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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
All textbooksKnight Calc 5th EditionCh 32: AC CircuitsProblem 12a
Chapter 32, Problem 12a

A capacitor has a peak current of 330 μA when the peak voltage at 250 kHz is 2.2 V. What is the capacitance?

Verified step by step guidance
1
Start by recalling the relationship between the peak current (I₀), the peak voltage (V₀), the angular frequency (ω), and the capacitance (C) for an AC circuit: I₀ = ωC * V₀.
Calculate the angular frequency (ω) using the formula ω = 2πf, where f is the frequency. Here, f = 250 kHz, so substitute this value into the formula.
Rearrange the formula I₀ = ωC * V₀ to solve for the capacitance (C): C = I₀ / (ω * V₀).
Substitute the given values into the formula: I₀ = 330 μA (convert to amperes: 330 × 10⁻⁶ A), V₀ = 2.2 V, and ω (calculated in the previous step).
Simplify the expression to find the capacitance (C). Ensure the units are consistent throughout the calculation.

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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 by the formula C = Q/V, where C is capacitance, Q is the charge stored, and V is the voltage across the capacitor. In AC circuits, capacitance also affects how the capacitor responds to changing voltages.
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Capacitors & Capacitance (Intro)

Peak Current and Peak Voltage

Peak current refers to the maximum instantaneous current flowing through a circuit, while peak voltage is the maximum voltage across a component. In the context of capacitors, these values are crucial for determining how the capacitor behaves under alternating current (AC) conditions, particularly at a specific frequency, such as 250 kHz in this case.
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RMS Current and Voltage

Reactance

Reactance is the opposition that a capacitor or inductor presents to the flow of alternating current due to its capacitance or inductance. For capacitors, the reactance (Xc) can be calculated using the formula Xc = 1/(2πfC), where f is the frequency and C is the capacitance. This relationship is essential for understanding how capacitors influence current and voltage in AC circuits.
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Related Practice
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

A low-pass filter consists of a 100 μF capacitor in series with a 159 Ω resistor. The circuit is driven by an AC source with a peak voltage of 5.00 V. What is the crossover frequency fc?

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Textbook Question

A 20 nF capacitor is connected across an AC generator that produces a peak voltage of 5.0 V. What is the instantaneous value of the emf at the instant when iC = IC?

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Textbook Question

A capacitor has a peak current of 330 μA when the peak voltage at 250 kHz is 2.2 V. If the peak voltage is held constant, what is the peak current at 500 kHz?

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Textbook Question

A high-pass RC filter with a crossover frequency of 1000 Hz uses a 100 Ω resistor. What is the value of the capacitor?

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Textbook Question

A 20 nF capacitor is connected across an AC generator that produces a peak voltage of 5.0 V. At what frequency f is the peak current 50 mA?