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Chapter 21: Problem 1

A lightbulb is connected to a \(120-\mathrm{V}\) (rms), \(60-\mathrm{Hz}\) source. How many times per second does the current reverse direction?

Short Answer

Expert verified
Answer: The current changes direction 120 times per second.

Step by step solution

01

Determine the frequency of the AC source

Given that the frequency of the source is 60 Hz, this means that there are 60 cycles per second. In each cycle, the voltage changes from a positive peak, through zero volts, to a negative peak, and back to zero volts. Thus, the current will change direction twice per cycle.
02

Calculate the number of times the current changes direction per second

As there are 60 cycles per second, and the current changes direction twice per cycle, we can calculate the number of times current changes direction per second by multiplying the frequency (in Hz) by 2: Number of direction changes per second = Frequency × 2 Number of direction changes per second = 60 Hz × 2
03

Present the final answer

Therefore, the current changes direction 120 times per second.

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Most popular questions from this chapter

The field coils used in an ac motor are designed to have a resistance of $0.45 \Omega\( and an impedance of \)35.0 \Omega$ What inductance is required if the frequency of the ac source is (a) \(60.0 \mathrm{Hz} ?\) and (b) $0.20 \mathrm{kHz} ?$
A \(0.400-\mu \mathrm{F}\) capacitor is connected across the terminals of a variable frequency oscillator. (a) What is the frequency when the reactance is \(6.63 \mathrm{k} \Omega ?\) (b) Find the reactance for half of that same frequency.
A \(0.250-\mu \mathrm{F}\) capacitor is connected to a \(220-\mathrm{V}\) rms ac source at \(50.0 \mathrm{Hz}\) (a) Find the reactance of the capacitor. (b) What is the rms current through the capacitor?
What is the rins current flowing in a \(4.50-\mathrm{kW}\) motor connected to a \(220-\mathrm{V}\) rms line when (a) the power factor is 1.00 and (b) when it is \(0.80 ?\)
A 6.20 -m \(H\) inductor is one of the elements in a simple RLC series circuit. When this circuit is connected to a \(1.60-\mathrm{kHz}\) sinusoidal source with an \(\mathrm{rms}\) voltage of \(960.0 \mathrm{V},\) an rms current of $2.50 \mathrm{A}\( lags behind the voltage by \)52.0^{\circ} .$ (a) What is the impedance of this circuit? (b) What is the resistance of this circuit? (c) What is the average power dissipated in this circuit?
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