Chapter 28: Problem 33
An electric drill draws \(4.6 \mathrm{A}\) rms at \(120 \mathrm{V}\) rms. If the current lags the voltage by \(25^{\circ},\) what's the drill's power consumption?
Chapter 28: Problem 33
An electric drill draws \(4.6 \mathrm{A}\) rms at \(120 \mathrm{V}\) rms. If the current lags the voltage by \(25^{\circ},\) what's the drill's power consumption?
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Get started for freeFor \(R L C\) circuits in which the resistance isn't too high, the \(Q\) factor may be defined as the ratio of the resonant frequency to the difference between the two frequencies where the power dissipated in the circuit is half the power dissipated at resonance. Using suitable approximations, show that this definition leads to \(Q=\omega_{0} L / R,\) with \(\omega_{0}\) the resonant frequency.
An AC current is given by \(I=495 \sin (9.43 t),\) with \(I\) in \(\mathrm{mA}\) and \(t\) in ms. Find (a) the rms current and (b) the frequency in \(\mathrm{Hz}\).
A step-up transformer increases voltage, or energy per unit charge. Why doesn't this violate energy conservation?
A series \(R L C\) circuit has \(R=18 \mathrm{k} \Omega, C=14 \mu \mathrm{F},\) and \(L=0.20 \mathrm{H}\) (a) At what frequency is its impedance lowest? (b) What's the impedance at this frequency?
Why does it make sense that inductive reactance increases with frequency?
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