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 7a

Chapter 32, Problem 7a

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?

Verified step by step guidance

Step 1: Understand the relationship between the peak current and the peak voltage in an AC circuit with a capacitor. The peak current (I₀) is given by the formula:

I

₀=V₀⁢X_c⁢-1, where

V

₀ is the peak voltage and

X

_c is the capacitive reactance.

Step 2: Recall the formula for capacitive reactance,

X

_c=(2πfC)-1, where

f

is the frequency of the AC generator and

C

is the capacitance of the capacitor.

Step 3: Substitute the given values into the formula for

X

_c. The capacitance is

0.30 μ F

(microfarads), which is equivalent to

0.30 \times 10 - 6 F

. The frequency is

100 Hz

. Plug these values into the formula:

X

_c=(2π⁢100⁢×⁢0.30⁢×⁢10⁢-6)-1.

Step 4: Once

X

_c is calculated, use the formula for peak current:

I

₀=V₀⁢÷⁢X_c. Substitute the peak voltage

V

₀=10⁢V and the calculated value of

X

_c into this formula.

Step 5: Simplify the expression to find the peak current

I

₀. This will give the peak current to and from the capacitor at the given frequency.

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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 electrical charge per unit voltage. It is measured in farads (F) and is defined as the ratio of the charge (Q) stored on one plate of the capacitor to the voltage (V) across the plates. In this question, the capacitor has a capacitance of 0.30 μF, which indicates how much charge it can store when connected to an AC voltage.

AC Voltage and Frequency

Alternating current (AC) voltage varies sinusoidally with time, characterized by its peak voltage and frequency. The frequency, measured in hertz (Hz), indicates how many cycles occur per second. In this case, the AC generator produces a peak voltage of 10 V at a frequency of 100 Hz, which is essential for calculating the current through the capacitor.

Peak Current in a Capacitor

The peak current (I_peak) through a capacitor in an AC circuit can be calculated using the formula I_peak = C * (dV/dt), where C is the capacitance and dV/dt is the rate of change of voltage. For a sinusoidal voltage, this can be simplified to I_peak = C * ω * V_peak, where ω (omega) is the angular frequency (2πf). This relationship shows how the current depends on both the capacitance and the frequency of the AC source.

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04:32

Phasors for Capacitors

Related Practice

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

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?

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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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