In a plasma globe, a hollow glass sphere is filled with low-pr...

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 V_o 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 R_G and R_P 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 X_C of C at this frequency in MΩ. (c) The voltage may be V_o = 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 R_G << X_C. Assuming also that R_P << X_C, 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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Accepted Answer

Welcome back everyone in this problem, an interactive light source consists of a thin glass layer coated with a transparent conductive material on both inner and outer surfaces. The glass acts as a dielectric with a thickness D equals 1.8 millimeters and a dielectric constant K equals seven. A user's fingertip with an area A equals one square centimeter touches the outer surface of the device. The system operates at an AC voltage V not equals 300 volts and frequency F equals 15 khz. The equivalent circuit includes a capacitor C and resistances rla representing the glass and our user representing the user first calculate the capacitance C of the glass layer assuming a parallel plate model. And the second determine the capacity reactant XC. Finally, third estimate the peak current assuming the resistance of the glass is much less than the capacity of reactants. And the resistance of the user is also much less than the capacity of reactants. Here we have a diagram of the scenario and for our answer choices, A says the capacitance is 3.4 peak of farads that the capacitive reactance is 3.1 mega ohms and the peak current is 39 micro amps B says they are 3.4 peak of farads, 3.1 one mega ohms and 97 micro amperes respectively. C 3.4 P of farads, 7.2 mega ohms and 39 micro amps and D 5.1 P of 3.1 mega ohms and 97 micros respectively. Now, before we try to figure out our three are the three parts of our question. Let's just make a note of all the information that we have here. So far, we know that our glass has a thickness D of 1.8 millimeters, the dielectric constant K equals seven. The user's fingertip has an area a of one square centimeter. The system operates at a voltage of 300 volts and a frequency F of 15 khz. Now, how can we use this information to help solve our problem? Well, let's start with the first part. We want to find the capacitance of the glass layer. What do we know about capacitance? Recall that capacitance is equal to the value of the dielectric constant multiplied by the permittivity of free space, epsilon knot multiplied by the area divided by the thickness D. Well, we have all of this information and we know the permittivity of free space as a constant. So if we substitute those values, then we should get C to be equal to seven multiplied by epsilon, not 8.85 multiplied by 10 to the negative 12 fites per meter multiplied by our area of one square centimeter which when we rewrite it, uh in square meters is one multiplied by 10 to the negative fourth square meters. And all of that is being divided by our thickness of 1.8 millimeters. Now, when we go ahead and divide here, we should get the capac the capacitance to be 3.4417 multiplied by 10 to the negative 12 fads. And if we write rewrite that in peak of fats to two significant figures, it's approximately equal to 3.4 peak of farads. Thus, this is the capacitance of our glass layer. Next, we can move on to the second part of our problem to determine the capacitive reactance. Now, what do we know about capacity reactants? Let me make some space down here. Well, recall, OK, that the capacity of reactants is equal to one divided by two pi FC. Now, we know our frequency, we've just found our capacitance. So again, we can go ahead and substitute the solve. So that's going to be one divided by two pi multiplied by 15 kilohertz, 15, multiplied by 10 to the third Hertz multiplied by our capacitance of 3.4417 multiplied by 10 to the negative 12 fat. Now, when we go ahead and solve here, OK, then we should get our capacity reactants to be 3.0829 multiplied by 10 to the sixth ohms or in mega ohms to two significant figures, approximately 3.1 mega ohms. Now let's move on to the final part of our problem. We would like to find the peak current assuming the resistance of the glass and the user are much smaller than the capacitive reactants. How can we figure out what that peak current is? Well, if those resistances are much smaller than the capacity reactants, then the peak current, OK, is going to be equal to V not OK. Initial volt is divided by the capacity reactants. We know that the knot is 300 volts and we've just figured out our capacity reactants as 3.0289 multiplied by 10 to the sixth ohms. So when we go ahead and solve here, we get our peak current to be 97 0.311 multiplied by 10 to the negative sixth amperes or to two significant figures in micro amperes approximately 97 micro amperes. Therefore, our peak current is 97 micro amperes capacity reactant is 3.1 mega ohms and our capacitance is 3.4 peak of fires. If we look back at our answer choices that tells us B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Capacitance

Reactance

Current and Safety

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