Question12-46 A 10 kHz voltage is applied to a $0.0047 \mu \mathrm{~F}$ capacitor, and 1 mA of rms current is measured. What is the value of the voltage? $\begin{array}{l} f=10 \mathrm{kHt}, C=0.0047 \mathrm{MF}, I_{\text {rms }}=1 \mathrm{~mA} \\ V_{\text {rms }}=I_{\text {rms }} x_{c}=\frac{I_{\text {rms }}}{2 \pi f c}=\frac{10^{-3} \mathrm{~A}}{2 \pi\left(10^{4} \mathrm{~Hz}\right)(0.0047 \mathrm{NF})} \\ \quad V_{\text {rms }}=3.4 \mathrm{~V} / \end{array}$

Studdy Solution
Use the capacitive reactance to find the rms voltage ( $V_{\text{rms}}$ ) using Ohm's Law for AC circuits:
$V_{\text{rms}} = I_{\text{rms}} \times X_c$
where: - $I_{\text{rms}} = 1 \, \text{mA} = 0.001 \, \text{A}$
Substitute the values:
$V_{\text{rms}} = 0.001 \times 3,389$
Calculate $V_{\text{rms}}$ :
$V_{\text{rms}} \approx 3.389 \, \text{V}$
The value of the rms voltage is approximately:
$\boxed{3.4 \, \text{V}}$

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