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A charged \(30 \mu \mathrm{F}\) capacitor is connected to a \(27 \mathrm{mH}\) inductor. What is the angular frequency of free oscillations of the circuit?

Short Answer

Expert verified
The angular frequency \( \omega \) of the free oscillations of the circuit is approximately \( 1.1223 \times 10^5 \) rad/s.

Step by step solution

01

Convert the values to correct units

Convert the given values to the correct units. The capacitance given is in \(\mu \mathrm{F}\) (microfarads), which needs to be converted to F (farads) and the inductor value is in \(\mathrm{mH}\) (millihenrys), which needs to be converted to H (henrys). To make the conversion, remember that 1F = \(10^6\) \(\mu \mathrm{F}\) and 1H = \(10^3\) \(\mathrm{mH}\). So, \(C = 30 \times 10^{-6} F\) and \(L = 27 \times 10^{-3} H\).
02

Calculation of angular frequency

Substitute the converted values of L and C into the formula \( \omega = \frac{1}{\sqrt{LC}} \) to get the angular frequency of the free oscillations of the circuit. So, \[\omega = \frac{1}{\sqrt{30 \times 10^{-6} \times 27 \times 10^{-3}}}\]. Calculating this will give the required angular frequency.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Angular Frequency in LC Circuits
In LC circuits, understanding angular frequency is crucial for predicting how these circuits behave in terms of oscillations. The angular frequency, usually denoted as \( \omega \), determines how quickly the circuit oscillates. It is related to the natural frequency of oscillation of the LC circuit.
Angular frequency is particularly important in contexts involving alternating current circuits and signals. Mathematically, it is given by the expression:
  • \( \omega = \frac{1}{\sqrt{LC}} \)
where \( L \) is the inductance and \( C \) is the capacitance.
This formula reflects the inverse relationship with both inductance and capacitance. Hence, a larger inductance or capacitance typically results in a lower angular frequency.
The unit of angular frequency is radians per second, which signifies how many cycles of oscillation happen within a second but expressed in radians, a unit of angular measure rather than pure cycles.
Capacitance Conversion Explained
In physics and engineering, capacitance must often be converted into standard units to ensure compatibility and consistency in calculations.
The given problem describes a capacitance of \(30 \, \mu \text{F} \), which is provided in microfarads (\( \mu \text{F}\)). However, in most scientific calculations, capacitance is expressed in farads (F).
To convert microfarads to farads, we recall:
  • 1 farad (F) = \(10^6 \) microfarads (\( \mu \text{F} \))
Thus, to perform the conversion:
  • \(30 \, \mu \text{F} = 30 \times 10^{-6} \text{F}\)
By converting to standard units, calculations can proceed accurately using formulas like the angular frequency formula previously mentioned.
Using standardized units reduces errors and simplifies mathematical operations, ensuring that the results are universally understood and accepted.
Inductance Conversion Details
Similarly, just like capacitance, inductance values are also often given in millihenrys (\( \text{mH} \)). For accurate scientific calculations, these values need to be converted into henrys (H).
Inductance, represented by \( L \), plays a central role in determining the behavior of LC circuits. It is measured in henrys, but small inductors are often specified in millihenrys.
The conversion is straightforward:
  • 1 henry (H) = \(10^3 \) millihenrys (\( \text{mH} \))
For example, in the problem, the inductor's value is 27 mH. The conversion to henrys would be:
  • \(27 \text{mH} = 27 \times 10^{-3} \text{H}\)
Accurate conversion is crucial to ensure the precision of results when calculating the angular frequency or other related electrical properties.
Applying these conversions helps to maintain consistency in scientific calculations and allows for seamless application of universal physics equations across various problem sets.

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Most popular questions from this chapter

(a) The peak voltage of an ac supply is \(300 \mathrm{~V}\). What is the rms voltage? (b) The rms value of current in an ac circuit is \(10 \mathrm{~A}\). What is the peak current?

A small town with a demand of \(800 \mathrm{~kW}\) of electric power at \(220 \mathrm{~V}\) is situated \(15 \mathrm{~km}\) away from an electric plant generating power at \(440 \mathrm{~V}\). The resistance of the two wire line carrying power is \(0.5 \Omega\) per \(\mathrm{km}\). The town gets power from the line through a \(4000-220 \mathrm{~V}\) step-down transformer at a sub-station in the town. (a) Estimate the line power loss in the form of heat. (b) How much power must the plant supply, assuming there is negligible power loss due to leakage? (c) Characterise the step up transformer at the plant.

Obtain the resonant frequency \(\omega_{r}\) of a series LCR circuit with \(L=2.0 \mathrm{H}, \mathrm{C}=32 \mu \mathrm{F}\) and \(R=10 \Omega\). What is the \(g\) -value of this circuit?

Answer the following questions: (a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage? (b) A capacitor is used in the primary circuit of an induction coil. (c) An applied voltage signal consists of a superposition of a de voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the de signal will appear across \(C\) and the ac signal across \(L\). (d) A choke coil in series with a lamp is connected to a de line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp's brightness. Predict the corresponding observations if the connection is to an ac line. (e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?

A circuit containing a \(80 \mathrm{mH}\) inductor and a \(60 \mu \mathrm{F}\) capacitor in series is connected to a \(230 \mathrm{~V}, 50 \mathrm{~Hz}\) supply. The resistance of the circuit is negligible. (a) Obtain the current amplitude and \(\mathrm{rms}\) values. (b) Obtain the rms values of potential drops across each element. (c) What is the average power transferred to the inductor? (d) What is the average power transferred to the capacitor? (e) What is the total average power absorbed by the circuit? ['Average implies 'averaged over one cycle'.]

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