An electric bulb is rated ‘$100{\rm{ W, }}250{\rm{ V}}$‘, What is the resistance of the bulb?
A. ${\rm{2}}{\rm{.5 ohm}}$
B. ${\rm{625 ohm}}$
C. ${\rm{25000 ohm}}$
D. ${\rm{1500 ohm}}$
Answer
609.9k+ views
Hint:In the solution we use the basic relation between power, voltage, and resistance to obtain the resistance from the given values of power and voltage of the bulb.
Complete Step by Step Answer:Given:
Power of the bulb $P = 100{\rm{ W}}$
Voltage of the bulb $V = 250{\rm{ V}}$
The relation between power, voltage, and resistance is given by
$R = \dfrac{{{V^2}}}{P}$
Where R is the resistance of the bulb.
Substituting the values of power and voltage in the above equation gives
$\begin{array}{c}
R = \dfrac{{{{\left( {250{\rm{ V}}} \right)}^2}}}{{100{\rm{ W}}}}\\
= \dfrac{{62500}}{{100}}\\
= 625{\rm{ ohm}}
\end{array}$
Therefore, the correct answer is option B, that is $625{\rm{ ohm}}$
Note: Make sure to get a strong grip on the fundamentals of the topic, so that this type of question can be solved easily and also make sure that you have some knowledge about domestic electric circuits.
Complete Step by Step Answer:Given:
Power of the bulb $P = 100{\rm{ W}}$
Voltage of the bulb $V = 250{\rm{ V}}$
The relation between power, voltage, and resistance is given by
$R = \dfrac{{{V^2}}}{P}$
Where R is the resistance of the bulb.
Substituting the values of power and voltage in the above equation gives
$\begin{array}{c}
R = \dfrac{{{{\left( {250{\rm{ V}}} \right)}^2}}}{{100{\rm{ W}}}}\\
= \dfrac{{62500}}{{100}}\\
= 625{\rm{ ohm}}
\end{array}$
Therefore, the correct answer is option B, that is $625{\rm{ ohm}}$
Note: Make sure to get a strong grip on the fundamentals of the topic, so that this type of question can be solved easily and also make sure that you have some knowledge about domestic electric circuits.
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