Power dissipated in an `L-C-R` series circuit connected to an `AC` source of emf `epsilon` is

Options

A. `(epsilon squareR)/(R^(2)+(Lomega-1/(Comega))^(2)]`
B. `sqrt(R^(2)+(Lomega-1/(Comega)^(2))/(R )`
C. `[R^(2)+(Lomega-1/(Comega))^(2)]/( R)`
D. `(omega^(2)R)/(sqrt(R^(2)+(Lomega+1/(Comega))^(2)`

Answer

To find the power dissipated in an L-C-R series circuit connected to an AC source of emf \( \epsilon \), we can follow these steps:

### Step 1: Understand the formula for power in an L-C-R circuit
The power dissipated in an L-C-R series circuit can be expressed using the formula:
\[ P = I_{\text{rms}}^2 \cdot R \]
where \( I_{\text{rms}} \) is the root mean square current and \( R \) is the resistance in the circuit.

### Step 2: Relate current to voltage and impedance
The root mean square current \( I_{\text{rms}} \) can be expressed in terms of the emf \( \epsilon \) and the impedance \( Z \) of the circuit:
\[ I_{\text{rms}} = \frac{\epsilon}{Z} \]
Substituting this into the power formula gives:
\[ P = \left(\frac{\epsilon}{Z}\right)^2 \cdot R \]

### Step 3: Substitute for impedance \( Z \)
The impedance \( Z \) for a series L-C-R circuit is given by:
\[ Z = \sqrt{(X_L - X_C)^2 + R^2} \]
where \( X_L = \omega L \) (inductive reactance) and \( X_C = \frac{1}{\omega C} \) (capacitive reactance). Therefore, we can rewrite \( Z \) as:
\[ Z = \sqrt{(\omega L - \frac{1}{\omega C})^2 + R^2} \]

### Step 4: Substitute \( Z \) back into the power formula
Now substituting \( Z \) back into the power equation:
\[ P = \frac{\epsilon^2 \cdot R}{Z^2} \]
Substituting for \( Z^2 \):
\[ P = \frac{\epsilon^2 \cdot R}{(\omega L - \frac{1}{\omega C})^2 + R^2} \]

### Final Result
Thus, the power dissipated in the L-C-R series circuit connected to an AC source of emf \( \epsilon \) is given by:
\[ P = \frac{\epsilon^2 \cdot R}{R^2 + (\omega L - \frac{1}{\omega C})^2} \]