In an L-R-C series circuit, the source has a voltage amplitude of 125V , R = 84.

Question

In an L-R-C series circuit, the source has a voltage amplitude of 125V , R = 84.0 Omega , and the reactance of the capacitor is 475 Omega . The voltage amplitude across the capacitor is 361V . Part A What is the current amplitude in the circuit? Part B What is the impedance? Part C What two values can the reactance of the inductor have? Enter your answers in ascending order separated by a comma. Part D For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? Multiple choice! A) the less value of X_L B) the larger value of X_L

Explanation / Answer

1 microfarads/1.000,000 = 1 Farad capacitive reactance = 1/wC (w used for angular frequency here; C for capacitance in Farads) inductive reactance = wL. (L is for inductance in henries) These two reactances must be equal at resonance, so wL = 1/wC. This gives w^2 = 1/(L*C) , or w = 1/[square root of L*C). Plug in your values and calculate what "w" must equal. This answer is in (rad)/sec if you Farads, not microfarads. (rad) means that radians are really dimensionless, as are degrees of angle. At the resonance frequency, the two reactances cancel each other's effects, and only the resistance limits the current. Divide the given voltage by the given resistance to find what the value of the maximum current equals (at resonance). This same current flows through L, C, and R. The voltage across the capacitor obeys Ohm's Law, so calculate its reactance from the 1/wC expression and divide it into the capacitor's voltage limit (rating) to get the maximum current allowed. Multiply that current value times the resistance, whose voltage drop equals the applied voltage at resonance, and the answer to the division is the maximum voltage which can be applied. NOTE: In a series circuit like this, the voltage across the capacitor can be greater at some frequencies than the voltage applied to the whole circuit by an electrical source. So your answer may be smaller than the voltage limit for the capacitor

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