A sereis R-C circuit is connected to AC voltage source. Consider two cases, (A) when C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current `I_(R)` through the resistor and voltage `V_(c)` across the capacitor are compared in the two cases. Which of the following is/ are true?

Options

A. `I_(R)^(A)gtI_(R)^(B)`
B. `I_(R)^(A)ltI_(R)^(B)`
C. `V_(C)^(A)ltV_(C)^(B)`
D. None of these

Explanation

To solve the problem, we need to analyze the behavior of the current through the resistor and the voltage across the capacitor in a series R-C circuit connected to an AC voltage source under two different conditions:

1. **Case A**: Capacitor without a dielectric medium.
2. **Case B**: Capacitor filled with a dielectric of constant 4.

### Step-by-Step Solution:

**Step 1: Understanding the Impedance in Both Cases**

In a series R-C circuit, the total impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + X_C^2} \]
where \( X_C \) is the capacitive reactance defined as:
\[ X_C = \frac{1}{\omega C} \]
In Case A, the capacitance is \( C \), and in Case B (with dielectric), the capacitance becomes \( C' = 4C \).

**Step 2: Calculate the Impedance for Each Case**

- **Case A (without dielectric)**:
\[ Z_A = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \]

- **Case B (with dielectric)**:
\[ Z_B = \sqrt{R^2 + \left(\frac{1}{\omega C'}\right)^2} = \sqrt{R^2 + \left(\frac{1}{\omega (4C)}\right)^2} = \sqrt{R^2 + \frac{1}{16\omega^2 C^2}} \]

**Step 3: Compare the Impedance**

Since \( C' = 4C \), we can see that:
\[ Z_B < Z_A \]
This is because the term \( \frac{1}{16\omega^2 C^2} \) is smaller than \( \frac{1}{\omega^2 C^2} \).

**Step 4: Calculate the Current Through the Resistor**

The current through the resistor \( I_R \) can be calculated using Ohm's law:
\[ I_R = \frac{V}{Z} \]

- For Case A:
\[ I_{R,A} = \frac{V}{Z_A} \]

- For Case B:
\[ I_{R,B} = \frac{V}{Z_B} \]

Since \( Z_B < Z_A \), it follows that:
\[ I_{R,B} > I_{R,A} \]

**Step 5: Calculate the Voltage Across the Capacitor**

The voltage across the capacitor \( V_C \) is given by:
\[ V_C = I_R \cdot X_C \]

- For Case A:
\[ V_{C,A} = I_{R,A} \cdot \frac{1}{\omega C} \]

- For Case B:
\[ V_{C,B} = I_{R,B} \cdot \frac{1}{\omega C'} \]

Substituting \( C' = 4C \):
\[ V_{C,B} = I_{R,B} \cdot \frac{1}{4\omega C} \]

**Step 6: Compare the Voltage Across the Capacitor**

Since \( I_{R,B} > I_{R,A} \) and \( V_{C,B} \) has a factor of \( \frac{1}{4} \):
\[ V_{C,B} < V_{C,A} \]

### Conclusion

From the analysis, we can conclude:
- The current through the resistor increases when the dielectric is introduced (Case B).
- The voltage across the capacitor decreases when the dielectric is introduced (Case B).

### Final Results
- **Current**: \( I_{R,B} > I_{R,A} \)
- **Voltage**: \( V_{C,B} < V_{C,A} \)