To solve the problem step by step, we will follow the outlined procedure to calculate the average power dissipated in the LCR circuit.
### Step 1: Calculate the Resonant Frequency
The resonant frequency (\( \omega_0 \)) of the LCR circuit is given by the formula:
\[
\omega_0 = \frac{1}{\sqrt{LC}}
\]
Given:
- \( L = 10 \, \text{mH} = 10 \times 10^{-3} \, \text{H} \)
- \( C = 1 \, \mu\text{F} = 1 \times 10^{-6} \, \text{F} \)
Substituting the values:
\[
\omega_0 = \frac{1}{\sqrt{(10 \times 10^{-3})(1 \times 10^{-6})}} = \frac{1}{\sqrt{10^{-8}}} = 10^4 \, \text{rad/s}
\]
### Step 2: Calculate the Frequency at 50% of Resonant Frequency
The frequency (\( \omega \)) at 50% of the resonant frequency is:
\[
\omega = \frac{\omega_0}{2} = \frac{10^4}{2} = 5 \times 10^3 \, \text{rad/s}
\]
### Step 3: Calculate the Impedance of the Circuit
The impedance (\( Z \)) of the LCR circuit is given by:
\[
Z = \sqrt{R^2 + \left( \omega L - \frac{1}{\omega C} \right)^2}
\]
Given:
- \( R = 150 \, \Omega \)
- \( L = 10 \times 10^{-3} \, \text{H} \)
- \( C = 1 \times 10^{-6} \, \text{F} \)
Calculating \( \omega L \) and \( \frac{1}{\omega C} \):
\[
\omega L = (5 \times 10^3)(10 \times 10^{-3}) = 50 \, \text{V}
\]
\[
\frac{1}{\omega C} = \frac{1}{(5 \times 10^3)(1 \times 10^{-6})} = 200 \, \text{V}
\]
Now, substituting these values into the impedance formula:
\[
Z = \sqrt{150^2 + (50 - 200)^2} = \sqrt{150^2 + (-150)^2} = \sqrt{22500 + 22500} = \sqrt{45000} = 150\sqrt{2} \, \Omega
\]
### Step 4: Calculate the Average Power Dissipated
The average power (\( P \)) dissipated in the circuit can be calculated using the formula:
\[
P = \frac{V_{\text{rms}}^2}{R} \cdot \cos \phi
\]
Where \( \cos \phi = \frac{R}{Z} \).
Given:
- The source voltage is \( V(t) = 150\sqrt{2} \cos(\omega t) \), thus \( V_{\text{rms}} = 150 \, \text{V} \).
Now, calculate \( \cos \phi \):
\[
\cos \phi = \frac{R}{Z} = \frac{150}{150\sqrt{2}} = \frac{1}{\sqrt{2}}
\]
Substituting into the power formula:
\[
P = \frac{(150)^2}{150} \cdot \frac{1}{\sqrt{2}} = 150 \cdot \frac{1}{\sqrt{2}} = \frac{150}{\sqrt{2}} = 75\sqrt{2} \, \text{W}
\]
### Final Answer
The average power dissipated per cycle is:
\[
P \approx 75 \, \text{W}
\]
