Question

A series RLC circuit has a source of time-varying emf providing \(12.0 \mathrm{~V}\) at a frequency \(f_{0}\), with \(L=7.00 \mathrm{mH}, R=100 . \Omega\), and \(C=0.0500 \mathrm{mF}\). a) What is the resonant frequency of this circuit? b) What is the average power dissipated in the resistor at this resonant frequency?

Short answer

Answer: The resonant frequency of the given RLC circuit is 84.7 Hz, and the average power dissipated in the resistor at this resonant frequency is 1.44 W.

Step-by-step solution

Find the resonant frequency

We will use the formula for the resonant frequency, \(f_0 = \frac{1}{2 \pi \sqrt{LC}}\). Given the values of L and C: L = 7.00 mH = \(7.00 \times 10^{-3}\) H C = 0.0500 mF = \(0.0500 \times 10^{-6}\) F Plug these values into the formula: \(f_0 = \frac{1}{2 \pi \sqrt{(7.00 \times 10^{-3})(0.0500 \times 10^{-6})}}\)

Calculate the resonant frequency

Compute the expression within the square root first: \((7.00 \times 10^{-3})(0.0500 \times 10^{-6}) = 3.50 \times 10^{-7}\) Now, compute the square root: \(\sqrt{3.50 \times 10^{-7}} = 1.87 \times 10^{-3}\) Finally, divide by \(2\pi\): \(f_0 = \frac{1}{2 \pi (1.87 \times 10^{-3})} = 84.7 Hz\)

Find the average power dissipated in the resistor

We'll use the average power formula, \(P = \frac{V^2}{R}\). Given the values of V and R: V = 12.0 V R = 100 \(\Omega\) Plug these values into the formula: \(P = \frac{(12.0)^2}{100}\)

Calculate the average power dissipated in the resistor

Compute the square of the voltage first: \((12.0)^2 = 144\) Now, divide by the resistance: \(P = \frac{144}{100} = 1.44 W\) The resonant frequency of the circuit is 84.7 Hz, and the average power dissipated in the resistor at this resonant frequency is 1.44 W.