Problem 36
Question
A Radio Tuning Circuit. The minimum capacitance of a variable capacitor in a radio is 4.18 \(\mathrm{pF}\) . (a) What is the inductance of a coil connected to this capacitor if the oscillation frequency of the \(L\) circuit is \(1600 \times 10^{3} \mathrm{Hz}\) , corresponding to one end of the AM radio broadcast band, when the capacitor is set to its minimum capacitance? (b) The frequency at the other end of the broadcast band is \(540 \times 10^{3}\) Hz. What is the maximum capacitance of the capacitor if the oscillation frequency is adjustable over the range of the broadcast band?
Step-by-Step Solution
Verified Answer
(a) Find \( L \), then (b) find max \( C \) using new \( f \).
1Step 1: Understand the Formula
The resonant frequency of an "L-C" circuit is given by the formula \( f = \frac{1}{2\pi\sqrt{LC}} \), where \( f \) is the frequency, \( L \) is the inductance, and \( C \) is the capacitance. We'll use this formula to find \( L \) for part (a) and the maximum capacitance for part (b).
2Step 2: Solve for Inductance - Setup Equation
For part (a), rearrange the formula to solve for \( L \):\[ L = \frac{1}{(2\pi f)^2 C} \] Substitute the given values: \( f = 1600 \times 10^3 \) Hz and \( C = 4.18 \times 10^{-12} \) F.
3Step 3: Solve for Inductance - Calculate
Plugging in the values:\[ L = \frac{1}{(2\pi \cdot 1600 \times 10^3)^2 \times 4.18 \times 10^{-12}} \]Calculate to find \( L \).
4Step 4: Solve for Maximum Capacitance - Setup Equation
For part (b), use the same formula rearranged to solve for \( C \):\[ C = \frac{1}{(2\pi f)^2 L} \]Now, substitute \( f = 540 \times 10^3 \) Hz and the value of \( L \) obtained from part (a).
5Step 5: Solve for Maximum Capacitance - Calculate
Substitute into the equation:\[ C = \frac{1}{(2\pi \cdot 540 \times 10^3)^2 \times L} \]Compute to find the maximum capacitance \( C \).
6Step 6: Conclusion
Based on the calculations:- For part (a), the inductance \( L \) is approximately found from the given frequency and capacitance.- For part (b), substituting the obtained inductance into the frequency equation gives the maximum capacitance.
Key Concepts
Resonant FrequencyInductanceCapacitance
Resonant Frequency
The resonant frequency is an essential aspect of radio tuning circuits. It represents the frequency at which the circuit naturally oscillates with the greatest amplitude. This is the key frequency where maximum energy transfer occurs between the inductor and the capacitor in the circuit. Mathematically, it is given by the formula:\[ f = \frac{1}{2\pi\sqrt{LC}} \]where:- \(f\) is the resonant frequency,- \(L\) is the inductance in henrys (H),- \(C\) is the capacitance in farads (F).This formula is crucial in determining how a radio tuning circuit can select different frequencies. By adjusting either the inductance or the capacitance, the circuit can tune into different stations on the AM band.
Inductance
Inductance is a critical component in radio tuning circuits. It refers to the property of a coil, which resists changes in electric current through it. Inductance is measured in henrys (H) and affects how the circuit interacts with the magnetic fields generated by alternating current.- The larger the inductance, the more the coil opposes the change in current, making the circuit resonate at a lower frequency if the capacitance remains constant.- The coil, often called an inductor, stores energy in the form of a magnetic field when electric current flows through it.In our exercise, finding the inductance was essential for setting the resonant frequency at the higher end of the AM band. By rearranging the resonant frequency formula to solve for \(L\):\[ L = \frac{1}{(2\pi f)^2 C} \]students learn how to calculate the inductance needed to achieve a specific resonant frequency.
Capacitance
Capacitance is another vital component in radio tuning circuits, defined as the ability of a system to store an electric charge. Capacitance is measured in farads (F), and in radio circuits, it plays a crucial role in determining resonant frequency alongside inductance.- A larger capacitance means more electric charge can be stored, which will lower the resonant frequency if the inductance remains constant.- Variable capacitors are used in tuning circuits to allow adjustment across a desired frequency range, such as the AM broadcast band.For example, the exercise required calculating the maximum capacitance of a capacitor to cover the entire range of the AM band. By using the formula rearranged for capacitance:\[ C = \frac{1}{(2\pi f)^2 L} \]students can understand how capacitance affects the ability of a circuit to select specific frequencies.
Other exercises in this chapter
Problem 34
A \(18.0-\mu \mathrm{F}\) capacitor is placed across a 22.5 \(\mathrm{V}\) batter for several seconds and is then connected across a \(12.0-\mathrm{mH}\) induct
View solution Problem 35
A capacitor with capacitance \(6.00 \times\) \(10^{-5} \mathrm{F}\) is charged by connecting it to a \(12.0-\mathrm{V}\) battery. The capacitor is disconnected
View solution Problem 37
An \(L-C\) circuit containing an \(80.0-\mathrm{mH}\) inductor and a \(1.25-\mathrm{nF}\) capacitor oscillates with a maximum current of 0.750 \(\mathrm{A}\) .
View solution Problem 38
In an \(L-C\) circuit, \(L=85.0 \mathrm{mH}\) and \(C=3.20 \mu \mathrm{F} .\) During the oscillations the maximum current in the inductor is 0.850 \(\mathrm{mA}
View solution