Question:

A 1-$\phi$, RC series circuit has \( R = 5\, \Omega \) and \( C = 10\, \mu F \). If the angular frequency of current is 20000 rad/sec, then the applied voltage

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For RC circuits, remember voltage lags current and use \( \theta = \tan^{-1}(1/\omega RC) \) for phase angle.
Updated On: May 23, 2025
  • Lags the current by \( \pi/4 \) radians
  • Leads the current by \( \pi/4 \) radians
  • Lags the current by \( \pi/2 \) radians
  • Leads the current by \( \pi/2 \) radians
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The Correct Option is A

Solution and Explanation

In an RC series circuit, the phase angle by which voltage lags the current is given by: \[ \theta = \tan^{-1}\left(\frac{1}{\omega RC}\right) \] Given: \[ R = 5\, \Omega, \ C = 10\, \mu F = 10 \times 10^{-6} F, \ \omega = 20000 \ \text{rad/sec} \] Now, \[ \omega RC = 20000 \times 5 \times 10 \times 10^{-6} = 1 \] So, \[ \theta = \tan^{-1}(1) = \frac{\pi}{4} \] Thus, applied voltage lags the current by \( \frac{\pi}{4} \) radians.
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