Problem 82 - Elementary Algebra

Question

The resonance frequency \(f\) in an electronic circuit containing inductance \(L\) and capacitance \(C\) in series is given by \(f=\frac{1}{2 \pi \sqrt{L C}}\) (a) Determine the resonance frequency in an electronic circuit if the inductance is 9 and the capacitance is 0.0001 . Use \(\pi=3.14\). (b) Determine the inductance in an electric circuit if the resonance frequency is 5.308 and the capacitance is 0.0001 . Use \(\pi=3.14\).

Step-by-Step Solution

Answer

**Question:** A resonance frequency in an electronic circuit containing inductance and capacitance can be found using the formula \(f=\frac{1}{2 \pi \sqrt{L C}}\). In the given electronic circuit, the inductance, L = 9 and the capacitance, C = 0.0001. Calculate the resonance frequency and find the value of inductance when the resonance frequency is 5.308, and the capacitance, C = 0.0001. **Answer:** The resonance frequency of the given electronic circuit is 5.308, and the inductance when the resonance frequency is 5.308 and the capacitance is 0.0001 is approximately 90.29.

Step 1: (a) Determine the resonance frequency

Given inductance L = 9 and capacitance C = 0.0001. Using the given formula for resonance frequency: \(f=\frac{1}{2 \pi \sqrt{L C}}\) We can now plug in the given values for L and C: \(f = \frac{1}{2 \times 3.14 \times \sqrt{9 \times 0.0001}}\) Now, calculate the numerator: 1. Calculate the product of L and C: \(9 \times 0.0001 = 0.0009\) 2. Calculate the square root of the product: \(\sqrt{0.0009} = 0.03\) 3. Calculate the denominator: \(2 \times 3.14 \times 0.03 = 0.1884\) 4. Finally, calculate the resonance frequency: \(f = \frac{1}{0.1884} = 5.308\) The resonance frequency in the electronic circuit is 5.308.

Step 2: (b) Determine the inductance

Given resonance frequency f = 5.308 and capacitance C = 0.0001. We need to find the value of inductance L using the given formula: \(f=\frac{1}{2 \pi \sqrt{L C}}\) To solve for L, first we will convert the formula into a form that isolates L: \(L=\frac{1}{4 \pi^2 f^2 C}\) Now, plug in the given values for f and C into the formula: \(L = \frac{1}{4 \times 3.14^2 \times 5.308^2 \times 0.0001}\) Now, calculate the denominator: 1. Calculate the square of π: \(3.14^2 = 9.8596\) 2. Calculate the square of f: \(5.308^2 = 28.1547\) 3. Calculate the product of all the variables: \(4 \times 9.8596 \times 28.1547 \times 0.0001 = 0.01107\) 4. Finally, calculate the value of inductance: \(L = \frac{1}{0.01107} = 90.290\) The inductance in the electric circuit is approximately 90.29.

Key Concepts

Resonance Frequency

Resonance frequency in an electronic circuit is the frequency at which the circuit naturally oscillates. In a circuit containing an inductor (L) and a capacitor (C) in series, this resonance frequency can be determined using the formula: \[ f = \frac{1}{2 \pi \sqrt{LC}} \]Here, \(f\) is the resonance frequency, \(L\) is the inductance measured in Henries (H), and \(C\) is the capacitance measured in Farads (F). The resonance frequency is especially important because at this frequency, the circuit effectively allows for maximum energy transfer and minimal impedance.

To calculate the resonance frequency:

• Firstly, multiply the inductance \(L\) and the capacitance \(C\) to find their product.
• Next, take the square root of the product \(LC\).
• Finally, use these values in the formula with \(\pi\) to find \(f\).

From the given example, with \(L = 9\, H\) and \(C = 0.0001\, F\), substituting into the equation produces a resonance frequency of 5.308 Hz.

Inductance

Inductance is a property of an electrical circuit or component that opposes changes in current. It is analogous to mass in mechanical systems, and it's measured in Henries (H). In a resonant circuit, inductance determines alongside capacitance the circuit's natural frequency of oscillation. To find inductance when given resonance frequency \(f\) and capacitance \(C\), we can rearrange the resonance frequency formula:\[ L = \frac{1}{4 \pi^2 f^2 C} \]Here's how to solve for \(L\):

• Square the resonance frequency \(f\).
• Multiply \(f^2\) with four times \(\pi^2\) and capacitance \(C\).
• Take the reciprocal of this product to find \(L\).

In the example provided, with a resonance frequency \(f = 5.308 \) Hz and capacitance \(C = 0.0001 \) F, the inductance calculated is approximately 90.29 H.

Capacitance

Capacitance is a measure of a component's ability to store charge, analogous to a reservoir in fluid systems. It is measured in Farads (F). In a resonant circuit, capacitance works with inductance to determine the circuit's natural frequency. The resonance frequency decreases with higher capacitance, affecting how circuits like radios and filters function. To understand its role in resonance, consider:

• Increased capacitance leads to a lower resonance frequency.
• Decreased capacitance results in a higher resonance frequency.

When designing circuits, selecting the appropriate capacitance value is crucial for achieving the desired operational frequency. In our exercise example, the given capacitance is 0.0001 F. This specific value helps set the resonance frequency together with the given values of inductance.