For a series L C R circuit, the power loss at resonance is

Options

A. `(V_(2))/(omegaL-(1)/(omegaC))`
B. `i^(2)Comega`
C. `i^(2)R`
D. `(V^(2))/(omegaC)`

Answer

To find the power loss at resonance in a series LCR circuit, we can follow these steps:

### Step 1: Understand the Impedance at Resonance
In a series LCR circuit, the total impedance (Z) is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Where:
- \( R \) is the resistance,
- \( X_L \) is the inductive reactance,
- \( X_C \) is the capacitive reactance.

At resonance, the inductive reactance equals the capacitive reactance:
\[ X_L = X_C \]
Thus, at resonance:
\[ Z = R \]

### Step 2: Determine the Current at Resonance
The current (I) in the circuit can be calculated using Ohm's law:
\[ I = \frac{V}{Z} \]
Substituting the impedance at resonance:
\[ I = \frac{V}{R} \]
Where \( V \) is the supply voltage.

### Step 3: Calculate the Power Loss
The power (P) in an AC circuit is given by:
\[ P = V \cdot I \cdot \cos(\theta) \]
At resonance, the current and voltage are in phase, which means the phase angle \( \theta = 0 \) and \( \cos(0) = 1 \). Thus, the power simplifies to:
\[ P = V \cdot I \]
Substituting the expression for current:
\[ P = V \cdot \left(\frac{V}{R}\right) = \frac{V^2}{R} \]

### Step 4: Express Power in Terms of Current
We can also express power in terms of current:
\[ P = I^2 R \]
Using the current we found earlier:
\[ P = \left(\frac{V}{R}\right)^2 R = \frac{V^2}{R} \]

### Conclusion
Thus, the power loss at resonance in a series LCR circuit is:
\[ P = \frac{V^2}{R} \]