In a given AC series circuit containing elements R, L and C & source voltage `= 220v`, and `omega =300 rad/s` it is known that if L alone is removed or if C alone is removed, phase difference between current & voltage remains `45^@` Find i_rms? `(R= 110 Omega)`

Options

A. `2A`
B. `2.5A`
C. `1A`
D. `1.5A`

Answer

To solve the problem step by step, we will analyze the given information and apply the relevant formulas for an AC series circuit containing resistance (R), inductance (L), and capacitance (C).

### Step 1: Understand the given values
We have:
- Source voltage, \( V_{rms} = 220 \, V \)
- Resistance, \( R = 110 \, \Omega \)
- Angular frequency, \( \omega = 300 \, rad/s \)

### Step 2: Analyze the phase difference conditions
According to the problem:
- When L is removed, the phase difference between current and voltage is \( 45^\circ \) (current lags).
- When C is removed, the phase difference is also \( 45^\circ \) (current leads).

This indicates that the circuit is at resonance when both L and C are present, leading to a balanced condition where the inductive reactance \( X_L \) equals the capacitive reactance \( X_C \).

### Step 3: Determine the impedance
In a series LCR circuit, the impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Since \( X_L = X_C \) at resonance, we have:
\[ Z = R \]
Substituting the value of R:
\[ Z = 110 \, \Omega \]

### Step 4: Calculate the current \( I_{rms} \)
The current in the circuit can be calculated using Ohm's law for AC circuits:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
Substituting the values:
\[ I_{rms} = \frac{220 \, V}{110 \, \Omega} = 2 \, A \]

### Final Answer
Thus, the root mean square current \( I_{rms} \) in the circuit is:
\[ \boxed{2 \, A} \]