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Laboratory experiments with series RLC circuits require some care, as these circuits can produce large voltages at resonance. Suppose you have a 1.00-H inductor (not difficult to obtain) and a variety of resistors and capacitors. Design a series RLC circuit that will resonate at a frequency (not an angular frequency) of 60.00 Hz and will produce at resonance a magnification of the voltage across the capacitor or the inductor by a factor of 10.9 times the input voltage or the voltage across the resistor. (a) What is the capacitance required for this circuit? (You may enter your calculation using scientific notation.)  F (b) And what is the resistance required for this circuit?  Ω

Question

Laboratory experiments with series RLC circuits require some care, as these circuits can produce large voltages at resonance. Suppose you have a 1.00-H inductor (not difficult to obtain) and a variety of resistors and capacitors. Design a series RLC circuit that will resonate at a frequency (not an angular frequency) of 60.00 Hz and will produce at resonance a magnification of the voltage across the capacitor or the inductor by a factor of 10.9 times the input voltage or the voltage across the resistor. (a) What is the capacitance required for this circuit? (You may enter your calculation using scientific notation.)  F (b) And what is the resistance required for this circuit?  Ω

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Solution

To solve this problem, we need to use the formulas for resonance frequency and quality factor in a series RLC circuit.

(a) The resonance frequency (f) in a series RLC circuit is given by the formula:

f = 1 / (2π√(LC))

where L is the inductance and C is the capacitance. We can rearrange this formula to solve for C:

C = 1 / (4π²f²L)

Substituting the given values:

C = 1 / (4π² * (60 Hz)² * 1 H) C = 7.07 x 10^-5 F

(b) The quality factor (Q) in a series RLC circuit, which is the ratio of the voltage across the inductor or capacitor to the voltage across the resistor at resonance, is given by the formula:

Q = 1/R * √(L/C)

We can rearrange this formula to solve for R:

R = √(L/C) / Q

Substituting the given values:

R = √(1 H / 7.07 x 10^-5 F) / 10.9 R = 3.77 Ω

So, the capacitance required for this circuit is approximately 7.07 x 10^-5 F and the resistance required for this circuit is approximately 3.77 Ω.

This problem has been solved

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