Question

An electric bulb rated for 500W at 100V is used in a circuit having a 200V supply. The resistance R that must be kept in series with the bulb so that the bulb draws 500W is -
A) 30 ohm B) 5 ohm C) 20 ohm
Give steps. BRAINLIEST SURE✨

Answer 1

See the attachment..................

Answer 2

Answer:

(c). The resistance will be 20 ohms.

Explanation:

Given that,

Power of bulb P= 500 W

Voltage V= 100 V

The resistance of the bulb is

The resistance is defined as:

$R= \dfrac{V^2}{P}$

Here, V= voltage

P = power

$R= \dfrac{(100)^2}{500}$

$R= 20\ Omega$

In case the voltage of the battery V = 200 V

Power = 500 W

Resistance R = 20 ohms

then, The current will be

$I = \sqrt{\dfrac{P}{R}}$

$I=\sqrt{\dfrac{500}{20}}$

$I= 5 A$

Using ohm's law

Voltage V= 200 V

The equivalent resistance of the circuit

$R=\dfrac{200}{5}$

$R= 40 \Omega$

The resistance of bulb is 20 ohms and extra resistance is 20 ohms.

Hence, The resistance will be 20 ohms.