question_answer

An electric bulb rated as 500W - 100V is used in a circuit having 200 V supply. The resistance R that must be put in series with the bulb, so that the bulb draws 500 W is

A) \[100\Omega \]           

B) \[50\Omega \]

C) \[20\Omega \]             

D) \[10\Omega \]

Correct Answer: C

Solution :

 We know that, \[p=Vi\] \[\therefore \]Current through the circuit, \[i=\frac{500}{100}=5A\] Potential difference across, \[R=100\,V\] \[\therefore \]Relation, \[V=IR\] \[100=5\times R\] \[\Rightarrow \] \[R=\frac{100}{5}=20\Omega \]