To solve the problem, we will follow these steps:
### Step 1: Determine the Current at Resonance
At resonance in an RLC circuit, the inductive reactance \(X_L\) equals the capacitive reactance \(X_C\). The formula for the current \(I\) in the circuit is given by:
\[
I = \frac{V}{\sqrt{R^2 + (X_C - X_L)^2}}
\]
Since \(X_C = X_L\) at resonance, the formula simplifies to:
\[
I_0 = \frac{V}{R}
\]
Given:
- Voltage \(V = 200 \, V\)
- Resistance \(R = 5 \, \Omega\)
Substituting the values:
\[
I_0 = \frac{200}{5} = 40 \, A
\]
### Step 2: Calculate the Inductive Reactance \(X_L\)
The inductive reactance \(X_L\) is given by:
\[
X_L = \omega L
\]
Where \(\omega = 2 \pi f\). To find \(X_L\), we first need to calculate the resonant frequency \(f\).
### Step 3: Calculate the Resonant Frequency \(f\)
The resonant frequency \(f\) is given by:
\[
f = \frac{1}{2 \pi \sqrt{LC}}
\]
Given:
- Inductance \(L = 0.6 \, H\)
- Capacitance \(C = 10 \, \mu F = 10 \times 10^{-6} \, F\)
Substituting the values:
\[
f = \frac{1}{2 \pi \sqrt{0.6 \times 10 \times 10^{-6}}}
\]
Calculating:
\[
f \approx \frac{1}{2 \pi \sqrt{6 \times 10^{-7}}} \approx \frac{1}{2 \pi \times 0.000775} \approx 65 \, Hz
\]
### Step 4: Calculate \(X_L\)
Now, substituting \(f\) into the equation for \(X_L\):
\[
X_L = 2 \pi f L = 2 \pi (65) (0.6) \approx 245 \, \Omega
\]
### Step 5: Calculate the Voltage Across the Inductor \(V_0\)
The voltage across the inductor \(V_0\) is given by:
\[
V_0 = I_0 \cdot X_L
\]
Substituting the values:
\[
V_0 = 40 \cdot 245 = 9800 \, V = 9.8 \, kV
\]
### Step 6: Calculate the Voltage Across the Capacitor \(V_1\)
The capacitive reactance \(X_C\) is given by:
\[
X_C = \frac{1}{\omega C}
\]
Substituting \(\omega\) and \(C\):
\[
X_C = \frac{1}{2 \pi (65) (10 \times 10^{-6})}
\]
Calculating \(X_C\):
\[
X_C \approx \frac{1}{2 \pi (65) (10^{-5})} \approx 245 \, \Omega
\]
Now, the voltage across the capacitor \(V_1\) is:
\[
V_1 = I_0 \cdot X_C = 40 \cdot 245 = 9800 \, V = 9.8 \, kV
\]
### Final Results
- \(I_0 = 40 \, A\)
- \(V_0 = 9.8 \, kV\)
- \(V_1 = 9.8 \, kV\)
