It is desired to set up an undriven L-C circuit in which the capacitor is originally charged to potential difference of 100.0 V . The maximum current is to be 1.0 A and the oscillation frequency is to be 1000 Hz . What are the required values of L and C ?

Answer

To solve the problem of finding the required values of inductance (L) and capacitance (C) for an undriven L-C circuit with given parameters, we can follow these steps:

### Step 1: Understand the given parameters
We have:
- Initial potential difference (V) = 100 V
- Maximum current (I₀) = 1 A
- Oscillation frequency (f₀) = 1000 Hz

### Step 2: Relate charge, capacitance, and potential
The charge (Q₀) on the capacitor can be calculated using the formula:
\[ Q₀ = C \cdot V \]
Where:
- \( C \) is the capacitance
- \( V \) is the voltage across the capacitor

### Step 3: Calculate the energy stored in the capacitor
The energy stored in the capacitor (U_C) can be expressed as:
\[ U_C = \frac{1}{2} C V^2 \]
Substituting the values:
\[ U_C = \frac{1}{2} C (100)^2 = 5000 C \]

### Step 4: Calculate the energy stored in the inductor
The energy stored in the inductor (U_L) is given by:
\[ U_L = \frac{1}{2} L I^2 \]
Substituting the maximum current:
\[ U_L = \frac{1}{2} L (1)^2 = \frac{1}{2} L \]

### Step 5: Set the energies equal
Since there is no resistance in the circuit, the maximum energy stored in the capacitor is equal to the maximum energy stored in the inductor:
\[ U_C = U_L \]
Thus:
\[ 5000 C = \frac{1}{2} L \]
Rearranging gives us:
\[ L = 10000 C \] (Equation 1)

### Step 6: Use the frequency formula
The oscillation frequency (f₀) is related to L and C by the formula:
\[ f₀ = \frac{1}{2\pi \sqrt{LC}} \]
Substituting the given frequency:
\[ 1000 = \frac{1}{2\pi \sqrt{LC}} \]
Rearranging gives:
\[ \sqrt{LC} = \frac{1}{2000\pi} \]
Squaring both sides:
\[ LC = \frac{1}{(2000\pi)^2} \] (Equation 2)

### Step 7: Substitute Equation 1 into Equation 2
From Equation 1, substitute \( L = 10000 C \) into Equation 2:
\[ (10000 C)C = \frac{1}{(2000\pi)^2} \]
This simplifies to:
\[ 10000 C^2 = \frac{1}{(2000\pi)^2} \]
\[ C^2 = \frac{1}{10000(2000\pi)^2} \]
\[ C^2 = \frac{1}{4 \times 10^8 \pi^2} \]
Taking the square root:
\[ C = \frac{1}{20000\pi} \]

### Step 8: Calculate the value of L
Now substitute the value of C back into Equation 1:
\[ L = 10000 C = 10000 \left(\frac{1}{20000\pi}\right) = \frac{10000}{20000\pi} = \frac{1}{2\pi} \]

### Final Results
Thus, the required values are:
- Capacitance \( C = \frac{1}{20000\pi} \) F
- Inductance \( L = \frac{1}{2\pi} \) H