A series resonant ac circuit contains a capacitance \(10^4\) F and an inductor of \(10^-4\) H. The frequency of electrical oscillations will be
- \(\frac{{10^5}}{{2\pi}}\)Hz
- \(10^3\) Hz
- \(\frac{{10}}{{2\pi}}\) Hz
- 10 Hz
The Correct Option is A
Approach Solution - 1
The resonance frequency (\( v \)) of an LC circuit is given by the formula:
\( v = \frac{1}{2 \pi \sqrt{L C}}\)
Substituting the values for the capacitance (\( C = 10^{-4} \)) and inductance (\( L = 10^{-6} \)):
\( v = \frac{1}{2 \pi \sqrt{10^{-6} \times 10^{-4}}} = \frac{1}{2 \pi \sqrt{10^{-10}}}\)
\(v = \frac{10^5}{2 \pi} \, \text{Hz}\)
Therefore, the resonance frequency of the circuit is: \( \frac{10^5}{2 \pi} \, \text{Hz} \).
Approach Solution -2
The resonant frequency \(f\) of a series resonant AC circuit containing an inductor \(L\) and a capacitor \(C\) is given by the formula:
\(f = \frac{1}{2\pi\sqrt{LC}}\)
Where:
- \(f\) is the resonant frequency in Hertz (Hz)
- \(L\) is the inductance in Henrys (H)
- \(C\) is the capacitance in Farads (F)
Given:
- \(C = 10^{-4} \, \text{F}\)
- \(L = 10^{-4} \, \text{H}\)
Substitute the given values into the formula:
\(f = \frac{1}{2\pi\sqrt{(10^{-4})(10^{-4})}}\)
\(f = \frac{1}{2\pi\sqrt{10^{-8}}}\)
\(f = \frac{1}{2\pi \times 10^{-4}}\)
\(f = \frac{10^4}{2\pi}\)
\(f = \frac{10000}{2\pi} \, \text{Hz}\)
It seems there might be a typo in the capacitance value given in the question. If the capacitance was \(10^{-4} F\) instead of \(10^4 F\), the correct resonant frequency would be \(\frac{10000}{2\pi}Hz\), but that is not among the options. Therefore, let's proceed with the assumption that the capacitance given is incorrect and intended to be \(10^{-4} F\).
If the capacitance is \(10^{-6} F\),
\(f = \frac{1}{2\pi\sqrt{(10^{-4})(10^{-6})}}\)
\(f = \frac{1}{2\pi\sqrt{10^{-10}}}\)
\(f = \frac{1}{2\pi \times 10^{-5}}\)
\(f = \frac{10^5}{2\pi} \, \text{Hz}\)
Thus, assuming a typo in the question, the closest answer would be: \(\frac{10^5}{2\pi} Hz\) is more correct if the capacitor is \(10^{-6}\)
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