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 voltage across a capacitor is described by the equation: \[ V = V_{\text{max}} \sin(\omega t) \] where \( V_{\text{max}} = 100 \sqrt{2} \) V represents the peak voltage and \( \omega \) is the angular frequency. In an AC circuit containing a capacitor, the RMS (Root Mean Square) current is related to the voltage by the formula: \[ I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} \] In this formula, \( V_{\text{rms}} \) is the RMS voltage, and \( X_C = \frac{1}{\omega C} \) is the capacitive reactance. The initial step involves calculating the RMS voltage: \[ V_{\text{rms}} = \frac{V_{\text{max}}}{\sqrt{2}} = \frac{100 \sqrt{2}}{\sqrt{2}} = 100 \, \text{V} \] Next, we determine the reactance \( X_C \). The formula for \( X_C \) is: \[ X_C = \frac{1}{\omega C} \] Substituting \( C = 2 \, \mu \text{F} = 2 \times 10^{-6} \, \text{F} \) and \( \omega = 2 \pi f \). For a standard mains frequency of \( f = 50 \, \text{Hz} \), the angular frequency is: \[ \omega = 2 \pi \times 50 = 314 \, \text{rad/s} \] The calculated reactance \( X_C \) is: \[ X_C = \frac{1}{314 \times 2 \times 10^{-6}} = 1592.5 \, \Omega \] The final calculation for the RMS current yields: \[ I_{\text{rms}} = \frac{100}{1592.5} \approx 0.0628 \, \text{A} \] This computed current does not align with the provided answer options. Given that currents in practical scenarios are often higher, an alternative \( \omega \) or frequency might be applicable. Further investigation into the given parameters is necessary to validate the result.