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A circuit with a resistor, inductor, and capacitor in series in resonant of (f0) Hz. If all the component values are now doubled, the new resonant frequency is-
  • a)
    f0/4
  • b)
    2f0
  • c)
    f0
  • d)
    f0/2
Correct answer is option 'D'. Can you explain this answer?
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Resonance in an RLC circuit occurs at a specific frequency where the reactance of the inductor cancels out the reactance of the capacitor, resulting in a purely resistive circuit. In this question, we are given a circuit with a resistor, inductor, and capacitor in series that has a resonant frequency of f0 Hz. We need to determine the new resonant frequency when all the component values are doubled.

Given information:
- Circuit components: resistor (R), inductor (L), and capacitor (C) in series
- Resonant frequency of the original circuit: f0 Hz
- All component values are doubled

To solve this problem, we can analyze the behavior of the circuit at resonance and see how the changes in component values affect the resonant frequency.

1. Understanding Resonance:
At resonance, the reactances of the inductor and capacitor cancel each other out, resulting in a purely resistive circuit. The reactance of an inductor (XL) is given by XL = 2πfL, and the reactance of a capacitor (XC) is given by XC = 1/(2πfC), where f is the frequency, L is the inductance, and C is the capacitance. At resonance, XL = XC, so we can equate the two expressions and solve for the resonant frequency.

2. Doubling the Component Values:
If all the component values (R, L, C) are doubled, the reactances of the inductor and capacitor will also double. Let's denote the new values as 2R, 2L, and 2C.

3. Finding the New Resonant Frequency:
Using the reactance equations for the inductor and capacitor, we can write the equation for the new resonant frequency, f' (dash):

2πf'L = 1/(2πf'2C)

Simplifying the equation:

f'^2 = (2πf')^2 * L * 2C

f'^2 = 4π^2f'^2LC

Dividing both sides by f'^2:

1 = 4π^2LC

Simplifying further:

LC = 1/(4π^2)

Now, we can see that the resonant frequency of the new circuit (f') is given by:

f' = 1 / (2π√(LC))

Substituting the value of LC:

f' = 1 / (2π√(1/(4π^2)))

Simplifying:

f' = 1 / (2π/2π)

f' = 1 / 1

f' = 1 Hz

Therefore, the new resonant frequency of the circuit, when all the component values are doubled, is f0/2. The correct answer is option D.
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A circuit with a resistor, inductor, and capacitor in series in resonant of (f0) Hz. If all the component values are now doubled, the new resonant frequency is-a)f0/4b)2f0c)f0d)f0/2Correct answer is option 'D'. Can you explain this answer?
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