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(C) The current $i$ in a series $R-C$ circuit is given by:$i = \frac{V_0}{Z} = \frac{V_0}{\sqrt{R^2 + X_C^2}} = \frac{V_0}{\sqrt{R^2 + (1/\omega C)^2}

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$V_a > V_b$

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(C) The current $i$ in a series $R-C$ circuit is given by:$i = \frac{V_0}{Z} = \frac{V_0}{\sqrt{R^2 + X_C^2}} = \frac{V_0}{\sqrt{R^2 + (1/\omega C)^2}}$The voltage across the capacitor is $V = i X_C = i \cdot \frac{1}{\omega C}$.Substituting the expression for $i$:$V = \frac{V_0}{\sqrt{R^2 + (1/\omega C)^2}} \cdot \frac{1}{\omega C} = \frac{V_0}{\sqrt{R^2 \omega^2 C^2 + 1}}$When the capacitor is filled with mica (dielectric constant $K > 1$),its capacitance increases,so $C_b > C_a$.$1$. For current: As $C$ increases,$X_C = 1/\omega C$ decreases,so the impedance $Z = \sqrt{R^2 + X_C^2}$ decreases. Thus,$i_b > i_a$.$2$. For voltage across the capacitor: From $V = \frac{V_0}{\sqrt{R^2 \omega^2 C^2 + 1}}$,as $C$ increases,the denominator increases,which means $V$ decreases. Therefore,$V_b  V_b$.

$A$ series $R-C$ circuit is connected to an alternating voltage source. Consider two situations:
$(a)$ When the capacitor is air-filled.
$(b)$ When the capacitor is mica-filled.
If the current through the resistor is $i$ and the voltage across the capacitor is $V$,then:

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$A$ series $R-C$ circuit is connected to an alternating voltage source. Consider two situations:$(a)$ When the capacitor is air-filled.$(b)$ When the capacitor is mica-filled.If the current through the resistor is $i$ and the voltage across the capacitor is $V$,then: