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Ch. 30 - Inductance, Electromagnetic Oscillations, and AC Circuits
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 29, Problem 91

For the circuit shown in Fig. 30–35, show that if the condition R₁ R₂ = L/C is satisfied then the potential difference between points a and b is zero for all frequencies.

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Step 1: Begin by analyzing the circuit. The circuit likely contains resistors (R₁ and R₂), an inductor (L), and a capacitor (C). The goal is to show that the potential difference between points a and b is zero when the condition R₁ R₂ = L/C is satisfied.
Step 2: Write the impedance of the inductor and capacitor. The impedance of the inductor is given by Z_L = iωL, and the impedance of the capacitor is Z_C = 1/(iωC), where ω is the angular frequency of the AC source.
Step 3: Use Kirchhoff's Voltage Law (KVL) to analyze the circuit. Write the voltage drops across R₁, R₂, L, and C in terms of their respective impedances and currents. Express the total voltage difference between points a and b in terms of these quantities.
Step 4: Substitute the condition R₁ R₂ = L/C into the expression for the potential difference. Simplify the resulting equation to show that the potential difference between points a and b becomes zero for all frequencies.
Step 5: Conclude that the condition R₁ R₂ = L/C ensures that the contributions of the inductor and capacitor cancel out the contributions of the resistors, resulting in zero potential difference between points a and b for all frequencies.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Resonance in RLC Circuits

In RLC circuits, resonance occurs when the inductive reactance (L) and capacitive reactance (C) are equal, leading to a condition where the circuit can oscillate at a specific frequency. This frequency is determined by the values of L and C, and at resonance, the impedance is minimized, allowing maximum current flow. Understanding this concept is crucial for analyzing how the circuit behaves at different frequencies.
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Resonance in Series LRC Circuits

Impedance and Phase Relationships

Impedance in an RLC circuit combines resistance (R) and reactance (X) and is represented as a complex number. The phase relationship between voltage and current is affected by the impedance, which varies with frequency. When the condition R₁ R₂ = L/C is met, it indicates a specific balance that results in zero potential difference between points a and b, regardless of frequency.
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Voltage Division in Circuits

Voltage division is a principle that describes how the total voltage across a series circuit is distributed among its components based on their resistances or impedances. In the context of the given circuit, if the condition R₁ R₂ = L/C holds, the voltage drop across the components leads to equal potential at points a and b, resulting in a net voltage of zero. This concept is essential for understanding how voltages behave in complex circuits.
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Related Practice
Textbook Question

In a plasma globe, a hollow glass sphere is filled with low-pressure gas and a small spherical metal electrode is located at its center. Assume an ac voltage source of peak voltage Vo and frequency f is applied between the metal sphere and the ground, and that a person is touching the outer surface of the globe with a fingertip, whose approximate area is 1.0 cm². The equivalent circuit for this situation is shown in Fig. 30–36, where RG and RP are the resistances of the gas and the person, respectively, and C is the capacitance formed by the gas, glass, and finger. (a) Determine C assuming it is a parallel-plate capacitor. The conductive gas and the person’s fingertip form the opposing plates of area A = 1.0 cm². The plates are separated by glass (dielectric constant K = 5.0) of thickness d = 2.0 mm. (b) In a typical plasma globe, f = 12 kHz. Determine the reactance XC of C at this frequency in MΩ. (c) The voltage may be Vo = 2500 V. With this high voltage, the dielectric strength of the gas is exceeded and the gas becomes ionized. In this “plasma” state, the gas emits light (“sparks”) and is highly conductive so that RG << XC. Assuming also that RP << XC, estimate the peak current that flows in the given circuit. Is this level of current dangerous? (d) If the plasma globe operated at f = 1.0 MHz, estimate the peak current that would flow in the given circuit. Is this level of current dangerous?


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Textbook Question

The RC circuit shown in Fig. 30–39 is a low-pass filter because it passes low-frequency ac signals with less attenuation than high-frequency ac signals. (a) Show that the voltage gain is A=Vout/Vin=1/(4π2f2R2C2+1)12A = V_{\(\text{out}\)}/V_{\(\text{in}\)} = 1/(4\(\pi\)^2 f^2 R^2 C^2 + 1)^{\(\frac{1}{2}\)} (b) Discuss the behavior of the gain A for f → 0 and f → ∞.

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Textbook Question

To detect vehicles at traffic lights, wire loops with dimensions on the order of 2 m are often buried horizontally under roadways. Assume the self-inductance of such a coil is L = 5.0 mH and that it is part of an LRC circuit as shown in Fig. 30–40 with C = 0.10 μF and R = 38 Ω. The ac voltage has frequency f and rms voltage Vrms. (a) The frequency f is chosen to match the resonant frequency f₀ of the circuit. Find f₀ and determine what the rms voltage (VR)rms across the resistor will be when f = f₀. (b) Assume that f, C, and R never change, but that, when a car is located above the buried coil, the coil’s self-inductance decreases by 10% (due to induced eddy currents in the car’s metal parts). Determine by what factor the voltage (VR)rms decreases in the presence of a car in comparison to no car above the loop and thus how it detects the presence of a car. (c) Describe how the eddy currents induced in the car reduce L. [Hint: Recall Eq. 30–4, the definition of inductance.]

Textbook Question

Show that if the inductor L in the filter circuit of Fig. 30–33 (Problem 87) is replaced by a large resistor R, there will still be significant attenuation of the ac voltage and little attenuation of the dc voltage if the input dc voltage is high and the current (and power) are low.

Textbook Question

Filter circuit. Figure 30–33 shows a simple filter circuit designed to pass dc voltages with minimal attenuation and to remove, as much as possible, any ac components (such as 60-Hz line voltage that could cause hum in an audio system, for example). Assume Vin = V1 + V2 where V1 is dc and V2 = V20 sin ωt, and that any resistance is very small. (a) Determine the current through the capacitor: give amplitude and phase (assume R = 0 and XL > XC). (b) Show that the ac component of the output voltage, V2out, equals (Q/C) - V1 where Q is the charge on the capacitor at any instant, and determine the amplitude and phase of V2out (c) Show that the attenuation of the ac voltage is greatest when XC << XL, and calculate the ratio of the output to input ac voltage in this case. (d) Compare the dc output voltage to input voltage.

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Textbook Question

Suppose a series LRC circuit has two resistors, R₁ and R₂, two capacitors, C₁ and C₂, and two inductors, L₁ and L₂ all in series. Calculate the total impedance of the circuit.

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