To solve the problem involving an L-C-R circuit connected to an AC source, we will follow these steps:
### Step 1: Identify Given Values
We have the following values from the problem statement:
- Voltage across the resistor, \( V_R = 40 \, V \)
- Voltage across the inductor, \( V_L = 40 \, V \)
- Voltage across the capacitor, \( V_C = 10 \, V \)
- Resistance, \( R = 4 \, \Omega \)
- The voltage source is given by \( V = V_0 \sin(100\pi t + \frac{\pi}{6}) \)
### Step 2: Calculate RMS Current
Using the formula for the voltage across the resistor:
\[
V_R = I_{\text{RMS}} \times R
\]
Substituting the known values:
\[
40 = I_{\text{RMS}} \times 4
\]
Solving for \( I_{\text{RMS}} \):
\[
I_{\text{RMS}} = \frac{40}{4} = 10 \, A
\]
### Step 3: Calculate Inductive Reactance
Using the formula for the voltage across the inductor:
\[
V_L = I_{\text{RMS}} \times X_L
\]
Substituting the known values:
\[
40 = 10 \times X_L
\]
Solving for \( X_L \):
\[
X_L = \frac{40}{10} = 4 \, \Omega
\]
### Step 4: Calculate Inductance \( L \)
From the relationship between inductive reactance and inductance:
\[
X_L = \omega L
\]
Where \( \omega = 100\pi \). Therefore:
\[
4 = 100\pi L
\]
Solving for \( L \):
\[
L = \frac{4}{100\pi} = \frac{1}{25\pi} \, H
\]
### Step 5: Calculate Capacitive Reactance
Using the formula for the voltage across the capacitor:
\[
V_C = I_{\text{RMS}} \times X_C
\]
Substituting the known values:
\[
10 = 10 \times X_C
\]
Solving for \( X_C \):
\[
X_C = \frac{10}{10} = 1 \, \Omega
\]
### Step 6: Calculate Capacitance \( C \)
From the relationship between capacitive reactance and capacitance:
\[
X_C = \frac{1}{\omega C}
\]
Substituting \( \omega = 100\pi \):
\[
1 = \frac{1}{100\pi C}
\]
Solving for \( C \):
\[
C = \frac{1}{100\pi} \, F
\]
### Conclusion
The values we calculated are:
- \( L = \frac{1}{25\pi} \, H \)
- \( C = \frac{1}{100\pi} \, F \)
### Final Answer
The correct option for the inductance \( L \) is \( \frac{1}{25\pi} \, H \), and for capacitance \( C \) is \( \frac{1}{100\pi} \, F \).
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