/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 A capacitor is connected across ... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
The value of the capacitance is approximately \(3.194 \times 10^{-7} \, \mathrm{F}\)

Step by step solution

01

Identify the given values

The values given in the problem are: Current, \(I = 23 \, \mathrm{mA} = 23 \times 10^{-3} \, \mathrm{A}\), and the rate of change of voltage, \(\frac{dV}{dt} = 7.2 \times 10^{4} \, \mathrm{V/s}\).
02

Rearrange the formula to solve for \(C\)

We start with the formula \(I = C \frac{dV}{dt}\). We need to find \(C\), so after rearranging the expression, we get \(C = \frac{I}{\frac{dV}{dt}}\).
03

Substitute the given values

Substitute the given values of \(I\) and \(\frac{dV}{dt}\) into the rearranged formula: \(C = \frac{23 \times 10^{-3}}{7.2 \times 10^{4}}\).
04

Calculate the value of \(C\)

The capacitance is obtained by performing the above division: \(C \approx 3.194 \times 10^{-7} \, \mathrm{F}\).

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Key Concepts

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

Understanding AC Circuits
Alternating Current (AC) circuits are fundamental to modern electrical systems, playing a crucial role in how power is delivered to homes and businesses. Unlike Direct Current (DC) which flows in one direction, AC reversibly changes its direction and magnitude cyclically. This feature is ideal for long-distance power transmission and for powering electric motors and appliances.

In AC circuits, the voltage across any component varies sinusoidally with time, and this variation leads to a corresponding cyclical change in current. These changes occur at a specific frequency, typically 50 or 60 Hz in most household systems. Understanding the behavior of components like capacitors within these circuits is essential for designing and troubleshooting electrical and electronic systems.
Capacitors in AC
Capacitors are passive electrical components that store and release electrical energy in the form of an electric field. In AC circuits, capacitors have unique properties compared to their behavior in DC circuits. When connected to an AC source, the capacitor will alternately charge and discharge in response to the alternating voltage.

The ability of a capacitor to store and discharge energy in an AC circuit is represented by its 'capacitive reactance', which depends on the frequency of the AC signal and the capacitor's value in Farads (F). Unlike in DC circuits where a steady state current is blocked once the capacitor is fully charged, in AC circuits the continuous voltage change allows capacitors to constantly charge and discharge, thereby allowing an alternating current to flow through them.
Current-Voltage Relationship in Capacitors
In the context of AC circuits, understanding the relationship between current and voltage across a capacitor is essential. The current flowing through a capacitor is proportional to the rate at which the voltage across it changes - a concept described by the equation: \( I = C \frac{dV}{dt} \), where \( I \) is the current, \( C \) is the capacitance, and \( \frac{dV}{dt} \) is the rate of change of voltage over time.

This relationship is critical because it shows that the peak current will occur when the rate of voltage change is the greatest. The current leads the voltage by 90 degrees in phase in a pure capacitive circuit, indicating that the current reaches its maximum value before the voltage does. This phase difference between the voltage and the current in AC circuits can significantly affect how the circuit functions, and thus the proper operation of devices that utilize capacitors.

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Most popular questions from this chapter

A series \(R L C\) circuit consists of a \(280 \Omega\) resistor, a \(25 \mu \mathrm{H}\) inductor, and an \(18 \mu \mathrm{F}\) capacitor. What is the rms current if the emf is supplied by a standard \(120 \mathrm{V}, 60 \mathrm{Hz}\) wall outlet?

A toaster oven is rated at \(1600 \mathrm{W}\) for operation at \(120 \mathrm{V}, 60 \mathrm{Hz}\). a. What is the resistance of the oven heater element? b. What is the peak current through it? c. What is the peak power dissipated by the oven?

An inductor is connected to a \(15 \mathrm{kHz}\) oscillator that produces an rms voltage of \(6.0 \mathrm{V}\). The peak current is \(65 \mathrm{mA}\). What is the value of the inductance \(L ?\)

Electricity is distributed to neighborhoods at a relatively high AC voltage, often \(7200 \mathrm{V}\). Transformers mounted on utility poles then transform this high voltage down to the \(120 \mathrm{V}\) used in homes. A typical transformer of this kind can handle as much as \(15 \mathrm{kW}\) of electric power flowing through it from its primary to its secondary. What is the primary current at this maximum power?

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?

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