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Step-by-Step Solution
Step 1: Understand the Problem
The question involves an LCR circuit with a resistance of 110β―Ξ©, operating at an angular frequency of 300β―rad/s and supplied by a 220β―V source. We have two scenarios:
When the capacitor is removed, the current lags the voltage by 45Β°.
When the inductor is removed, the current leads the voltage by 45Β°.
We want to find the RMS current flowing in the fully assembled LCR circuit (with inductor, capacitor, and resistor connected).
Step 2: Key Observation β Resonance Condition
In an AC circuit containing an inductor, resistor, and capacitor, resonance occurs when the inductive reactance ($X_L$) and the capacitive reactance ($X_C$) exactly cancel each other out. Mathematically, this is when
$$X_L = X_C.$$
When the circuit is in resonance, the net reactance is zero, and the current is in phase with the voltage.
The question states that removing the capacitor causes the current to lag by 45Β°, and removing the inductor causes the current to lead by 45Β°. This symmetric lead-lag behavior suggests that, in the presence of both inductor and capacitor, their reactances balance each other. Therefore, the circuit is effectively in resonance.
Step 3: Write the Resonance Condition for Impedance
At resonance, the impedance $Z$ of the circuit is purely resistive, i.e.,
$$Z = R.$$
Hence, the total impedance of the LCR circuit in resonance is:
$$Z = 110\,\Omega.$$
Step 4: Calculate the RMS Current
The RMS current $I_{\mathrm{rms}}$ in a series AC circuit is given by:
$$I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{Z}.$$
Since the circuit is in resonance and $Z = R = 110\,\Omega$,
$$I_{\mathrm{rms}} = \frac{220}{110} = 2\,\text{A}.$$
Step 5: Final Answer
The RMS current flowing in the circuit is
$$\boxed{2\,\text{A}}.$$
