Power in an AC Circuit 🔋

Apply a sinusoidal voltage \( v = v_m \sin \omega t \) to a series RLC circuit and the current becomes \( i = i_m \sin(\omega t + \phi) \). The amplitude ratio is \( \dfrac{v_m}{i_m} = Z \) and the phase angle is \( \phi = \tan^{-1}\!\Bigl(\dfrac{X_C – X_L}{R}\Bigr) \). :contentReference[oaicite:0]{index=0}

Instantaneous and Average Power ⚡

The instantaneous power simply multiplies voltage and current:

\( p = v\,i = v_m i_m \sin \omega t \,\sin(\omega t + \phi) \)

Use the product-to-sum trick and you get

\( p = \dfrac{v_m i_m}{2}\Bigl[\cos \phi – \cos(2\omega t + \phi)\Bigr] \). :contentReference[oaicite:1]{index=1}

Over one full cycle the oscillating term averages out, leaving

\( P = \dfrac{v_m i_m}{2}\cos\phi = VI\cos\phi = I^{2}Z\cos\phi \). :contentReference[oaicite:2]{index=2}

The Power Factor 🌟

The multiplier \( \cos\phi \) is the power factor. * \( \cos\phi = 1 \) means every ampere does useful work. * \( \cos\phi = 0 \) means all current is “reactive” or wattless—it just sloshes energy back and forth. :contentReference[oaicite:3]{index=3}

Four Key Situations 🧐

  • Pure resistor – \( \phi = 0 \Rightarrow \cos\phi = 1\). Maximum power! 🔥 :contentReference[oaicite:4]{index=4}
  • Pure inductor or capacitor – \( \phi = 90^\circ \Rightarrow \cos\phi = 0\). Power drops to zero even though current flows. 🌀 :contentReference[oaicite:5]{index=5}
  • General RLC – Only the resistor burns energy; \( X_L \) and \( X_C \) just shift the phase. :contentReference[oaicite:6]{index=6}
  • Resonance – Set \( X_C = X_L \) and you force \( \phi = 0\). Power shoots up to \( P = I^{2}R \). 🎯 :contentReference[oaicite:7]{index=7}

Why a Low Power Factor Hurts 🚚

To deliver power \( P \) at voltage \( V \) you need \( I = \dfrac{P}{V\cos\phi} \). A small \( \cos\phi \) forces a larger current and boosts the line loss \( I^{2}R \). :contentReference[oaicite:8]{index=8}

Boosting the Power Factor 💡

Connect a capacitor in parallel. The capacitor draws a leading reactive current that cancels the lagging current from inductive loads. After cancellation, only the in-phase component \( I_p \) remains, driving \( \cos\phi \) toward 1. :contentReference[oaicite:9]{index=9}

Worked Example 📝

  • Given \( R = 3\,\Omega,\; L = 25.48\text{ mH},\; C = 796\,\mu\text{F} \) and a 50 Hz supply (peak 283 V).
  • Calculate \( X_L = 8\,\Omega,\; X_C = 4\,\Omega \).
  • Impedance \( Z = 5\,\Omega \) and \( \phi = -53.1^\circ \) (current lags). :contentReference[oaicite:10]{index=10}
  • RMS current \( I = \dfrac{283/\sqrt{2}}{5} \approx 40\text{ A} \).
  • Power \( P = I^{2}R \approx 4.8\text{ kW} \) and power factor \( \cos\phi = 0.6 \). 💪

Resonance Super-Charge 🚀

Tune the same circuit to its resonant frequency \( \nu_r = 35.4\text{ Hz} \). Then \( Z = R = 3\,\Omega \) and the rms current jumps to 66.7 A, raising the power to about 13.3 kW. :contentReference[oaicite:11]{index=11}

High-Yield Ideas for NEET 🎯

  1. Remember \( P = VI\cos\phi \) and call \( \cos\phi \) the power factor. :contentReference[oaicite:12]{index=12}
  2. Zero average power in purely inductive or capacitive circuits—even with current flowing. :contentReference[oaicite:13]{index=13}
  3. Phase angle formula \( \phi = \tan^{-1}\!\bigl(\tfrac{X_C – X_L}{R}\bigr) \) lets you tell who leads or lags. :contentReference[oaicite:14]{index=14}
  4. Low power factor → bigger transmission losses; high factor saves energy. :contentReference[oaicite:15]{index=15}
  5. Use a parallel capacitor to cancel wattless current and improve \( \cos\phi \). :contentReference[oaicite:16]{index=16}

Keep practicing and you’ll power through these questions! ⚡🤓