An L-C circuit contains a 0.60 H inductor and a `25muF` capacitor. What is rate of change of current when the charges on the capacitor is`3.0xx10^(-5)C?`

Answer

To solve the problem, we need to find the rate of change of current (di/dt) in an LC circuit when the charge (Q) on the capacitor is given. We will use the relationship between the EMF across the inductor and the EMF across the capacitor.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Inductance (L) = 0.60 H
- Capacitance (C) = 25 μF = \(25 \times 10^{-6}\) F
- Charge on the capacitor (Q) = \(3.0 \times 10^{-5}\) C

2. **Use the Formula for EMF Across the Inductor:**
The EMF (E) across the inductor is given by:
\[ E_L = L \frac{di}{dt} \]

3. **Use the Formula for EMF Across the Capacitor:**
The EMF across the capacitor is given by:
\[ E_C = \frac{Q}{C} \]

4. **Set the EMFs Equal:**
In an LC circuit, the EMF across the inductor is equal to the EMF across the capacitor:
\[ L \frac{di}{dt} = \frac{Q}{C} \]

5. **Rearranging the Equation:**
We can rearrange this equation to solve for \(\frac{di}{dt}\):
\[ \frac{di}{dt} = \frac{Q}{LC} \]

6. **Substituting the Known Values:**
Substitute the values of Q, L, and C into the equation:
\[ \frac{di}{dt} = \frac{3.0 \times 10^{-5}}{0.60 \times 25 \times 10^{-6}} \]

7. **Calculating the Denominator:**
Calculate \(0.60 \times 25 \times 10^{-6}\):
\[ 0.60 \times 25 = 15 \quad \text{(in henry-farads)} \]
\[ 15 \times 10^{-6} = 15 \times 10^{-6} \text{ H·F} \]

8. **Final Calculation:**
Now, substitute back into the equation:
\[ \frac{di}{dt} = \frac{3.0 \times 10^{-5}}{15 \times 10^{-6}} = \frac{3.0}{15} \times 10^{1} = 0.2 \times 10^{1} = 2 \text{ A/s} \]

### Final Answer:
The rate of change of current \(\frac{di}{dt}\) is \(2 \text{ A/s}\).

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