Reactance: If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the frequency is closest to

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Quizplus · Accepted Answer

The reactance $X_C$ of a capacitor is given by $X_C = \frac{1}{2\pi fC}$, where $f$ is the frequency in hertz, and $C$ is the capacitance in farads. The voltage $V$ across the capacitor is related to the current $I$ through it by $V = IX_C$. Rearranging the formula for $X_C$ to solve for $f$ and substituting $V/I$ for $X_C$, we get $f = \frac{1}{2\pi C(V/I)}$. Substituting the given values $V = 12.0\,V$, $I = 3.33\,mA = 3.33 \times 10^{-3}\,A$, and $C = 8.50\,nF = 8.50 \times 10^{-9}\,F$, we find $f \approx 5.20\,MHz$.

Reactance: If the Voltage Amplitude Across an 8