In an L-R-C series circuit, the resistance is 460 ohms, the inductance is 0.400

Question

In an L-R-C series circuit, the resistance is 460 ohms, the inductance is 0.400 henrys, and the capacitance is 2.00x102 microfarads. What is the resonance angular frequency wo of the circuit? Express your answer in radians per second to three significant figures Hints wo = 1.12-104 Submit rad/s Correct Part B The capacitor can withstand a peak voltage of 580 volts. If the voltage source operates at the resonance frequency, what maximum voltage amplitude Vmax can the source have if the maximum capacitor voltage is not exceeded? Express your answer in volts to three significant figures. Hints Vmax

Explanation / Answer

1. = 1/(LC) = 1/(0.40×2E-8) = 11.180E3 = 11.180 × 10^3 rad/s.

2. At this angular frequency, inductive reactance and capacitive reactance mutually cancel. That is, L = 1/. (. In these circumstances, circuit behaves as if the series circuit consisted on resistance R only.

On the other hand, voltage across capacitor is given by V = I / C. Assuming V = 580 V, and solving for I,

I = C V = 11.180 × 10^3 × 2 × 10^-8 × 580 = 0.1296 A

This is the peak value for current. If you compute voltage across the inductor (as V = I XL = I L, you'll find that it is also 580 V, the same as for the capacitor. However, there's a 180º phase shift between these voltages, so they cancel out each other. Across the entire series circuit, the only measurable voltage drop is that across the resistor. Now, which voltage should there be across a 460 resistor in order for current to be 0.1296A? By Ohm's law,

V = I R = 0.1296 A × 460 = 59.616 V.

Again, since I is a peak (maximum) value, this should be the peak value for the voltage source.

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