Two resistors of $200\Omega$ and $400\Omega$ are connected in series with a battery of 100 V . A bul

Question

Two resistors of $200\Omega$ and $400\Omega$ are connected in series with a battery of 100 V . A bulb rated at $200	ext{ }V,100	ext{ }W$ is connected across the $400\Omega$ resistance. The potential drop across the bulb is $\_\_\_\_$ V.

Accepted Answer

Sol. [Key Concept: Bulb resistance $R = V^{2}/P$; then analyse the series/parallel network.]For resistance of bulb$$R = \frac{V^{2}}{P} = \frac{200 	imes 200}{100} = 400\Omega $$Now circuit will look like this$V_{1}\&{	ext{ }V}_{2}$ will be same as resistances are same$${V_{1} = V_{2} = 50	ext{ }V }{V_{	ext{bulb}	ext{~}} = V_{2} = 50	ext{ }V }$$

Question

Options

Solution

More Current Electricity and Electrical Instrument Questions