An AC source is connected to a capacitor. The current in the current is I. Now a dielectric slab is inserted into the capacitor , then the new current is

Options

A. equal I
B. more than I
C. less than I
D. may be more than or less than I

Explanation

To solve the problem step by step, we will analyze the effect of inserting a dielectric slab into a capacitor connected to an AC source.

### Step-by-Step Solution:

1. **Understanding Initial Conditions**:
- We have an AC source connected to a capacitor with an initial capacitance \( C \).
- The initial current flowing through the circuit is \( I \).

2. **Effect of Inserting a Dielectric**:
- When a dielectric slab is inserted into the capacitor, the capacitance increases. The new capacitance \( C' \) can be expressed as:
\[ C' = K \cdot C \]
where \( K \) is the dielectric constant of the material.

3. **Capacitive Reactance**:
- The capacitive reactance \( X_C \) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
- Since the capacitance increases to \( C' \), the new capacitive reactance \( X_C' \) becomes:
\[ X_C' = \frac{1}{\omega C'} = \frac{1}{\omega (K \cdot C)} = \frac{1}{K} \cdot \frac{1}{\omega C} = \frac{X_C}{K} \]
- This shows that \( X_C' \) decreases because \( K > 1 \).

4. **Current in the Circuit**:
- The current \( I \) in the circuit is related to the voltage \( V \) and the capacitive reactance \( X_C \) by the formula:
\[ I = \frac{V}{X_C} \]
- After inserting the dielectric, the new current \( I' \) is given by:
\[ I' = \frac{V}{X_C'} = \frac{V}{\frac{X_C}{K}} = K \cdot \frac{V}{X_C} = K \cdot I \]
- Since \( K > 1 \), it follows that \( I' > I \).

5. **Conclusion**:
- Therefore, the new current \( I' \) after inserting the dielectric slab is greater than the initial current \( I \).

### Final Answer:
The new current \( I' \) is greater than the initial current \( I \).

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