Comparison of Current and Capacitor Voltage in RC Circuit with Dielectric

To compare the current $I_R$ through the resistor and the voltage $V_C$ across the capacitor in the two cases, let the RMS voltage of the AC source be $V$ and its angular frequency be $\omega$.

1. Capacitive Reactance and Impedance

The capacitive reactance of a capacitor with capacitance $C'$ is given by:
$X_C = \frac{1}{\omega C'}$

The total impedance $Z$ of a series $R-C$ circuit is:
$Z = \sqrt{R^2 + X_C^2}$

• Case (A): The capacitor has no dielectric, so its capacitance is $C_A = C$.
  $X_C^A = \frac{1}{\omega C}$
  $Z_A = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}$

• Case (B): The capacitor is filled with a dielectric of constant $K = 4$, so its capacitance becomes $C_B = 4C$.
  $X_C^B = \frac{1}{4\omega C}$
  $Z_B = \sqrt{R^2 + \left(\frac{1}{4\omega C}\right)^2}$

2. Comparison of Current ($I_R$)

The RMS current through the resistor is equal to the total current in the series circuit:
$I_R = \frac{V}{Z}$

Since $X_C^B < X_C^A$, it follows that:
$Z_B < Z_A$

Therefore, comparing the currents in both cases:
$I_R^A = \frac{V}{Z_A} < \frac{V}{Z_B} = I_R^B \implies I_R^A < I_R^B$

This shows that Option (B) is correct.

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3. Comparison of Voltage across Capacitor ($V_C$)

The RMS voltage across the capacitor is given by:
$V_C = I_R \cdot X_C = \frac{V}{\sqrt{R^2 + X_C^2}} \cdot X_C = \frac{V}{\sqrt{\frac{R^2}{X_C^2} + 1}}$

• For Case (A):
  $V_C^A = \frac{V}{\sqrt{\omega^2 R^2 C^2 + 1}}$

• For Case (B):
  $V_C^B = \frac{V}{\sqrt{\omega^2 R^2 (4C)^2 + 1}} = \frac{V}{\sqrt{16\omega^2 R^2 C^2 + 1}}$

Comparing the denominators, we have:
$\sqrt{16\omega^2 R^2 C^2 + 1} > \sqrt{\omega^2 R^2 C^2 + 1}$

Therefore:
$V_C^A > V_C^B$

This shows that Option (C) is correct.

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Conclusion

• $I_R^A < I_R^B$ (Option B)
• $V_C^A > V_C^B$ (Option C)

Correct Options: B and C