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Unformatted text preview: CHAPTER 23 ALTERNATING CURRENT CIRCUITS CONCEPTUAL QUESTIONS ____________________________________________________________________________________________ 1. REASONING AND SOLUTION A light bulb and a parallel plate capacitor (including a dielectric material between the plates) are connected in series to the 60Hz ac voltage present at a wall outlet. Since the capacitor and the light bulb are in series, the rms current at any instant is the same through each element, and is given by Equation 23.6: , where Z is the impedance of the circuit. The impedance of the circuit is given by Equation 23.7 with X L = 0 (since there is no inductance in the circuit): I V rms rms = / Z Z R X R X = + = + 2 2 2 2 (– ) C C . According to Equation 19.10, if the dielectric between the plates of the capacitor is removed, the capacitance decreases by a factor of κ , where κ is the dielectric constant. From Equation 23.2, X C f C = 1 2 / ( ) π , we see that decreasing the capacitance increases the capacitive reactance . Therefore, the impedance of the circuit, X C Z R X = + 2 2 C , increases. The rms current is I V rms rms Z = / and will, therefore, be less than it was before the dielectric was removed. Thus, the brightness will decrease . ____________________________________________________________________________________________ 2 . SSM REASONING AND SOLUTION The ends of a long straight wire are connected to the terminals of an ac generator, and the current is measured. The wire is then disconnected, wound into the shape of a multipleturn coil, and reconnected to the generator. After the wire is wound into a coil, it has a greater inductance. When the generator is turned on, the coil will develop a voltage that opposes a change in the current according to Faraday's law. Since the induced voltage opposes the rise in current, the rms current in the circuit will be less than it was before the wire was wound into a coil. This can also be seen by considering Equations 23.6 and 23.7. Before the wire was wound into a coil, its primary property was that of resistance, and the current through the wire was given by with Z = R , or I V rms rms = / Z R I V rms rms = / . After the wire is wound into a coil, the wire possesses both resistance and inductance. Now the current in the coil is given by , where I V rms rms = / Z Z R X = + 2 2 L . Since Z is necessarily greater than R , the rms current is less when the wire is wound into a coil. ____________________________________________________________________________________________ 3 . REASONING AND SOLUTION An aircore inductor is connected in series with a light bulb of resistance R . This circuit is plugged into an electrical outlet. The current in the circuit is given by Equation 23.6, I V rms rms Z = / , where, from Equation 23.7, Z R X = + 2 2 L . When a piece of iron is inserted in the inductor, the magnetic field in the inductor is enhanced relative to that in air, and the inductance increases. Equation 23.4, X L = 2 f L...
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 Spring '08
 holland
 Current, Light, Alternating Current, Inductor, Electrical impedance

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