Question

A \[100W/200V\]bulb and an inductor are connected in series to a \[220V/50Hz\]supply.
Find the power consumed by the bulb.
A. \[100W\]
B. \[92W\]
C. \[84W\]
D. \[74W\]

Answer 1

Hint:The rate of dissipation of electric energy is called electric power. When a current \[I\] is set up in conductor across which a potential difference \[V\] is applied, the energy dissipated in time
\[t\] is given by \[w = VIt\]. Therefore, the electric power dissipated will be
\[P = \dfrac{w}{t} = \dfrac{{VIt}}{t} = VI\]

According to Ohm’s law \[V = IR\], therefore
\[P = \left( {IR} \right)I = {I^2}R\]
Formula Used: The electric power dissipated is given by: \[P = {\left( {\dfrac{V}{{\left| Z \right|}}}
\right)^2}R\]

where, \[P\]is the electric power dissipated

\[I\] is the current
\[R\]is the resistance
\[V\] is the voltage
\[Z\] is the impedance

Impedance is given by :\[Z = \sqrt {{R^2} + {X^2}} \] where, \[R\]is the resistance and \[X\]is the inductive reactance.

The inductive reactance is given as:\[X = 2\pi f\]where, \[f\]is the frequency.

Complete step by step solution:
The electric power dissipated is also given by
\[P = {I^2}R = {\left( {\dfrac{V}{{\left| Z \right|}}} \right)^2}R\] \[ \to (1)\]
where, \[P\]is the electric power dissipated
\[I\] is the current
\[R\]is the resistance

\[V\] is the voltage
\[Z\] is the impedance
The value of resistance will be \[R = \dfrac{{{{\left( {200} \right)}^2}}}{{100}} = 400\Omega \]
Impedance is given by
\[Z = \sqrt {{R^2} + {X^2}} \] \[ \to (2)\]

where, \[R\]is the resistance and \[X\]is the inductive reactance.

The inductive reactance is given as
\[X = 2\pi f\]
where, \[f\]is the frequency.

The frequency of the supply is given as \[50Hz\]. Therefore, \[X = 2\pi (50) = 100\pi \]
Substituting the values of \[R\] and \[X\] in equation (2)
Thus, \[Z = \sqrt {{R^2} + {X^2}} = \sqrt {{{\left( {400} \right)}^2} + {{\left( {100\pi } \right)}^2}} =
\sqrt {160000 + 98696.044} = \sqrt {258696.044} = 508.6217\]

Substituting this value of \[Z\] in equation (1)
\[P = {\left( {\dfrac{V}{{\left| Z \right|}}} \right)^2}R = {\left( {\dfrac{{220}}{{508.6217}}} \right)^2}
\times 400 = 0.1870 \times 400 = 74.8368Watt \approx 74W\]

Hence, option (D) is the correct answer.

Note:The reactance is actually of two types. The circuit in which an inductor is used is called as inductive reactance and depends on the inductance being used. It has a certain DC or AC frequency. On the other hand, the circuit in which a capacitor is used is called a capacitive reactance and depends on the capacitance being used. The total reactance is the subtraction of both the reactance.

In the given problem, only the inductor is connected in series with a bulb. Therefore, only inductive reactance is used.