Question

A three way light bulb has three brightness setting (low, medium, high) but only two filaments. A particular light bulb (of the said descripition) when connected across a 120 V line can with higher resistance burns out. The two filaments are arranged in three settings.
(i) higher resistance filament only working for 60 W
(ii) low resistance filament working for 120 W
(iii) low resistance filament working for 60 W
(iv) high resistance filament working for 120 W
(v) low and high resistance filaments in parallel for 180 W
(vi) low and high resistance filament in series for 180 W

• A. (i), (ii) and (v) are correct
• B. (i), (ii) and (vi) are correct
• C. (iii), (iv) and (v) are correct
• D. (iii), (iv) and (vi) are correct

Answer 1

The correct option is A (i), (ii) and (v) are correct

Since higher resistance filament burns out. It must work for $60W$. Lower resistance doesn't burn out and hence it works for $120W$. Now since in parallel voltage across two remains same (which we are supplying), the powers of two resistors must add up to $60+120=180W$ in parallel.

Hence, option A is correct.