Question

Given the voltage equation \( V = 100 \sqrt{2} \sin(\omega t) \) and capacitance \( C = 2 \, \mu \text{F} \), calculate the RMS current.

• A. \( 10 \, \text{A} \)
• B. \( 20 \, \text{A} \)
• C. \( 50 \, \text{A} \)
• D. \( 100 \, \text{A} \)

Hint:

In AC circuits involving capacitors, the RMS voltage and current are important for understanding power. The current through a capacitor depends on the frequency of the AC signal and the capacitance.

Answer 1

The Correct Option is A

The equation for the voltage across a capacitor is: \[ V = V_{\text{max}} \sin(\omega t) \] where \( V_{\text{max}} = 100 \sqrt{2} \) V is the maximum voltage, and \( \omega \) is the angular frequency. For an AC circuit with a capacitor, the RMS (Root Mean Square) current is related to the voltage by: \[ I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} \] where: - \( V_{\text{rms}} \) is the RMS voltage, - \( X_C = \frac{1}{\omega C} \) is the reactance of the capacitor. First, calculate the RMS voltage: \[ V_{\text{rms}} = \frac{V_{\text{max}}}{\sqrt{2}} = \frac{100 \sqrt{2}}{\sqrt{2}} = 100 \, \text{V} \] Now, calculate the reactance \( X_C \). The reactance \( X_C \) is given by: \[ X_C = \frac{1}{\omega C} \] Substitute \( C = 2 \, \mu \text{F} = 2 \times 10^{-6} \, \text{F} \) and \( \omega = 2 \pi f \). For a typical mains frequency of \( f = 50 \, \text{Hz} \), we have: \[ \omega = 2 \pi \times 50 = 314 \, \text{rad/s} \] Now, calculate \( X_C \): \[ X_C = \frac{1}{314 \times 2 \times 10^{-6}} = 1592.5 \, \Omega \] Finally, calculate the RMS current: \[ I_{\text{rms}} = \frac{100}{1592.5} \approx 0.0628 \, \text{A} \] However, this current does not match the provided answer choices. Given that the current is usually much larger in real-world problems, there might be a different \( \omega \) or frequency to account for. We might need to consider another approach based on the given details. Please verify the assumptions to confirm the result.