Question

An electric bulb of resistance 500 ohm, draws a current of 0.4 A. Calculate the power of the bulb and the potential difference at its end.

Kindly don't spam. Irrelevant answer will be reported.​

Answer 1

Explanation:

P=V×I

I=

V

P

I=

220 V

3000 W

I=13.63 A

So, the current rating of the motor is 13.63 A. Hence, this is the required solution.

Learn more,

Answer 2

S O L U T I O N :

$\underline{\bf{Given\::}}$

• An electric bulb of resistance, (R) = 500 ohm
• Current drawn, (i) = 0.4 Ampere

$\underline{\bf{Explanation\::}}$

As we know that formula of the power;

$\boxed{\bf{P = \frac{i^{2} Rt}{t} }}$

Rate of heat = Power

$\mapsto\tt{P= i^{2} R}$

$\mapsto\tt{P = (0.4)^{2} \times 500}$

$\mapsto\tt{P = 0.16 \times 500}$

$\mapsto\bf{P = 80\:watt}$

∴ The power of the bulb will be 80 watt .

Now,

Using by another formula from electric power;

$\boxed{\bf{P=\frac{V^{2} }{R} }}$

A/q

$\mapsto\tt{PR = V^{2} }$

$\mapsto\tt{80 \times 500 = V^{2}}$

$\mapsto\tt{40000 = V^{2}}$

$\mapsto\tt{V = \sqrt{40000} }$

$\mapsto\bf{V = 200\:volt}$

Thus,

The potential difference at it's end will be 200 volt .