Problem 30 - University Physics with Modern Physics

Question

An \(L-R-C\) series circuit consists of a source with voltage amplitude 120 \(\mathrm{V}\) and angular frequency \(50.0 \mathrm{rad} / \mathrm{s},\) a resistor with \(R=400 \Omega,\) an inductor with \(L=9.00 \mathrm{H},\) and a capacitor with capacitance \(C .\) (a) For what value of \(C\) will the current amplitude in the circuit be a maximum? (b) When \(C\) has the value calculated in part (a), what is the amplitude of the voltage across the inductor?

Step-by-Step Solution

Answer

(a) C = 4.44 × 10^-5 F; (b) Voltage across inductor = 135 V

Step 1: Understanding Resonance in L-R-C Circuit

In an L-R-C series circuit, the current amplitude is at its maximum when the circuit is at resonance. Resonance occurs when the inductive reactance \(X_L = \omega L\) is equal to the capacitive reactance \(X_C = \frac{1}{\omega C}\). At resonance, these reactances cancel each other, minimizing the impedance.

Step 2: Setting Reactances Equal for Resonance

To find the value of the capacitance \(C\) for maximum current amplitude, set the inductive reactance equal to the capacitive reactance:\[\omega L = \frac{1}{\omega C}\]Here \(\omega = 50.0 \text{ rad/s}\) and \(L = 9.00 \text{ H}\). Solving for \(C\), we get:\[C = \frac{1}{\omega^2 L}\]

Step 3: Calculating Capacitance

Substitute the given values into the formula for \(C\):\[C = \frac{1}{(50)^2 \times 9.00}\]\[C = \frac{1}{2500 \times 9.00}\]\[C = \frac{1}{22500}\]\[C = 4.44 \times 10^{-5} \text{ F}\]

Step 4: Calculating Voltage across the Inductor at Resonance

At resonance, the total impedance of the circuit is equal to just the resistance \(R\). The current in the circuit at resonance can be calculated as:\[I = \frac{V}{R} = \frac{120}{400} = 0.3 \text{ A}\]The amplitude of the voltage across the inductor \(V_L\) is given by:\[V_L = I \cdot X_L\]Substituting \(X_L = \omega L\), we have:\[V_L = I \cdot \omega L = 0.3 \times 50 \times 9 = 135 \text{ V}\]

Key Concepts

Resonance

Resonance in an L-R-C circuit is a fascinating phenomenon where the circuit's current amplitude reaches its peak. This happens when the circuit is tuned such that the inductive reactance matches the capacitive reactance. This clever balance minimizes the overall impedance, allowing the current to flow unhindered. At resonance, the inductive reactance, which depends on the inductor's characteristics, is the same as the capacitive reactance, which relies on the capacitor's characteristics. One might wonder why this is important. Resonance leads to maximum energy transfer from the power source to the circuit. This is critical in designing circuits for applications like radio tuners, where you want peak efficiency.

Inductive Reactance

Inductive reactance is a measure of an inductor's opposition to changes in current. It plays a crucial role in determining how a circuit behaves, especially at different frequencies. Mathematically, it is expressed as:\[X_L = \omega L\]where - \(X_L\) is the inductive reactance,- \(\omega\) is the angular frequency of the source, - and \(L\) is the inductance of the coil. An important characteristic of inductive reactance is its direct proportionality to frequency. As frequency increases, \(X_L\) also increases, meaning that at higher frequencies, an inductor will provide more resistance to the flow of current.

Capacitive Reactance

Capacitive reactance is the measure of a capacitor's opposition to changes in voltage, and it's crucial for understanding how capacitors influence a circuit. The formula for capacitive reactance is:\[X_C = \frac{1}{\omega C}\]where - \(X_C\) is the capacitive reactance, - \(\omega\) is the angular frequency, - and \(C\) is the capacitance. Unlike inductive reactance, capacitive reactance decreases with an increase in frequency, indicating that capacitors allow more current to pass through at higher frequencies. This characteristic is why capacitors and inductors can balance each other out in a circuit, achieving resonance.

Impedance

Impedance in an L-R-C circuit combines both resistive elements and reactive elements (inductive and capacitive reactance). It determines how much the circuit resists the flow of alternating current. When looking at impedance, remember:- At resonance: - Inductive and capacitive reactances are equal and cancel each other. - Impedance reduces to just the resistive component \(R\). Impedance is key to understanding the energy efficiency of a circuit. Lower impedance means less energy loss and higher efficiency. The unit of impedance is ohms \(\Omega\). It's crucial for circuit designers to balance all these elements to achieve desired performance.

Current Amplitude

Current amplitude reflects the peak value of the current in an AC circuit. In an L-R-C circuit, this amplitude can vary greatly with the circuit's impedance. At resonance, the current amplitude is at its maximum because the impedance is solely due to the resistor ( \(R\) ), and the reactances have canceled each other out. To find the current amplitude at resonance, use:\[I = \frac{V}{R}\]where- \(I\) is the current amplitude,- \(V\) is the voltage amplitude,- and \(R\) is the resistance.Understanding current amplitude is essential for ensuring that the circuit operates within safe and efficient limits.