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Step-by-Step Solution
Step 1: Understand the Circuit Configurations
We have a series LCR circuit with resistance R = 200 Ω, supplied by an AC source of
220 V (rms) at 50 Hz. We analyze two specific situations:
(i) When the capacitor is removed, the circuit becomes an LR circuit, and the current lags the voltage by 30°.
(ii) When the inductor is removed, the circuit becomes a CR circuit, and the current leads the voltage by 30°.
From these phase relationships, we will determine the values of ωL and 1/(ωC).
Step 2: Phase Angle in the LR Circuit
When the circuit is LR only, the current lags behind the voltage by 30°. This implies the phase angle φ = 30°.
For an LR circuit,
$ \tan \phi = \frac{\omega L}{R} $.
Substituting φ = 30° and R = 200 Ω:
$ \tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{\omega L}{200} $.
Therefore,
$ \omega L = 200 \times \frac{1}{\sqrt{3}} = \frac{200}{\sqrt{3}}. $
Step 3: Phase Angle in the CR Circuit
When the circuit is CR only, the current leads the voltage by 30°. This again implies the same magnitude of phase angle φ = 30° (but leading).
For a CR circuit,
$ \tan \phi = \frac{1}{\omega CR}. $
Substituting φ = 30° and R = 200 Ω:
$ \tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{1}{\omega C \times 200}. $
Hence,
$ \frac{1}{\omega C} = 200 \times \frac{1}{\sqrt{3}} = \frac{200}{\sqrt{3}}. $
Step 4: Determine the Net Reactance of the LCR Circuit
In the full LCR circuit, the inductive reactance is $ \omega L = \frac{200}{\sqrt{3}} $
and the capacitive reactance is $ \frac{1}{\omega C} = \frac{200}{\sqrt{3}} $.
The difference between the inductive and capacitive reactances is:
$ \left| \frac{1}{\omega C} - \omega L \right| = \left| \frac{200}{\sqrt{3}} - \frac{200}{\sqrt{3}} \right| = 0. $
This implies that the net reactance of the circuit is zero, so the circuit is purely resistive. Therefore, the total impedance Z is:
$ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{200^2 + 0^2} = 200 \,\Omega. $
Step 5: Calculate the Power Dissipated in the LCR Circuit
Since the circuit is effectively purely resistive (net reactance = 0), the power factor is 1.
The power dissipated in a resistive circuit is given by
$ P = I_{\text{rms}}^2 R = V_{\text{rms}} \, I_{\text{rms}} \cos \phi. $
Alternatively, using
$ P = \frac{V_{\text{rms}}^2 \, R}{Z^2}, $
we have:
$ P = \frac{(220)^2 \times 200}{(200)^2} = \frac{220 \times 220}{200} = 242 \, W. $
Final Answer
The power dissipated in the LCR circuit is 242 W.
