1. Given: A series $RC$ circuit is connected to an AC source.

   Two cases:

   • Case $A$: capacitor without dielectric, capacitance $C$
   • Case $B$: capacitor filled with dielectric constant $K=4$, so new capacitance $C' = 4C$

We compare:

• Current through resistor $I_R$ (same as circuit current in series)
• Voltage across capacitor $V_C$

2. Impedance of a series RC circuit

For a series $RC$ circuit, capacitive reactance is $X_C = \frac{1}{\omega C}$

Total impedance magnitude is $Z = \sqrt{R^2 + X_C^2}$

Current is $I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + X_C^2}}$

Since the resistor is in series, $I_R = I$

3. Compare current in the two cases

Case A:

$X_C^A = \frac{1}{\omega C}$

Case B:

Since $C' = 4C$, $X_C^B = \frac{1}{\omega (4C)} = \frac{1}{4\omega C} = \frac{X_C^A}{4}$

So reactance decreases in case $B$. Hence impedance becomes smaller: $Z_B = \sqrt{R^2 + (X_C^B)^2} < \sqrt{R^2 + (X_C^A)^2} = Z_A$

Therefore current becomes larger in case $B$: $I_R^B > I_R^A$

So, $I_R^A < I_R^B$

• Option A: $I_R^A > I_R^B$ → False
• Option B: $I_R^A < I_R^B$ → True

4. Voltage across capacitor

Voltage across capacitor is $V_C = I X_C$

Also, $V_C = \frac{V X_C}{\sqrt{R^2 + X_C^2}}$

Let $f(X_C)=\frac{V X_C}{\sqrt{R^2+X_C^2}}$

As $X_C$ decreases, $V_C$ decreases.

Now in case $B$, $X_C^B = \frac{X_C^A}{4} < X_C^A$

Hence, $V_C^B < V_C^A$

Therefore, $V_C^A > V_C^B$

• Option C: $V_C^A > V_C^B$ → True
• Option D: $V_C^A < V_C^B$ → False

5. Final correct options

The true statements are: $\boxed{B,\ C}$

6. Comparison with stored correct answer

Stored correct answer: $B, C$

My derived answer matches the stored answer exactly.