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Math  /  Algebra

Question12-46 A 10 kHz voltage is applied to a 0.0047μ F0.0047 \mu \mathrm{~F} capacitor, and 1 mA of rms current is measured. What is the value of the voltage? f=10kHt,C=0.0047MF,Irms =1 mAVrms =Irms xc=Irms 2πfc=10−3 A2π(104 Hz)(0.0047NF)Vrms =3.4 V/\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

STEP 1

1. The voltage across the capacitor is alternating current (AC) with a frequency of 10 kHz.
2. The capacitor value is given in microfarads (μF\mu \mathrm{F}).
3. The current measured is the root mean square (rms) value.
4. We need to find the rms voltage across the capacitor.

STEP 2

1. Calculate the capacitive reactance (XcX_c).
2. Use XcX_c to find the rms voltage (VrmsV_{\text{rms}}).

STEP 3

Calculate the capacitive reactance (XcX_c) using the formula:
Xc=12πfC X_c = \frac{1}{2 \pi f C}
where: - f=10 kHz=10,000 Hz f = 10 \, \text{kHz} = 10,000 \, \text{Hz} - C=0.0047 μF=0.0047×10−6 F C = 0.0047 \, \mu\text{F} = 0.0047 \times 10^{-6} \, \text{F}
Substitute the values into the formula:
Xc=12π(10,000)(0.0047×10−6) X_c = \frac{1}{2 \pi (10,000) (0.0047 \times 10^{-6})}
Calculate Xc X_c :
Xc=12π×10,000×0.0047×10−6 X_c = \frac{1}{2 \pi \times 10,000 \times 0.0047 \times 10^{-6}} Xc≈3,389 Ω X_c \approx 3,389 \, \Omega

STEP 4

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

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