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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 93

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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Step 1: To determine the capacitance (C) in part (a), use the formula for a parallel-plate capacitor: C=K𝔻Ad, where K is the dielectric constant (5.0), 𝔻 is the permittivity of free space (8.85×10-12 F/m), A is the area of the fingertip (1.0 cm² = 1.0×10-4), and d is the thickness of the glass (2.0 mm = 2.0×10-3 m). Substitute these values into the formula to calculate C.
Step 2: For part (b), calculate the capacitive reactance (XC) using the formula: XC=1(2πfC), where f is the frequency (12 kHz = 12×103 Hz) and C is the capacitance calculated in step 1. Substitute the values to find XC in ohms, then convert to MΩ.
Step 3: For part (c), estimate the peak current (Ipeak) using Ohm's law: Ipeak=VoXC, where Vo is the peak voltage (2500 V) and XC is the capacitive reactance calculated in step 2. Substitute the values to find Ipeak. Then, compare the current to the threshold for danger (typically 10 mA for humans) to determine if it is dangerous.
Step 4: For part (d), repeat the calculation of Ipeak using the same formula as in step 3, but with a new frequency f = 1.0 MHz = 1.0×106 Hz. Substitute the new frequency into the formula for XC (step 2) and then calculate Ipeak. Again, compare the current to the danger threshold to assess safety.
Step 5: Summarize the results for parts (c) and (d), discussing whether the current levels at the two frequencies (12 kHz and 1.0 MHz) are dangerous. Highlight the role of capacitive reactance in limiting current flow and how it changes with frequency.

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Key Concepts

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

Capacitance

Capacitance is the ability of a system to store electric charge per unit voltage. In the context of a parallel-plate capacitor, it is determined by the formula C = (K * ε₀ * A) / d, where K is the dielectric constant, ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between them. This concept is crucial for calculating the capacitance of the gas and fingertip in the plasma globe.
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Reactance

Reactance is the opposition that a capacitor presents to alternating current (AC) due to its capacitance. It is given by the formula XC = 1 / (2πfC), where f is the frequency of the AC signal and C is the capacitance. Understanding reactance is essential for determining how the capacitor behaves at different frequencies, particularly in the context of the plasma globe's operation.
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Current and Safety

Current is the flow of electric charge, measured in amperes (A). In the context of the plasma globe, the peak current can be estimated using Ohm's law and the circuit's resistances. Evaluating whether this current level is dangerous involves understanding the thresholds for human safety, as currents above certain levels can cause harm or be lethal, especially in high-voltage scenarios.
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Related Practice
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

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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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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