What will be the equivalent resistance if four bulbs of 12 V - 3 W are joined in series?

Correct answer: d 192 Ω

Each has resistance R = V²/P = 144 ÷ 3 = 48 Ω; four in series give 4 × 48 = 192 Ω.

Why the other options are wrong

• a 0.021 Ω: 0.021 Ω comes from the inverse P/V² (3 ÷ 144); the right form is R = V²/P = 48 Ω for each.
• b 0.25 Ω: 0.25 is the current of each (3 ÷ 12 A), not a resistance; the equivalent resistance is 192 Ω.
• c 48 Ω: 48 Ω is the resistance of one bulb; four in series add up to 192 Ω.