Question 70: AC fundamentals

Question

A series RC-current 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 IR through the resistor and voltage VC across the capacitor are compared in the two cases. Which of the following is/are true?

Accepted Answer

In case A,
The capacitive reactance is
$$X_C^A = {1 \over {\omega C}}$$
Impedance of the circuit is
$${Z_A} = \sqrt {{{(R)}^2} + {{\left( {{1 \over {\omega C}}} \right)}^2}} $$
$$I_R^A = {V \over {\sqrt {{{(R)}^2} + {{\left( {{1 \over {\omega C}}} \right)}^2}} }}$$ ...... (i)
$$V_C^A = {{I_R^A} \over {\omega C}} = {V \over {\sqrt {{{(R\omega C)}^2} + 1} }}$$ ...... (ii)
In case B,
The capacitive reactance is
$$X_C^B = {1 \over {\omega (4C)}} = {1 \over {4\omega C}}$$
Impedance of the circuit is
$${Z_B} = \sqrt {{R^2} + {{\left( {{1 \over {4\omega C}}} \right)}^2}} $$
$$I_R^B = {V \over {\sqrt {{R^2} + {{\left( {{1 \over {4\omega C}}} \right)}^2}} }}$$ ..... (iii)
$$V_C^B = {V \over {\sqrt {{{(4R\omega C)}^2} + 1} }}$$ ..... (iv)
From (i) and (iii), we conclude that
$$I_R^A  V_C^B$$

Answer

$$I_R^A > I_R^B$$

Answer

$$I_R^A < I_R^B$$

Answer

$$V_C^A > V_C^B$$

Answer

$$V_C^A < V_C^B$$

AC fundamentals

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