An AC circuit consists of a resistance and a choke coil in series . The resistance is of 220 `Omega` and choke coils is of 0.7 H . The power abosorbed from 220 V and 50 Hz , source connected with the circuit , is

Options

A. 55 W
B. 110 W
C. 220 W
D. 440 W

Explanation

To solve the problem step by step, we will follow the outlined procedure to calculate the power absorbed in the AC circuit consisting of a resistance and a choke coil in series.

### Step 1: Calculate the Inductive Reactance (X_L)
The formula for inductive reactance is given by:
\[ X_L = 2 \pi f L \]
Where:
- \( f \) = frequency (50 Hz)
- \( L \) = inductance (0.7 H)

Substituting the values:
\[ X_L = 2 \pi (50) (0.7) = 2 \pi \times 35 = 70 \pi \approx 219.91 \, \Omega \]

### Step 2: Calculate the Impedance (Z)
The impedance in a series circuit with resistance (R) and inductive reactance (X_L) is calculated using:
\[ Z = \sqrt{R^2 + X_L^2} \]
Where:
- \( R = 220 \, \Omega \)
- \( X_L \approx 220 \, \Omega \) (as calculated above)

Substituting the values:
\[ Z = \sqrt{(220)^2 + (220)^2} = \sqrt{48400 + 48400} = \sqrt{96800} = 220 \sqrt{2} \approx 311.13 \, \Omega \]

### Step 3: Calculate the RMS Current (I_rms)
The RMS current can be calculated using:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
Where:
- \( V_{rms} = 220 \, V \)

Substituting the values:
\[ I_{rms} = \frac{220}{220 \sqrt{2}} = \frac{1}{\sqrt{2}} \, A \approx 0.707 \, A \]

### Step 4: Calculate the Power (P)
The power absorbed in the circuit can be calculated using:
\[ P = I_{rms}^2 \times R \]
Substituting the values:
\[ P = \left(\frac{1}{\sqrt{2}}\right)^2 \times 220 = \frac{1}{2} \times 220 = 110 \, W \]

### Final Answer
The power absorbed from the 220 V and 50 Hz source connected with the circuit is **110 watts**.

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