel out: \[ f = \frac{100}{2} = 50 \text{ Hz} \] ### Conclusion The frequency of the alternating current is: \[ \boxed{50 \text{ Hz}} \]

To find the frequency of the alternating current (A.C.) generated by the A.C. generator, we can follow these steps: ### Step 1: Identify the given equation The equation for the electromotive force (e.m.f) generated by the A.C. generator is given as: \[ \epsilon = 300 \sin(100\pi t) \] ...
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