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The resonant frequency of an RCL circuit is \(1.3 \mathrm{kHz},\) and the value of the inductance is \(7.0 \mathrm{mH}\). What is the resonant frequency (in \(\mathrm{kHz}\) ) when the value of the inductance is 1.5 \(\mathrm{mH} ?\)

Short Answer

Expert verified
The resonant frequency with 1.5 mH inductance is approximately 3.12 kHz.

Step by step solution

01

Understand the formula for resonant frequency

The resonant frequency \( f_0 \) of an RCL circuit is given by the formula: \( f_0 = \frac{1}{2\pi\sqrt{LC}} \), where \( L \) is the inductance and \( C \) is the capacitance.
02

Use given information

We know the resonant frequency with an inductance \( L_1 = 7.0 \ \mathrm{mH} \) is \( f_1 = 1.3 \ \mathrm{kHz}\). We need to find the new resonant frequency \( f_2 \) when the inductance is \( L_2 = 1.5 \ \mathrm{mH} \).
03

Express capacitance in terms of resonant frequency and inductance

From the formula \( f_0 = \frac{1}{2\pi\sqrt{LC}} \), rearrange to express \( C = \frac{1}{(2\pi f_0)^2 L} \).
04

Find the capacitance using initial conditions

Use the initial condition to find capacitance: \[ C = \frac{1}{(2\pi \times 1.3 \times 10^3)^2 \times 7.0 \times 10^{-3}} \].
05

Calculate the new resonant frequency using \( C \) and new inductance

Once \( C \) is calculated, use it to find the new frequency with \( L_2 = 1.5 \ \mathrm{mH} \) as follows: \[ f_2 = \frac{1}{2\pi\sqrt{L_2C}} \].
06

Simplify and calculate the new frequency

Note that the frequency is inversely proportional to the square root of inductance, so: \[ f_2 = f_1 \times \sqrt{\frac{L_1}{L_2}} \ = 1.3 \times \sqrt{\frac{7.0}{1.5}} \]. Calculate \( f_2 \).

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Key Concepts

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

Understanding Resonant Frequency
Resonant frequency is a crucial concept in the study of RLC circuits, representing the frequency at which the circuit naturally oscillates with the least impedance. This particular frequency is dependent on both the inductance (L) and capacitance (C) of the circuit. It is given by the equation: \[ \text{Resonant Frequency} (f_0) = \frac{1}{2\pi\sqrt{LC}} \]. This expression indicates that the resonant frequency is inversely proportional to the square root of both the inductance and the capacitance.

When the circuit reaches its resonant frequency, its impedance is minimized, and the circuit can efficiently transfer energy between the inductor and capacitor. This property is widely used in various applications such as radio receivers and transmission antennas.
Explaining Inductance
Inductance is a fundamental property of circuits, it describes the ability of an inductor to store energy in a magnetic field when electrical current flows through it. Measured in henrys (H), inductance is crucial in determining the behavior of alternating current (AC) circuits.

It is an inherent property of coils, represented mathematically by the letter L. In an RLC circuit, the inductance affects how the circuit reacts to changes in current. Particularly, in the context of resonant frequency, higher inductance will reduce the resonant frequency, as seen from the formula \( f_0 = \frac{1}{2\pi\sqrt{LC}} \).

Thus, when inductance values change, we can expect a corresponding impact on the resonant frequency of the circuit. Inductors are key components in tuning frequencies for filtering and oscillation in myriad electronics.
Understanding Capacitance
Capacitance is the ability of a component or circuit to store an electrical charge, and it plays a crucial role in tuning the frequency of an RLC circuit. Measured in farads ( F ), capacitance defines how much charge a capacitor can store per unit voltage.

Capacitance directly influences the resonant frequency alongside inductance. According to the formula for resonant frequency, increasing the capacitance of an RLC circuit decreases its resonant frequency. This makes capacitors indispensable for controlling signal frequencies within electronic devices.

In practical terms, capacitors are the components that hold energy within an RLC circuit, oscillating charge at a specific frequency that defines the resonant frequency the circuit will naturally settle on.
Calculating Frequency
Frequency calculation in RLC circuits involves understanding the relationship between frequency and the circuit components: inductance and capacitance. The resonant frequency formula \( f_0 = \frac{1}{2\pi\sqrt{LC}} \) provides the foundation for these calculations.

When tasked with determining a new resonant frequency after modifying inductance, it's essential to maintain the capacitance constant. We use the known resonant frequency and parameters to first solve for the unknown capacitance. From there, employing the formula again allows recalculating the resonant frequency for changed inductance.

Specifically, to exemplify this calculation, if one reduces the inductance from 7.0 \( ext{mH}\) to 1.5 \( ext{mH}\) and knows the initial frequency, we can determine that: \[ f_2 = f_1 \times \sqrt{\frac{L_1}{L_2}} \]

This approach illustrates the inverse square root relationship between inductance and frequency, ensuring precise adjustments in real-world applications.

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Most popular questions from this chapter

A light bulb has a resistance of \(240 \Omega\). It is connected to a standard wall socket (120 V, \(60.0 \mathrm{~Hz}\) ). (a) Determine the current in the bulb. (b) Determine the current in the bulb after a \(10.0-\mu \mathrm{F}\) capacitor is added in series in the circuit. (c) It is possible to return the current in the bulb to the value calculated in part (a) by adding an inductor in series with the bulb and the capacitor. What is the value of the inductance of this inductor?

The resonant frequency of a series \(\mathrm{RCL}\) circuit is \(9.3 \mathrm{kHz}\). The inductance and capacitance of the circuit are each tripled. What is the new resonant frequency?

A circuit consists of a \(3.00-\mu F\) and a \(6.00-\mu F\) capacitor connected in series across the terminals of a 510 -Hz generator. The voltage of the generator is \(120 \mathrm{~V}\). (a) Determine the equivalent capacitance of the two capacitors. (b) Find the current in the circuit.

Concept Questions (a) An inductance \(L_{1}\) is connected across the terminals of an ac generator, which delivers a current to it. Then a second inductance \(L_{2}\) is connected in parallel with \(L_{1}\). Does the presence of \(L_{2}\) alter the current in \(L_{1} ?\) (b) The generator delivers a current to \(L_{2}\) as well as \(L_{1} .\) Would the removal of \(L_{1}\) alter the current in \(L_{2} ?\) (c) Is the current delivered to the parallel combination greater or smaller than the current to either inductance alone? Give your reasoning. Problem An ac generator has a frequency of \(2.2 \mathrm{kHz}\) and a voltage of \(240 \mathrm{~V}\). An inductance \(L_{1}=6.0 \mathrm{mH}\) is connected across its terminals. Then a second inductance \(L_{2}=\) \(9.0 \mathrm{mH}\) is connected in parallel with \(L_{1} .\) Find the current that the generator delivers to \(L_{1}\) and to the parallel combination. Check to see that your answers are consistent with your answers to the Concept Questions.

A capacitor is attached to a 5.00 -Hz generator. The instantaneous current is observed to reach a maximum value at a certain time. What is the least amount of time that passes before the instantaneous voltage across the capacitor reaches its maximum value?

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