To solve the problem step by step, we will follow the information provided in the question and the video transcript.
### Step 1: Calculate the capacitance (C) using the energy stored in the capacitor
The average electric field energy stored in the capacitor is given by:
\[
U_C = \frac{1}{2} C V_{rms}^2
\]
Given \( U_C = 25 \, \text{mJ} = 25 \times 10^{-3} \, \text{J} \).
Rearranging the formula to find \( C \):
\[
C = \frac{2 U_C}{V_{rms}^2}
\]
We will need to calculate \( V_{rms} \) later, but we can express it in terms of \( I_{rms} \) and \( R \):
\[
V_{rms} = I_{rms} \cdot R
\]
Given \( I_{rms} = 0.10 \, \text{A} \) and \( R = 300 \, \Omega \):
\[
V_{rms} = 0.10 \times 300 = 30 \, \text{V}
\]
Now substituting \( V_{rms} \) back into the capacitance equation:
\[
C = \frac{2 \times 25 \times 10^{-3}}{30^2} = \frac{50 \times 10^{-3}}{900} = \frac{50}{900} \times 10^{-3} = \frac{1}{18} \times 10^{-3} \approx 0.0556 \, \text{F} \approx 20 \, \mu\text{F}
\]
### Step 2: Calculate the inductance (L) using the energy stored in the inductor
The average magnetic energy stored in the inductor is given by:
\[
U_L = \frac{1}{2} L I_{rms}^2
\]
Given \( U_L = 5 \, \text{mJ} = 5 \times 10^{-3} \, \text{J} \):
\[
L = \frac{2 U_L}{I_{rms}^2}
\]
Substituting the known values:
\[
L = \frac{2 \times 5 \times 10^{-3}}{(0.10)^2} = \frac{10 \times 10^{-3}}{0.01} = 1 \, \text{H}
\]
### Step 3: Calculate the voltage across the inductor (V_L)
The voltage across the inductor is given by:
\[
V_L = I_{rms} \cdot X_L
\]
Where \( X_L = \omega L \) and \( \omega = 2 \pi f \). Given \( f = \frac{50}{\pi} \):
\[
\omega = 2 \pi \left(\frac{50}{\pi}\right) = 100 \, \text{rad/s}
\]
Thus,
\[
X_L = 100 \cdot 1 = 100 \, \Omega
\]
Now substituting back to find \( V_L \):
\[
V_L = 0.10 \cdot 100 = 10 \, \text{V}
\]
### Step 4: Calculate the voltage across the capacitor (V_C)
The voltage across the capacitor is given by:
\[
V_C = I_{rms} \cdot X_C
\]
Where \( X_C = \frac{1}{\omega C} \):
\[
X_C = \frac{1}{100 \cdot 20 \times 10^{-6}} = \frac{1}{0.002} = 500 \, \Omega
\]
Thus,
\[
V_C = 0.10 \cdot 500 = 50 \, \text{V}
\]
### Step 5: Calculate the total RMS voltage (V_rms)
Using the formula for total voltage in an LCR circuit:
\[
V_{rms} = \sqrt{V_R^2 + V_L^2 + V_C^2}
\]
Where \( V_R = 30 \, \text{V}, V_L = 10 \, \text{V}, V_C = 50 \, \text{V} \):
\[
V_{rms} = \sqrt{30^2 + 10^2 + 50^2} = \sqrt{900 + 100 + 2500} = \sqrt{3500} \approx 59.16 \, \text{V}
\]
### Final Answer
The RMS voltage \( V_{rms} \) in the circuit is approximately \( 59.16 \, \text{V} \).
