Question

An electric bulb is made of tungsten filament of resistance \[R\Omega \]. It is marked \[100W\]and \[230V\]. Then, the value of \[R\] is:
                           \[\begin{align}
  & A.300\Omega \\
 & B.529\Omega \\
 & C.739\Omega \\
 & D.100\Omega \\
\end{align}\]

Answer 1

Hint: We can use the basic equation of electric power to solve this question. Power is given as the product of electric voltage and current. Here what we need to find is resistance of the bulb. By sung ohm’s law and substituting it we will get the result.
      Formula used:
                               \[\begin{align}
  & P=\dfrac{{{V}^{2}}}{R} \\
 & \\
\end{align}\]
Complete solution: As mentioned in the hint, the formula we use is derived from the basic power equation. Power is given in watts. So to find resistance we will just need to re arrange the power equation. For those who don’t know, power is the product of current and voltage.
         i.e , \[P=V\times I\] - - -(1)
 But, by ohm's law, we know \[I=\dfrac{V}{R}\]
 By substituting this in the power equation (1),
We get, \[P=\dfrac{{{V}^{2}}}{R}\] - - - -(2)
   In the question it is given that,
       The maximum voltage for the given bulb, \[V=230V\]
        Power of the given bulb, \[P=100W\]
   So now, to find the resistance \[R\] of the bulb, we rearrange the equation (2)
     \[\Rightarrow R=\dfrac{{{V}^{2}}}{P}\]
            \[R=\dfrac{{{(230)}^{2}}}{100}=\dfrac{52900}{100}\]
              \[\begin{align}
  & R=529\Omega \\
 & \\
\end{align}\]

So, we found the resistance of the given bulb as \[529\Omega \]. That indicates the correct answer as option b.

Note: This question can be easily done if we are aware of the direct relationship between power, voltage and resistance. So, we could easily skip the step where we use ohm's law. So it will always help you if you remember the direct equation to solve the problem faster. So always keep in mind that \[P=\dfrac{{{V}^{2}}}{R}\].