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Chapter 20: Problem 34

The rms current in a \(47-\Omega\) resistor is 0.50 A. What is the peak value of the voltage across this resistor?

Short Answer

Expert verified
The peak voltage across the resistor is approximately 33.2 V.

Step by step solution

01

Identify Known Values

We are given the resistance of the resistor as \( R = 47 \, \Omega \) and the RMS current as \( I_{\text{rms}} = 0.50 \, A \).
02

Calculate RMS Voltage

Use Ohm's law for RMS values to find the RMS voltage: \( V_{\text{rms}} = I_{\text{rms}} \times R \). Substitute the known values: \( V_{\text{rms}} = 0.50 \, A \times 47 \, \Omega = 23.5 \, V \).
03

Calculate Peak Voltage

The peak voltage \( V_{\text{peak}} \) is related to the RMS voltage by the equation \( V_{\text{peak}} = \sqrt{2} \times V_{\text{rms}} \). Substitute the RMS voltage value: \( V_{\text{peak}} = \sqrt{2} \times 23.5 \, V \approx 33.2 \, V \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resistor
A resistor is an essential component in electrical circuits. Its primary role is to limit the flow of electrical current. Measured in ohms (惟), its value dictates how much it resists the flow of electric current. This resistance is crucial in determining how much voltage is required for a certain amount of current to pass through the resistor, based on Ohm's Law.

  • Ohm's Law states that the voltage across a resistor is directly proportional to the current flowing through it, expressed as \( V = I imes R \).
  • In a circuit, resistors ensure that electronic components receive the required amount of current without being damaged.
If we take the example from the given exercise, a 47-ohm resistor allows an RMS current of 0.50 A to flow through it. This example highlights how resistors manage current flow, making them a vital part of almost every electronic device.
RMS Current
RMS (Root Mean Square) current is a statistical measure of the magnitude of a varying current. It provides a value that represents the effective or "average" amount of current that flows through a circuit in one complete cycle of a waveform, like a sine wave.

  • RMS current is used because direct current (DC) measurements can't directly apply to alternating current (AC), where the current changes direction.
  • It reflects the amount of work done by the AC current, similar to how a DC current would perform in the same conditions.
In our exercise, the RMS current through the resistor is 0.50 A. This value is crucial in calculating the RMS voltage that can later be translated into peak voltage, showcasing its importance in AC circuits.
Peak Voltage
Peak voltage represents the maximum voltage value in an AC cycle. While RMS values give us the effective voltage, the peak voltage demonstrates how high the waveform reaches in one cycle.

  • The relationship between RMS and peak voltage in a sine wave can be given by: \( V_{\text{peak}} = \sqrt{2} \times V_{\text{rms}} \).
  • This conversion is necessary because many electrical devices use RMS values for operation, but peak values for handling maximum load.
For the exercise, using the rms voltage of 23.5 V, the peak voltage calculates to approximately 33.2 V. This conversion showcases how understanding different voltage measures is essential for designing and analyzing AC electrical systems.

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Most popular questions from this chapter

According to Equation \(20.7,\) an ac voltage \(V\) is given as a function of time \(t\) by \(V=V_{0} \sin 2 \pi f t,\) where \(V_{0}\) is the peak voltage and \(f\) is the frequency (in hertz). For a frequency of \(60.0 \mathrm{Hz}\), what is the smallest value of the time at which the voltage equals one-half of the peak value?

You and your team are designing a small tube heater that consists of a small ceramic tube wrapped with a special heater wire composed of Nichrome. When you run an electrical current through the wire, the wire (and therefore the tube) heats up through resistive heating. Nichrome is an alloy composed of \(80 \%\) nickel and \(20 \%\) chromium, and has a resistivity of \(\rho=1.25 \times 10^{-6} \Omega \cdot \mathrm{m} .\) (a) What is the resistance per centimeter of 28-gauge Nichrome wire? ("28 gauge" means the wire has a diameter \(D=0.320 \mathrm{mm}\) (b) You wrap the tube with \(45.0 \mathrm{cm}\) of the Nichrome wire, and you want the power output of your heater to be \(120 \mathrm{W}\). What current is needed? (c) What voltage will provide the current calculated in (b)?

An \(86-\Omega\) resistor and a \(67-\Omega\) resistor are connected in series across a battery. The voltage across the \(86-\Omega\) resistor is 27 V. What is the voltage across the \(67-\Omega\) resistor?

The current in a series circuit is 15.0 A. When an additional \(8.00-\Omega\) resistor is inserted in series, the current drops to \(12.0 \mathrm{A}\). What is the resistance in the original circuit?

Two scales on a nondigital voltmeter measure voltages up to 20.0 and \(30.0 \mathrm{V},\) respectively. The resistance connected in series with the galvanometer is \(1680 \Omega\) for the \(20.0-\mathrm{V}\) scale and \(2930 \Omega\) for the \(30.0-\mathrm{V}\) scale. Determine the coil resistance and the full-scale current of the galvanometer that is used in the voltmeter.
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