Ch 32: AC Circuits

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 32: AC Circuits

Problem 7b

Chapter 32, Problem 7b

A 0.30 μF capacitor is connected across an AC generator that produces a peak voltage of 10 V. What is the peak current to and from the capacitor if the emf frequency is 100 kHz?

Verified step by step guidance

Step 1: Recall the formula for capacitive reactance, which is the opposition to current flow in a capacitor in an AC circuit. The formula is:

X c_{c} = \frac{1}{π}

, where

f

is the frequency of the AC source and

C

is the capacitance.

Step 2: Substitute the given values into the formula for capacitive reactance. Here,

f

= 100 kHz =

100000

Hz and

C

= 0.30 μF =

0.30 \times 10^{-6}

Step 3: Use Ohm's law for AC circuits to find the peak current. The formula is:

I peak = \frac{V}{peak}

, where

V peak

is the peak voltage and

X c

is the capacitive reactance.

Step 4: Substitute the given peak voltage,

V peak = 10

V, and the calculated value of

X c

from Step 2 into the formula for peak current.

Step 5: Simplify the expression to find the peak current, ensuring that all units are consistent. The result will give the peak current to and from the capacitor in amperes.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Capacitance

Capacitance is the ability of a capacitor to store charge per unit voltage, measured in farads (F). In this case, the capacitor has a capacitance of 0.30 μF, which indicates how much electric charge it can hold at a given voltage. The relationship between charge (Q), voltage (V), and capacitance (C) is given by the formula Q = C * V.

AC Voltage and Current

In an alternating current (AC) circuit, the voltage and current vary sinusoidally with time. The peak voltage (V_peak) is the maximum voltage reached, which in this scenario is 10 V. The peak current can be calculated using the capacitive reactance, which depends on the frequency of the AC signal and the capacitance of the capacitor.

Capacitive Reactance

Capacitive reactance (X_C) is the opposition that a capacitor offers to the flow of AC current, calculated using the formula X_C = 1 / (2πfC), where f is the frequency and C is the capacitance. At a frequency of 100 kHz and a capacitance of 0.30 μF, this reactance determines how much current can flow through the capacitor when subjected to the peak voltage.

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Capacitors & Capacitance (Intro)

Related Practice

Textbook Question

A 0.30 μF capacitor is connected across an AC generator that produces a peak voltage of 10 V. What is the peak current to and from the capacitor if the emf frequency is 100 Hz?

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Textbook Question

A capacitor is connected to a 15 kHz oscillator. The peak current is 65 mA when the rms voltage is 6.0 V. What is the value of the capacitance C?

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Textbook Question

The peak current to and from a capacitor is 10 mA. What is the peak current if the emf frequency is doubled?

Textbook Question

The emf phasor in FIGURE EX32.1 is shown at t = 2.0 ms. What is the peak value of the emf?

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Textbook Question

The peak current to and from a capacitor is 10 mA. What is the peak current if the emf peak voltage is doubled (at the original frequency)?

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Textbook Question

The emf phasor in FIGURE EX32.1 is shown at t = 2.0 ms. What is the angular frequency ω? Assume this is the first rotation.

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