An inductor 20 mH, a capacitor 50 μF, and a resistor 40 Ω are connected in series across a source of emf V = 1

Question

An inductor 20 mH, a capacitor 50 μF, and a resistor 40 Ω are connected in series across a source of emf V = 10 sin 340t. The power loss in the AC circuit is:

Accepted Answer

The power loss in a series LCR circuit occurs only across the resistor and is given by $P = I_{rms}^2 R$ . 
1. **Identify parameters:** $L = 20 	ext{ mH} = 20 	imes 10^{-3} 	ext{ H}$, $C = 50 \mu	ext{F} = 50 	imes 10^{-6} 	ext{ F}$, $R = 40 \Omega$. From $V = 10 \sin 340t$, peak voltage $V_m = 10 	ext{ V}$ and angular frequency $\omega = 340 	ext{ rad/s}$ .
2. **Calculate Reactances:** 
   $X_L = \omega L = 340 	imes 20 	imes 10^{-3} = 6.8 \Omega$ .
   $X_C = \frac{1}{\omega C} = \frac{1}{340 	imes 50 	imes 10^{-6}} \approx 58.82 \Omega$ .
3. **Calculate Impedance ($Z$):** 
   $Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{40^2 + (58.82 - 6.8)^2} = \sqrt{1600 + (52.02)^2} \approx \sqrt{1600 + 2706} \approx 65.62 \Omega$.
4. **Calculate RMS Current:** 
   $I_{rms} = \frac{V_{rms}}{Z} = \frac{V_m}{\sqrt{2} Z} = \frac{10}{1.414 	imes 65.62} \approx 0.108 	ext{ A}$ .
5. **Calculate Power:** 
   $P = I_{rms}^2 R \approx (0.108)^2 	imes 40 \approx 0.46 	ext{ W}$.
   Among the given options, $0.51 	ext{ W}$ is the closest value.

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