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Chapter 30: Problem 37

A series circuit contains a \(100.0-\Omega\) resistor, a \(0.500-\mathrm{H}\) inductor, a 0.400 - \(\mu\) F capacitor, and a time-varying source of emf providing \(40.0 \mathrm{~V}\). a) What is the resonant angular frequency of the circuit? b) What current will flow through the circuit at the resonant frequency?

Short Answer

Expert verified
Question: Calculate the resonant angular frequency and the current flowing through the circuit at the resonant frequency for a series RLC circuit with an inductor of 0.500 H, a capacitor of 0.400 μF, and a resistance of 100.0 Ω connected to an emf source with a voltage of 40.0 V. Answer: The resonant angular frequency of the circuit is 1432.31 rad/s and the current flowing through the circuit at the resonant frequency is 0.40 A.

Step by step solution

01

Calculate the resonant angular frequency

We need to calculate the resonant angular frequency, \(\omega_{0}\), using the formula \(\omega_{0} = \frac{1}{\sqrt{LC}}\). We are given the values of L (\(0.500 H\)) and C (\(0.400 \times 10^{-6} F\)). So, plug these values into the formula: \(\omega_{0} = \frac{1}{\sqrt{(0.500)(0.400\times10^{-6})}}\) Now, calculate \(\omega_{0}\): \(\omega_{0} = 1432.31 \, rad/s\) The resonant angular frequency of the circuit is \(1432.31 \, rad/s\).
02

Calculate the current

Next, we have to calculate the current flowing through the circuit at the resonant frequency. We'll use the formula \(I = \frac{V}{Z}\), where \(V = 40.0V\). At resonance, the impedance is equal to the resistance of the circuit, so we will have \(Z = R = 100.0\Omega\). Now, plug these values into the formula: \(I = \frac{40.0V}{100.0\Omega}\) Now, calculate the current: \(I = 0.40 A\) The current flowing through the circuit at the resonant frequency is \(0.40 A\). To summarize: a) The resonant angular frequency of the circuit is \(\omega_{0} =\) \(1432.31 \, rad/s\). b) The current flowing through the circuit at the resonant frequency is \(I = 0.40 A\).

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Most popular questions from this chapter

The AM radio band covers the frequency range from \(520 \mathrm{kHz}\) to \(1610 \mathrm{kHz}\). Assuming a fixed inductance in a simple LC circuit, what ratio of capacitance is necessary to cover this frequency range? That is, what is the value of \(C_{\mathrm{h}} / C_{\mathrm{l}}\) where \(C_{\mathrm{h}}\) is the capacitance for the highest frequency and \(C_{1}\) is the capacitance for the lowest frequency? a) 9.59 b) 0.104 c) 0.568 d) 1.76

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