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 $10^6Hz$ 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. $45^\circ$
  2. $90^\circ$
  3. $75^\circ$
  4. $60^\circ$
  1. One microfarad capacitor is joined to a 200V, 50Hz alternator. The rrns current through capacitor is:
  1. $6.28 \times 10^{-2}A$
  2. $7.5 \times 10^{-4}A$
  3. $10.52 \times 10^{-2}A$
  4. $15.25 \times 10^{-2}A$

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) $90^\circ$
  2. (a) $6.28 \times 10^{-2}A$
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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