Question

An electric bulb is designed to consume 55 W when operated at 110 volts. It is connected to a 220 V, 50 Hz line through a choke coil in series. What should be the inductance of the coil for which the bulb gets the correct voltage?

Answer 1

Hint: Inductance is property of conductor virtue of which prevents the conductor to change electric current flowing through it. Self-induction is in case when induction occurs in the same circuit through which current flows. Knowing the basic formulas and theoretical details of inductance will lead to the answer.

Formulas Used:
$I = \dfrac{P}{V}$
$V = I \times R$

Complete answer:
When bulb was given direct current then according to relation between power, current and voltage:
$P = VI$
Applying the value of power and voltage from question to get value of current.
$I = \dfrac{P}{V} = \dfrac{{55}}{{110}} = \dfrac{1}{2}A$
Using ohm’s law to yield value of resistance:
$V = I \times R$
$R = \dfrac{V}{I} = \dfrac{{110}}{{\dfrac{1}{2}}} = 220\,\Omega $
On application of A.c. source of current will bring impedance in circuit in place of resistance,
Impedance $\left( Z \right) = \sqrt {{R^2} + {x_1}^2} = \sqrt {{R^2} + {\omega _1}^2} \Omega $
Therefore ohm’s law will be replaced as:
$V = I \times Z$
Putting values of voltage (from question) and current (evaluated above):
$220 = \dfrac{1}{2} \times \sqrt {{R^2} + {\omega _1}^2} $
$\omega = 2\pi fL$
Where, $f$ = frequency and $L$ = Inductance
So we can write above equation as,
$220 = \dfrac{1}{2} \times \sqrt {{{\left( {220} \right)}^2} + \left( {2\pi \times 50 \times L} \right)} $
Evaluating above equation will yield,
Inductance $(L) = 1.2\,H$

Note:
Impedance generated in coils is only present in case of AC current hence it is also known as AC resistance or Reactance. AC current continuously changes its direction and magnitude with time from zero to max and then again to initial zero. This makes AC energy loss minimum and less than DC hence, it is a better option over DC.