An electric bulb of 500 W at 100 V is used in a circuit having a 200 V supply. Calculate the resistance R to be connected in series with the bulb, so that the power delivered by the bulb is 500 W.

Given, power rating of bulb, P_B = 500W

Voltage across bulb, V_B = 100V

Supply voltage, V_S = 200V

If a resistance R is attached in series with the bulb, then the voltage across resistance will be 100 V.

Now, current flowing in circuit when bulb delivers power of 500 W is given as

       $\begin{array}{l} {\mathrm{P}}_{\mathrm{B}}={\mathrm{V}}_{\mathrm{B}}\mathrm{I}\\ \Rightarrow 500=100\times \mathrm{I}\\ \Rightarrow \mathrm{I}=5\mathrm{A}\end{array}$

Same amount of current will flow from the resistance as it is connected in series.

Using Ohm’s law,

      $\begin{array}{l} \mathrm{V}=\mathrm{IR}\\ \Rightarrow 100=5\times \mathrm{R}\\ \Rightarrow \mathrm{R}=20\mathrm{\Omega }\end{array}$

Thus, the resistance connected in series is 20Ω.