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The \(12 \mathrm{V}\) rating of the bulb refers to the rms voltage. What is the peak voltage across the bulb? A. \(8.5 \mathrm{V}\) B. \(12 \mathrm{V}\) C. \(17 \mathrm{V}\) D. \(24 \mathrm{V}\)

Short Answer

Expert verified
The peak voltage across the bulb is approximately \(C. 17 \mathrm{V}\).

Step by step solution

01

Recall the relationship between rms voltage and peak voltage

The rms voltage for a sinusoidal waveform can be related to the peak voltage using the formula \(V_p = \sqrt{2} * V_{rms}\). 'Vp' stands for peak voltage and 'Vrms' is the rms voltage.
02

Substitute the given Vrms into the formula

Substitute the given Vrms that is 12V into the formula to calculate the peak voltage. \(V_p = \sqrt{2} * 12V\)
03

Calculate the peak voltage

Perform the multiplication to find the peak voltage: \(V_p = 16.97V ≈ 17V\).

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Key Concepts

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

Peak Voltage
When dealing with AC (alternating current) circuits, understanding peak voltage is crucial. Peak voltage represents the maximum voltage value reached by a sinusoidal waveform in one cycle.
For example, in the context of the exercise, when we have an RMS voltage of 12V, the peak voltage will be higher. This is because RMS (Root Mean Square) value is like the "average" amount of electricity delivered by AC power circuits.
To find the peak voltage from a known RMS value, you can use the formula:
  • \(V_p = \sqrt{2} \times V_{rms}\)
Here, \(V_p\) is the peak voltage, and \(V_{rms}\) is the RMS voltage.
In the original exercise example, since the RMS voltage is 12V, the calculation would be \(V_p = \sqrt{2} \times 12V = 16.97V\), which can be approximated to 17V.
This shows the importance of peak voltage in understanding how intense the actual voltage can get at maximum. It helps in designing and checking how safe a circuit is in its worst-case scenario.
Voltage Calculations
Voltage calculations in the context of AC circuits help us gauge how electrical systems perform. When handling AC voltages, we rarely deal with one static number because the voltage constantly changes. This leads to two main types of values we calculate: peak voltage and RMS voltage.
  • **RMS Voltage**: Gives an "average" value of AC voltage concerning power consumption, allowing for comparison with DC voltage. When you see typical voltage ratings, such as 120V or 230V, these are RMS values.
  • **Peak Voltage**: The maximum voltage in a cycle. Calculated using the formula \(V_p = \sqrt{2} \times V_{rms}\).
Knowing how to convert between these values is essential. It ensures that devices can operate safely under varying electrical conditions. For example, integrating this into device design, engineers must consider both these calculations to prevent potential overloads.
Additionally, these calculations are pivotal in designing electrical systems and ensuring they operate effectively under specified conditions. Understanding these intricacies helps avoid equipment failure and enhances efficiency.
Sinusoidal Waveform
The sinusoidal waveform is fundamental to understanding AC circuits. It represents how voltage or current alternates over time, swinging systematically between a maximum and minimum value. Most AC power supplied to homes and industries is sinusoidal.
  • **Characteristics**: A sinusoidal waveform is smooth and periodic. It can be described using amplitude (the peak value), frequency (how often it cycles per second), and phase (the time shift from a reference point).
  • **RMS and Peak Values**: While peak values reflect the waveform's maximum point, RMS provides a more useful average for practical power calculation. The sinusoidal nature of AC allows for efficient energy distribution in these terms.
Sinusoidal waveforms make complex electrical circuits predictable and manageable. Engineers use these waveforms to design systems that ensure constant and reliable power delivery.
Understanding sinusoidal behavior is vital for anyone working with AC circuits, showing why it's a standard for efficiently supplying electricity.

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Most popular questions from this chapter

A \(60 \mathrm{W}\) ( \(120 \mathrm{V}\) ) night light is turned on for an average of \(12 \mathrm{h}\) a day year round. What is the annual cost of electricity at a billing rate of \(\$ 0.10 / \mathrm{kWh} ?\)

\(\mathrm{A} 20 \mu \mathrm{F}\) capacitor is connected across an \(\mathrm{AC}\) generator that produces a peak voltage of \(6.0 \mathrm{V}\). The peak current is \(0.20 \mathrm{A}\). What is the oscillation frequency in Hz?

A capacitor is connected across an AC source. At one instant in time, the current through the capacitor is \(23 \mathrm{mA}\) when the voltage across the capacitor is changing at a rate of \(7.2 \times 10^{4} \mathrm{V} / \mathrm{s} .\) What is the value of the capacitance?

The girl in Figure 26.12 of the chapter has her hand on the sphere of a Van de Graaff generator that is at a potential of \(400,000 \mathrm{V}\). She is standing on an insulating platform, so no current flows through her. But what happens if she touches something that is grounded? Is she still safe? Figure \(P 26.57\) shows the equivalent circuit. \(C_{\mathrm{VDG}}=20 \mathrm{pF}\) represents the capacitance of the Van de Graaff sphere, \(C_{\text {girl }}=100 \mathrm{pF}\) is the capacitance of the girl's body, and \(R=5 \mathrm{k} \Omega\) is the resistance of her body, from one hand to the other. The switch is closed when she touches ground. What is the initial current at the instant the switch is closed? b. What is the time constant for the current to decay? c. Occupational safety experts have found that a shock is safe if the product of the voltage, the current, and the time the current is delivered is less than \(13.5 \mathrm{V} \cdot \mathrm{A} \cdot \mathrm{s}=13.5 \mathrm{J}\) (see Problem 26.28). Estimate this product. Is this shock safe?

An FM radio station broadcasts at a frequency of \(100 \mathrm{MHz}\). What inductance should be paired with a 10 pF capacitor to build a receiver circuit for this station?

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