An AC voltage is applied across a capacitor. If the frequency of the supply voltage is doubled, then the capacitive reactance will become:

  1. Double
  2. Half
  3. Remain unchanged
  4. Can't say

Answer (Detailed Solution Below)

Option 2 : Half
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Detailed Solution

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CONCEPT:

Capacitive reactance:

  • The capacitive reactance is the opposition offered by the capacitor in an AC circuit to the flow of ac current.
  • Its SI unit is Ohm(Ω).
  • The capacitive reactance is given as,

\(⇒ X_C=\frac{1}{ω C}=\frac{1}{2π fC}\)

Where ω = angular frequency, f = frequency of ac current and C = capacitance

Impedance:

  • Impedance is essentially everything that obstructs the flow of electrons within an electrical circuit.
  • For a pure capacitor, the capacitive reactance is equal to the impedance.

AC voltage applied to a capacitor:

  • When an AC voltage is applied to a capacitor, the current in the circuit is given as,

\(⇒ I=\frac{V}{X_C}\)

  • In a pure capacitor circuit, the current is ahead of the voltage by one-fourth of a period.

F1 Shraddha Prabhu 09.07.2021 D1

F1 Shraddha Prabhu 09.07.2021 D2

CALCULATION:

  • We know that the capacitive reactance is given as,

\(⇒ X_C=\frac{1}{2π fC}\)     -----(1)

Where f = frequency of ac current and C = capacitance

For case 1:

\(⇒ X_C=\frac{1}{2π fC}\)     -----(2)

For case 2: (f' = 2f)

\(⇒ X_C^{'}=\frac{1}{2π f'C}\)

\(⇒ X_C^{'}=\frac{1}{2π \times2fC}\)

\(⇒ X_C^{'}=\frac{1}{2}\times\frac{1}{2πfC}\)     -----(3)

By equation 2 and equation 3,

\(⇒ X_C^{'}=\frac{X_C}{2}\)

  • Hence option 2 is correct.
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