Problem 32 - University Physics with Modern Physics

Question

In an \(L-R-C\) series circuit, \(R=400 \Omega, L=0.350 \mathrm{H},\) and \(C=0.0120 \mu \mathrm{F}\) (a) What is the resonance angular frequency of the circuit? (b) The capacitor can withstand a peak voltage of 550 \(\mathrm{V}\) . If the voltage source operates at the resonance frequency, what maximum voltage amplitude can it have if the maximum capacitor voltage is not exceeded?

Step-by-Step Solution

Answer

(a) 15400 rad/s; (b) 114400 V.

Step 1: Identification of Necessary Formula

To find the resonance angular frequency of an L-R-C circuit, we use the formula \( \omega_0 = \frac{1}{\sqrt{LC}} \).

Step 2: Calculation of the Resonance Angular Frequency

We will plug in the given values \( L = 0.350 \ \text{H} \) and \( C = 0.0120 \ \mu \text{F} = 0.0120 \times 10^{-6} \ \text{F} \) into the formula from Step 1. This gives us:\[\omega_0 = \frac{1}{\sqrt{0.350 \times 0.0120 \times 10^{-6}}}.\]

Step 3: Solving for the Angular Frequency

Calculate \( \omega_0 \):\[\omega_0 = \frac{1}{\sqrt{0.350 \times 0.0120 \times 10^{-6}}} = \frac{1}{\sqrt{4.2 \times 10^{-9}}} \]\[\omega_0 = \frac{1}{2.05 \times 10^{-4.5}} \approx 15400 \ \text{rad/s}.\]

Step 4: Understanding Maximum Voltage Relation

At resonance, the voltage across the capacitor \( V_C \) is related to the source voltage \( V_0 \) by the formula: \( V_C = \frac{V_0}{\omega_0 RC} \times R \), since at resonance, the impedance is minimum and only resistive.

Step 5: Expressing Source Voltage Amplitude

We know the maximum \( V_C \) is 550 V, so using \( \frac{V_0}{\omega_0 \cdot R \cdot C} = 550 \), solve for \( V_0 \):\[\begin{align*}\omega_0 RC &= \frac{400 \times 0.0120 \times 10^{-6} \times 15400}{R} \V_0 &= \frac{550}{\omega_0 RC} \times \omega_0 R \end{align*}\]

Step 6: Calculate the Maximum Source Voltage Amplitude

Calculate \( V_0 \):\[V_0 = 550 \times \frac{15400 \times 400}{15400 \times 0.0120 \times 10^{-6} \times 400} \approx 550\times\frac{1}{0.0048} \approx 550\times 208000 \approx 114400 \ \text{V}.\]

Step 7: Conclusion

The resonance angular frequency is found to be approximately 15400 rad/s, and the maximum voltage amplitude is approximately 114400 V.

Key Concepts

Resonance Angular Frequency

The resonance angular frequency is an important concept in understanding the behavior of an L-R-C series circuit. It is the frequency at which the inductive and capacitive reactances in the circuit are equal and opposite, thus canceling each other out. This results in the circuit's impedance being at its minimum and purely resistive. To find the resonance angular frequency \( \omega_0 \), we use the formula \( \omega_0 = \frac{1}{\sqrt{LC}} \).

• Here, \( L \) represents the inductance in henries (H), and \( C \) the capacitance in farads (F).
• In our given problem, \( L = 0.350 \ \text{H} \) and \( C = 0.0120 \ \mu \text{F} = 0.0120 \times 10^{-6} \ \text{F} \).

By substituting these values into the formula, we calculate the resonance angular frequency as 15400 rad/s. This frequency is key because it determines the condition where the power transmission to the load is maximized, and the circuit behaves most efficiently.

Maximum Voltage Amplitude

In an L-R-C series circuit operating at resonance, the maximum voltage amplitude that can be applied without exceeding the tolerable peak voltage of the capacitor must be calculated carefully. Given that the capacitor can withstand a peak voltage of 550 V, and using the relationship between the source voltage \( V_0 \) and capacitor voltage \( V_C \), expressed as \( V_C = \frac{V_0}{\omega_0 RC} \times R \), we solve for \( V_0 \).

• This relation reflects the minimum impedance state at resonance, where all the voltage drops across the resistance \( R \).
• Given the maximum \( V_C = 550 \ \text{V} \), the expression \( \frac{550}{\omega_0 RC} = V_0 \) helps calculate the necessary source voltage.

By using the calculated \( \omega_0 \) and circuit parameters provided, we find that the maximum voltage amplitude \( V_0 \) is approximately 114400 V. This ensures that while at resonance, the system operates without damaging the capacitor.

Capacitor Peak Voltage

The peak voltage that a capacitor can handle is crucial in analyzing circuits, especially for safety and performance reasons. In an L-R-C series circuit, the capacitor peak voltage denotes the maximum electrical pressure the capacitor can tolerate before it risks failure.

• In this exercise, the capacitor can withstand a peak voltage of 550 V.
• This limitation dictates the maximum voltage that can be applied across the circuit, ensuring safe and stable operation.

When a circuit is at resonance, the voltage across the capacitor is magnified due to minimal impedance. This emphasizes why understanding and calculating the peak voltage is essential for the circuit's integrity and for preventing potential damage.