A capacitor has a peak current of 330 μA when the peak voltage at 250 kHz is 2.2 V. What is the capacitance?
Ch 32: AC Circuits
32장, 문제 10
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?
검증된 단계별 안내1
Step 1: Recall the formula for capacitive reactance, which is given by \( X_C = \frac{1}{2 \pi f C} \), where \( X_C \) is the capacitive reactance, \( f \) is the frequency, and \( C \) is the capacitance. This formula relates the frequency and capacitance to the opposition to current flow in a capacitor.
Step 2: Use Ohm's Law for AC circuits, which states \( I = \frac{V}{X_C} \), where \( I \) is the current, \( V \) is the voltage, and \( X_C \) is the capacitive reactance. Rearrange this formula to solve for \( X_C \): \( X_C = \frac{V}{I} \).
Step 3: Substitute the given values for the rms voltage (\( V = 6.0 \; \text{V} \)) and the peak current (\( I_{\text{peak}} = 65 \; \text{mA} \)). Note that the rms current \( I_{\text{rms}} \) is related to the peak current by \( I_{\text{rms}} = \frac{I_{\text{peak}}}{\sqrt{2}} \). Calculate \( I_{\text{rms}} \) first, then use it to find \( X_C \).
Step 4: Once \( X_C \) is determined, use the formula \( X_C = \frac{1}{2 \pi f C} \) to solve for the capacitance \( C \). Rearrange the formula to \( C = \frac{1}{2 \pi f X_C} \). Substitute the frequency \( f = 15 \; \text{kHz} = 15,000 \; \text{Hz} \) and the calculated \( X_C \) into this equation.
Step 5: Perform the calculations to find the value of \( C \). Ensure that the units are consistent throughout the calculation (e.g., convert mA to A and kHz to Hz). The final result will give the capacitance in Farads (F).

비슷한 문제에 대한 검증된 영상 답변:
이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
도움이 되었나요?
주요 개념
질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.
Capacitance
Capacitance is the ability of a capacitor to store electrical energy in an electric field, measured in farads (F). It is defined as the ratio of the electric charge stored on one plate of the capacitor to the voltage across the plates. In AC circuits, capacitance affects how the capacitor reacts to changing voltages and currents, influencing the overall impedance of the circuit.
추천 영상:
가이드 코스
Capacitors & Capacitance (Intro)
RMS Voltage
RMS (Root Mean Square) voltage is a statistical measure of the magnitude of a varying voltage. It represents the equivalent DC voltage that would deliver the same power to a load. In AC circuits, RMS voltage is crucial for calculating power and current, as it provides a consistent value for analysis, especially when dealing with sinusoidal waveforms.
추천 영상:
가이드 코스
RMS Current and Voltage
Peak Current
Peak current refers to the maximum instantaneous current flowing through a circuit during one cycle of an AC waveform. It is important for understanding the behavior of components like capacitors in AC circuits, as it helps determine how much current the capacitor can handle at its maximum charge. The relationship between peak current, RMS voltage, and capacitance is essential for calculating the value of the capacitor.
추천 영상:
가이드 코스
RMS Current and Voltage
관련 실천
교과서 질문
153
views
교과서 질문
The peak current to and from a capacitor is 10 mA. What is the peak current if the emf frequency is doubled?
110
views
교과서 질문
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)?
108
views
교과서 질문
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?
126
views
교과서 질문
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?
161
views
교과서 질문
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?
109
views
