Question

When an A.C. voltage of V = 330 sin(100πt) is applied to a capacitor, it produces a current of 1 = 1.5 cos(100πt). The capacitive reactance of the capacitor is

• A. 120Ω
• B. 180Ω
• C. 200Ω
• D. 220Ω
• E. 280Ω

Answer 1

The Correct Option is D

The relationship between the voltage and the current in an AC circuit with a capacitor is given by: \[ V = I \cdot X_C \] where:
\( V \) is the voltage across the capacitor,
\( I \) is the current through the capacitor,
\( X_C \) is the capacitive reactance. We are given:
\( V = 330 \sin(100\pi t) \), - \( I = 1.5 \cos(100\pi t) \).

The RMS value of the voltage \( V_{\text{rms}} \) and current \( I_{\text{rms}} \) are: \[ V_{\text{rms}} = \frac{330}{\sqrt{2}} = 233.2 \, \text{V} \] \[ I_{\text{rms}} = \frac{1.5}{\sqrt{2}} = 1.061 \, \text{A} \] Now, using the formula \( V_{\text{rms}} = I_{\text{rms}} \cdot X_C \), we can solve for \( X_C \): \[ 233.2 = 1.061 \cdot X_C \] \[ X_C = \frac{233.2}{1.061} \approx 220 \, \Omega \]

Thus, the capacitive reactance of the capacitor is \( 220 \, \Omega \), which is option (D).

Answer 2

Given: 
Voltage: V = 330 sin(100πt)
Current: I = 1.5 cos(100πt)

First, we write current in sine form to compare phase:
Since cos(θ) = sin(θ + π/2),
I = 1.5 sin(100πt + π/2)
This phase shift indicates a capacitor (current leads voltage). ✔️

Peak voltage, V₀ = 330 V
Peak current, I₀ = 1.5 A

Capacitive reactance is given by:
X_C = V₀ / I₀ = 330 / 1.5 = 220 Ω

Final Answer: 220 Ω