Question

electric bulb 250W and 230V 1) find the resistamce and 2)
safe limit of the current that can flow through the bulb​

Answer 1

S O L U T I O N :

Given :

• Power (P) = 250 w
• Potential difference (V) = 230 v

Case (I), Find the resistance.

We know that,

P = V²/R ★

[ Put the values ]

⟶ 250 = 230²/R

⟶ 250R = 52900

⟶ R = 52900/250

⟶ R = 211.6 ohm ★

Case (II), Find safe limit of the current that can flow through the bulb.

According to ohms law,

★ V = IR

[ Put the values ]

⟶ 230 = I × 211.6

⟶ I = 230/211.6

⟶ I = 1.08 A ★

Hence Solved!!