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Chapter 28: Problem 53

An RLC circuit includes a 1.5 -H inductor and a \(250-\mu \mathrm{F}\) capacitor rated at 400 V. The circuit is connected across a sine-wave generator with \(V_{\mathrm{p}}=32 \mathrm{V} .\) What minimum resistance will ensure that the capacitor voltage does not exceed its rated value when the circuit is at resonance?

Short Answer

Expert verified
The minimum resistance to ensure that the capacitor voltage does not exceed its rated value when the circuit is at resonance is \(R\) ohms.

Step by step solution

01

Calculate the Resonant Frequency

The resonant frequency is given by the formula: \(f_{\mathrm{r}}=\frac{1}{2 \pi \sqrt{L C}}\) where \(L = 1.5 H\) is the inductance and \(C = 250 \mu F\) is the capacitance. Substituting these values in, we calculate the resonant frequency.
02

Calculate the RMS Voltage

The RMS (root mean square) voltage is \(V_{\mathrm{rms}}=\frac{V_{\mathrm{p}}}{\sqrt{2}}\), where \(V_{\mathrm{p}} = 32 V\) is the peak voltage. Solve for \(V_{\mathrm{rms}}\).
03

Calculate the Minimum Resistance

The voltage across the capacitor at resonance is given by \(V_{\mathrm{rms}}=\frac{I_{\mathrm{rms}}}{\omega_{\mathrm{r}} C}\). To find the minimum resistance, \(R\), we rearrange the terms and use the definition of the RMS current, \(I_{\mathrm{rms}} = V_{\mathrm{rms}}/R\), to obtain the following form: \(R=\frac{1}{\omega_{\mathrm{r}} C}\). The angular frequency, \(\omega_{\mathrm{r}}\), is \(2\pi f_{\mathrm{r}}\). Substituting our calculated values in, we can solve for the minimum resistance to prevent the capacitor voltage from exceeding its rated value.

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