Question
Let a source of alternating e.m.f. $\text{E} = \text{E}_\circ\sin\omega\text{t}$ be connected to a capacitor of capacitance C. If 'I' is the instantaneous value of current in the circuit at instant t, then $\text{I}=\frac{\text{E}_0}{\frac{1}{\omega\text{C}}}\sin\Big(\omega\text{t}+\frac{\pi}{2}\Big).$ The capacitive reactance limits the amplitude of current in a purely capacitive circuit and it is given by $\text{X}_\text{C}=\frac{1}{\omega\text{C}}.$

  1. What is the unit of capacitive reactance?
  1. Farad
  2. Ampere
  3. Ohm
  4. Ohm-1
  1. The capacitive reactance of a $5\mu\text{F}$ capacitor for a frequency of 106Hz is:
  1. $0.032\Omega$
  2. $2.52\Omega$
  3. $1.25\Omega$
  4. $4.51\Omega$
  1. In a capacitive circuit, resistance to the flow of current is offered by:
  1. Resistor
  2. Capacitor
  3. Inducto
  4. Frequency
  1. In a capacitive circuit, by what value of phase angle does alternating current leads the e.m.f?
  1. 45º
  2. 90º
  3. 75º
  4. 60º
  1. One microfarad capacitor is joined to a 200V, 50Hz alternator. The rrns current through capacitor is:
  1. 6.28 × 10-2A
  2. 7.5 × 10-4A
  3. 10.52 × 10-2A
  4. 15.25 × 10-2A

Answer

  1. (c) Ohm

Solution:

Ohm is the unit of capacitive reactance.

  1. (a) $0.032\Omega$

Solution:

Capacitive reactance, $\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\mu\text{C}}$

$=\frac{1}{2\pi\times10^6\times5\times10^{-6}}$

$=0.032\Omega$

  1. (b) Capacitor

Solution:

In capacitive circuit, resistance to the flow of current is offered by the capacitor.

  1. (b) 90º
  2. (a) 6.28 × 10-2A

Solution:

Current, $\text{l}_\text{v}=\frac{\text{E}_\text{V}}{\text{X}_\text{C}}=\frac{1}{\frac{1}{2\pi\mu\text{C}}}=(2\pi\mu\text{C})\text{E}_\text{V}$

$\text{I}_\text{v}=2\times3.14\times50\times10^{-6}\times200$

$=6.28 × 10^{-2}\text{A}$

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