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Chapter 21: Problem 53

A series \(R L C\) circuit has \(R=500.0 \Omega, L=35.0 \mathrm{mH},\) and $C=87.0 \mathrm{pF} .$ What is the impedance of the circuit at resonance? Explain.

Short Answer

Expert verified
In an RLC circuit with a resistance of 500.0 惟, an inductance of 35 mH, and a capacitance of 87 pF, the impedance at the resonant frequency is purely resistive and equal to the resistance value, which is 500.0 惟.

Step by step solution

01

Calculate the resonant frequency

To find the resonant frequency, we use the formula: $$f_{res} = \frac{1}{2\pi\sqrt{LC}}$$ where L is the inductance and C is the capacitance.
02

Plug in given values and solve

In our case, \(L=35\times 10^{-3} H\) and \(C=87\times 10^{-12} F\). Substituting these values in the formula, we get $$f_{res} = \frac{1}{2\pi\sqrt{(35\times 10^{-3})(87\times 10^{-12})}}$$ Now, calculate the value of \(f_{res}\).
03

Determine inductive and capacitive reactance at resonance

At resonance, inductive reactance (XL) and capacitive reactance (XC) are equal. $$X_{L} = X_{C}$$ $$2\pi f_{res} L = \frac{1}{2\pi f_{res} C}$$
04

Calculate the impedance of the circuit at resonance

At resonance, the impedance is purely resistive and equal to the resistance value. So, the impedance at resonance, Z, is given by the resistance value. $$Z = R = 500.0 \Omega$$ The impedance of the RLC circuit at resonance is 500.0 \(\Omega\).

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Most popular questions from this chapter

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