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Svetllana [295]

2 years ago

6

If I connect an inductor (L) to a capacitor (C), I will get an LC oscillator circuit with some natural frequency omega. If I wer

e to quadruple the capacitance of the circuit, what would be the new natural frequency

1 answer:

mart [117]2 years ago

8 0

The new natural frequency would be ω/2.

we know that,

$f = \frac{1}{2 pi \sqrt{LC} }$ = ω. -> equation 1

now, when capacitance is quadrupled,

$f' = \frac{1}{2 pi \sqrt{L ( 4C )} }$

$f' = \frac{1}{2 pi (2)\sqrt{LC} }$ . -> equation 2

substituting value of equation 1 in equation 2 , we get,

$f' = \frac{w}{2}$

Hence, the new natural frequency of the circuit is ω/2.

what do you mean by frequency ?

The resonant frequency for a particular circuit is the frequency at which this equality stands true. Where L is the inductance in henries and C is the capacitance in farads, this is the LC circuit's resonant frequency.

Learn more about frequency here:-

brainly.com/question/12530980

#SPJ4

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2 years ago

Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c

Travka [436]

Answer:

22.1 V

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$A_4=3 cm^2=3\times 10^{-4} m^2$

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$R=\frac{\rho l}{A}$

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$R=R_1+R_2+R_3+R_4$

$R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}$

$R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}$

Substitute the values

$R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})$

$R=\rho l(2.62\times 10^4)$

$V=145 V$

$I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}$

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Voltage across the 2.5 square cm wire= $\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V$

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6 0

3 years ago

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s344n2d4d5 [400]

Answer:

At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

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At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

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Hence, the car would catch up with the truck at $t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}$ .

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