Question

A bulb is rated at 100V, 100W. It can be treated as a resistor. If the inductance of an inductor (called choke coil) that should be connected in series with the bulb to operate the bulb at its rated power with the help of an ac source of 200V and 50Hz is $\dfrac{{\sqrt x }}{\pi }H$. Find x.

Answer 1

Hint: Induction is defined as the magnetic field which is proportional to the rate of change of the magnetic field. Induction is also known as inductance. Inductance is expressed as ‘L’ and its SI unit is Henry. To find the solution of the given question write down all the given physical properties and apply the
Formula Used:
$P{\text{ }} = {\text{ }}\dfrac{{{V^2}}}{R}$
$V = {V_ \circ }\sin \omega t$

Complete answer:
In order to find the solution of the given question, first we will find the resistance ‘R’ of the bulb.
As we know, mathematically the power dissipated in a resistor is stated by $P{\text{ }} = {\text{ }}\dfrac{{{V^2}}}{R}$ which means power decreases if resistance increases.
$\eqalign{
  & \dfrac{{{V^2}}}{R} = 100 \cr
  & \Rightarrow R = \dfrac{{{{100}^2}}}{{100}} = 100\Omega \cr} $
Now, we know that an external AC that is an alternating current voltage source is driven by the function $V = {V_ \circ }\sin \omega t$. Where ‘V’ is the instantaneous potential difference in the circuit, ‘${V_ \circ }$’ ​ is the maximum value in the oscillating potential difference, ‘$\sin \omega t$’ is the graph which governs the oscillating nature, and ‘$\omega $’ is the angular frequency.
Here, V = $200\sqrt 2 \sin \left( {2\pi 50t} \right)V$
$\omega = 2\pi 50$ rad/sec
In phasor notation, the magnitude of the current or voltage are shown only with their root mean square or RMS value.
So, here in phasor,
${I_{rms}} = \dfrac{{200}}{{\sqrt {{R^2} + {L^2}{\omega ^2}} }}A$
Therefore,
$\eqalign{
  & {I^2}_{rms}R = 100 \cr
  & \Rightarrow \dfrac{{{{200}^2}}}{{{R^2} + {L^2}\omega }}R = 100 \cr
  & \Rightarrow \dfrac{{{{200}^2}}}{{{R^2} + {L^2}{{\left( {2\pi 50} \right)}^2}}}100 = 100 \cr
  & \Rightarrow \dfrac{1}{{{{100}^2}}}\dfrac{{{{200}^2}}}{{{R^2} + {L^2}{\pi ^2}}}100 = 100 \cr
  & \Rightarrow \dfrac{4}{{1 + {L^2}{\pi ^2}}} = 1 \cr
  & \therefore L = \dfrac{{\sqrt 3 }}{\pi }H \cr} $
Thus, according to the question the inductance of an inductor which should be connected in series with the bulb to operate the bulb at its rated power with the help of an ac source of 200V and 50Hz is given by $\dfrac{{\sqrt x }}{\pi }H$.
Hence, the value of x will be 3.

Note:
Induction is further classified in two types, self-induction and mutual induction.
When there occurs a change in the current or magnetic flux of the coil, an opposing induced electromotive force is produced, this phenomenon is known as Self Induction.
Mathematically it is given as, $E = - L\dfrac{{dI}}{{dt}}$
When the current in one coil induces emf in the other coil is defined as the Mutual inductance. Mathematically, mutual inductance ${M_{21}}$ of coil 2 with respect to coil 1 is given by,
$\eqalign{
  & \phi = I \cr
  & \Rightarrow \phi = MI \cr} $
Where ‘M’ is expressed as the mutual inductance of the two coils.