Problem 28
Question
An \(L-R-C\) series circuit is constructed using a \(175-\Omega\) resistor, a \(12.5-\mu \mathrm{F}\) capacitor, and an \(8.00-\mathrm{mH}\) inductor, all connected across an ac source having a variable frequency and a voltage amplitude of 25.0 \(\mathrm{V}\) (a) At what angular frequency will the impedance be smallest, and what is the impedance at this frequency? (b) At the angular frequency in part (a), what is the maximum current through the inductor? (c) At the angular frequency in part (a), find the potential difference across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one-half its greatest positive value. (d) In part \((\mathrm{c}),\) how are the potential differences across the resistor, inductor, and capacitor related to the potential difference across the ac source?
Step-by-Step Solution
VerifiedKey Concepts
Angular Frequency
For the circuit to be at resonance, the inductive reactance \( X_L \) must be equal to the capacitive reactance \( X_C \). This means \( X_L = X_C \) or \( \omega L = \frac{1}{\omega C} \).
- \( X_L = \omega L \) represents the opposition to the current in an inductor. It depends on the inductance \( L \) and angular frequency \( \omega \).
- \( X_C = \frac{1}{\omega C} \) represents the opposition to the current in a capacitor. It depends on the capacitance \( C \) and angular frequency \( \omega \).
Impedance
- At any point, the total impedance can be calculated as: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
- At resonance, the circuit has its minimum impedance because \( X_L = X_C \). This results in the inductive and capacitive reactances cancelling each other out. Hence, \( Z \) simplifies to the resistance \( R \), which in this case is 175 \( \Omega \).
- This is significant because it indicates that the circuit will allow maximum current flow when the impedance is at its lowest.
Maximum Current
- \( I_{max} = \frac{V}{Z} \)
- \( I_{max} = \frac{25.0}{175} \approx 0.143 \text{ A} \)