Question

An electric bulb of resistance 500 12 draws current
0.4 A from the source. Calculate : (a) the power of
bulb and (b) the potential difference at its end.
Ans. (a) 80 W, (b) 200 ​

Answer 1

Answer :-

a) Power of the bulb is 80 W.

b) Potential difference at its end is 200 V.

Explanation :-

Given :

Resistance (R) = 500 Ω

Electric current (I) = 0.4 A

To find :

Power (P)

Potential difference (V)

Solution :

➤ From Ohm's law :

⇒ V = IR

⇒ V = 500 × 0.4

⇒ V = 200 V

➤ Power :

⇒ P = V²/R

⇒ P = 200²/500

⇒ P = 40000/500

⇒ P = 80 W

∴,

a) Power of the bulb is 80 W.

b) Potential difference at its end is 200 V.

Answer 2

Answer:

Power of bulb is 80 w

potential difference at its end is 200W