A 10 $\mu$F capacitor is connected to a 210 V, 50 Hz source. The peak current in the circuit is nearly ($\pi =

Question

A 10 $\mu$F capacitor is connected to a 210 V, 50 Hz source. The peak current in the circuit is nearly ($\pi = 3.14$):

Accepted Answer

To find the peak current ($i_m$), we first calculate the peak voltage ($V_m$) and the capacitive reactance ($X_C$). 
1. **Peak Voltage ($V_m$):** The given source voltage is the RMS value ($V_{rms} = 210$ V). According to the sources, $V_m = V_{rms}\sqrt{2} = 210 	imes 1.414 \approx 296.94$ V . 
2. **Capacitive Reactance ($X_C$):** $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$. Substituting the values: $X_C = \frac{1}{2 	imes 3.14 	imes 50 	imes 10 	imes 10^{-6}} = \frac{1}{314 	imes 10^{-5}} \approx 318.47 \ \Omega$ . 
3. **Peak Current ($i_m$):** $i_m = \frac{V_m}{X_C} = V_m \omega C = 296.94 	imes (314 	imes 10^{-5}) \approx 0.932$ A. Thus, the peak current is nearly **0.93 A**.

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