In an `LR`-circuit, the inductive reactance is equal to the resistance `R` of the circuit. An e.m.f `E=E_(0)cos (omegat)` applied to the circuit. The power consumed in the circuit is

Options

A. `E_(0)^(2)/(sqrt(2R)`
B. `(E_(0)^(2))/(4R)`
C. `(E_(0)^(2))/(2R)`
D. `(E_(0)^(2))/(8R)`

Explanation

To solve the problem, we need to find the power consumed in an LR circuit where the inductive reactance \( X_L \) is equal to the resistance \( R \) of the circuit. The applied e.m.f. is given by \( E = E_0 \cos(\omega t) \).

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Inductive reactance: \( X_L = R \)
- E.M.F: \( E = E_0 \cos(\omega t) \)

2. **Calculate the RMS Values:**
- The RMS value of the e.m.f. is given by:
\[ E_{\text{rms}} = \frac{E_0}{\sqrt{2}} \]

3. **Calculate the Impedance \( Z \):**
- The impedance in an LR circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
- Since \( X_L = R \), we substitute:
\[ Z = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} \]

4. **Calculate the RMS Current \( I_{\text{rms}} \):**
- The RMS current can be calculated using Ohm's law for AC circuits:
\[ I_{\text{rms}} = \frac{E_{\text{rms}}}{Z} \]
- Substituting the values we have:
\[ I_{\text{rms}} = \frac{\frac{E_0}{\sqrt{2}}}{R\sqrt{2}} = \frac{E_0}{2R} \]

5. **Calculate the Power Factor \( \cos(\phi) \):**
- The power factor is given by:
\[ \cos(\phi) = \frac{R}{Z} \]
- Substituting \( Z = R\sqrt{2} \):
\[ \cos(\phi) = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}} \]

6. **Calculate the Average Power \( P \):**
- The average power consumed in the circuit is given by:
\[ P = E_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos(\phi) \]
- Substituting the values we calculated:
\[ P = \left(\frac{E_0}{\sqrt{2}}\right) \cdot \left(\frac{E_0}{2R}\right) \cdot \left(\frac{1}{\sqrt{2}}\right) \]
- Simplifying this expression:
\[ P = \frac{E_0^2}{\sqrt{2} \cdot 2R \cdot \sqrt{2}} = \frac{E_0^2}{4R} \]

### Final Result:
The power consumed in the circuit is:
\[ P = \frac{E_0^2}{4R} \]