Question

Катушка индуктивностью L замыкают на конденсатор емкостью С1, при этом в полученном колебательном контуре возникают колебания с частотой υ1 = 50 Гц. Если эту же катушку замкнуть на конденсатор с емкостью С2, то частота возникающих колебаний станет υ2 = 100 Гц. Какой частоты возникнут колебания в контуре, если конденсаторы емкостью С1 и С2 соединить последовательно, и подключить к используемой катушке индуктивности?

Answer 1

Ответ:

В контуре возникнут колебания частотой приблизительно 111,8 Гц

Объяснение:

Дано:

$L$

$C_{1}$

$\nu_{1} =$ 50 Гц

$C_{2}$

$\nu_{2} =$ 100 Гц

Найти:

$\nu_{12} \ - \ ?$

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Решение:

По формуле Томсона:

$\boxed{T = 2\pi \sqrt{LC} }$ - период колебаний

$\nu = \dfrac{1}{T} = \dfrac{1}{2\pi \sqrt{LC} }$

$\bigg(\nu \bigg)^{2} = \bigg( \dfrac{1}{2\pi \sqrt{LC} } \bigg)^{2}$

$\nu^{2} = \dfrac{1}{4\pi ^{2}LC} \Longrightarrow \boxed{ C = \dfrac{1}{4\pi ^{2}\nu^{2}L} }$

При последовательном соединении конденсаторов:

$\dfrac{1}{C_{12}} = \dfrac{1}{C_{1}} +\dfrac{1}{C_{2}}$

$\dfrac{1}{C_{12}} = \dfrac{C_{1} +C_{2}}{C_{1}C_{2}} \Longrightarrow C_{12} = \dfrac{C_{1}C_{2}}{C_{1} +C_{2}}$

$C_{12} = \dfrac{C_{1}C_{2}}{C_{1} +C_{2}} = \dfrac{\dfrac{1}{4\pi ^{2}\nu_{1}^{2}L} \cdot \dfrac{1}{4\pi ^{2}\nu_{2}^{2}L}}{\dfrac{1}{4\pi ^{2}\nu_{1}^{2}L} + \dfrac{1}{4\pi ^{2}\nu_{2}^{2}L}} = \dfrac{\dfrac{1}{4\pi ^{2}\nu_{1}^{2}L} \cdot \dfrac{1}{4\pi ^{2}\nu_{2}^{2}L}}{\dfrac{1}{4\pi ^{2}L} \bigg(\dfrac{1}{\nu_{1}^{2}} + \dfrac{1}{\nu_{2}^{2}} \bigg)} =$$= \dfrac{\dfrac{1}{4\pi ^{2}\nu_{1}^{2}L} \cdot \dfrac{1}{4\pi ^{2}\nu_{2}^{2}L}}{\dfrac{1}{4\pi ^{2}L} \bigg(\dfrac{\nu_{2}^{2} + \nu_{1}^{2}}{\nu_{1}^{2}\nu_{2}^{2}} \bigg)} = \dfrac{4\pi ^{2}L\nu_{1}^{2}\nu_{2}^{2}}{4\pi ^{2}\nu_{1}^{2}L \cdot 4\pi ^{2}\nu_{2}^{2}L(\nu_{1}^{2} + \nu_{2}^{2})} =\dfrac{1}{4\pi ^{2}L(\nu_{1}^{2} + \nu_{2}^{2})}$

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Частота контура состоящего из конденсаторов ёмкостью $C_{1}$ и $C_{2}$.

$\nu_{12} = \dfrac{1}{2\pi \sqrt{LC_{12}} } = \dfrac{1}{2\pi \sqrt{L \cdot\dfrac{1}{4\pi ^{2}L(\nu_{1}^{2} + \nu_{2}^{2})}} } = \dfrac{1}{\dfrac{2 \pi}{2 \pi}\sqrt{\dfrac{1}{\nu_{1}^{2} + \nu_{2}^{2}} } } =$

$= \dfrac{\dfrac{1}{1} }{\dfrac{1}{\sqrt{\nu_{1}^{2} + \nu_{2}^{2}} } } =\sqrt{\nu_{1}^{2} + \nu_{2}^{2}}$.

$\boxed{ \boldsymbol{ \nu_{12}= \sqrt{\nu_{1}^{2} + \nu_{2}^{2}} } }$

$\nu_{12} =$ √(2500 Гц² + 10 000 Гц²) $\approx$ 111,8 Гц

Ответ: $\nu_{12} \approx$ 111,8 Гц.

Answer 2

Solving for the Frequency of the Oscillations

When a coil with inductance L is connected to a capacitor with capacitance C1, oscillations with a frequency of υ1 = 50 Hz occur. When the same coil is connected to a capacitor with capacitance C2, the frequency of the oscillations becomes υ2 = 100 Hz. We are asked to find the frequency of the oscillations when the capacitors C1 and C2 are connected in series and connected to the coil.

To solve for the frequency of the oscillations in the circuit, we can use the formula for the resonant frequency of an LC circuit:

Resonant frequency (υ) = 1 / (2π√(LC))

Where: - L = inductance of the coil - C = capacitance of the capacitor

Calculating the Resonant Frequency

First, let's calculate the resonant frequency when the coil is connected to the capacitor with capacitance C1:

Given: - υ1 = 50 Hz - C1 = capacitance of the first capacitor

Using the formula for the resonant frequency, we can solve for the inductance L:

L = (1 / (4π^2C1υ1^2))

Next, we can calculate the inductance L using the given values.

Finding the Inductance (L) for C1

Given: - υ1 = 50 Hz - C1 = capacitance of the first capacitor

Using the formula: L = (1 / (4π^2C1υ1^2))

Substitute the given values: L = (1 / (4π^2 * C1 * (50)^2))

Calculating the Resonant Frequency for C2

Next, let's calculate the resonant frequency when the coil is connected to the capacitor with capacitance C2:

Given: - υ2 = 100 Hz - C2 = capacitance of the second capacitor

Using the formula for the resonant frequency, we can solve for the inductance L:

L = (1 / (4π^2C2υ2^2))

Next, we can calculate the inductance L using the given values.

Finding the Inductance (L) for C2

Given: - υ2 = 100 Hz - C2 = capacitance of the second capacitor

Using the formula: L = (1 / (4π^2C2υ2^2))

Substitute the given values: L = (1 / (4π^2 * C2 * (100)^2))

Calculating the Combined Capacitance

Now, we are asked to find the frequency of the oscillations when the capacitors C1 and C2 are connected in series and connected to the coil. When capacitors are connected in series, their combined capacitance (Cs) can be calculated using the formula:

1 / Cs = 1 / C1 + 1 / C2

We can then use the combined capacitance Cs to calculate the frequency of the oscillations in the circuit using the formula for the resonant frequency.

Finding the Combined Capacitance (Cs)

Given: - C1 = capacitance of the first capacitor - C2 = capacitance of the second capacitor

Using the formula: 1 / Cs = 1 / C1 + 1 / C2

Substitute the given values to find Cs.

Calculating the Frequency with Combined Capacitance

Finally, we can calculate the frequency of the oscillations in the circuit using the combined capacitance Cs and the inductance L:

Resonant frequency (υ) = 1 / (2π√(LCs))

Substitute the calculated values of L and Cs to find the frequency of the oscillations in the circuit.

By following these steps, we can determine the frequency of the oscillations in the circuit when the capacitors C1 and C2 are connected in series and connected to the coil.