Question

An inductor of 1 henry is connected across a$220{\text{V}}$,$50Hz$ supply. The peak value of the current is approximate:
A) $0.5A$.
B) $0.7A$.
C) $1A$.
D) $1.4A$.

Answer 1

Hint:The formula of peak voltage and also the formula for peak current can be used to solve this problem. An inductor is a device which is used to get the desired value of frequency. When current is passed through the inductor then there is formation of a magnetic field which opposes the flow of current. The inductor is used as an energy storing device.

Formula used: The formula for peak current is given by ${\text{peak current}} = \dfrac{{{\text{peak voltage}}}}{{{\text{inductive reactance}}}}$Here inductive reactance is resistance offered by the inductor in an AC circuit. The formula for peak voltage is given by${\text{peak voltage}} = {\text{rms voltage}} \times \sqrt 2 $, also the inductive reactance is given by\[{\text{inductive reactance}} = \omega L\].

Step by step solution:
As it is given as that root mean square voltage is given as$220{\text{V}}$
So let us calculate the value of peak voltage.
$
  {\text{peak voltage}} = {\text{rms voltage}} \times \sqrt 2 \\
  {\text{peak voltage}} = 220 \times \sqrt 2 \\
  {\text{peak voltage}} = 220\sqrt 2 {\text{V}} \\
 $
Now let us calculate the value of peak current,
$
  {\text{peak current}} = \dfrac{{{\text{peak voltage}}}}{{{\text{inductive reactance}}}} \\
  {\text{peak current}} = \dfrac{{220\sqrt 2 }}{{\omega L}} \\
  {\text{peak current}} = \dfrac{{220\sqrt 2 }}{{\omega \cdot 1}} \\
  {\text{peak current}} = \dfrac{{220\sqrt 2 }}{\omega } \\
 $………eq. (1)
The value of $\omega $ is equal to.
$
  \omega = 2\pi f \\
  \omega = 2\pi \cdot \left( {50} \right) \\
  \omega = 100 \cdot \pi \\
 $
Let us put the value of $\omega $ in equation (1)
$
  {\text{peak current}} = \dfrac{{220\sqrt 2 }}{\omega } \\
  {\text{peak current}} = \dfrac{{220\sqrt 2 }}{{100 \cdot \pi }} \\
  {\text{peak current}} = 0.9{\text{A}} \\
 $
So the peak current is ${\text{peak current}} \approx 1{\text{A}}$.

So the correct answer for this problem is option C.

Note: The maximum amount of current that can flow through the circuit is called peak current, similarly the maximum value of voltage in waveform of voltage is called peak voltage. It is important for students to remember the formula for the calculation of peak current and also the formula for calculation of peak voltage.