Bulbs in series and parallel for CET aspirants

Combination of Bulbs

Bulbs in Series

(i) Total power consumed 1/P_total = 1/P₁ + 1/P₂ +...

(ii) If ‘n’ bulbs are identical, P_total = P/N

(iii) P_consumed (Brightness) ∝ V ∝ R ∝ 1/P_rated i.e. in series combination bulb of lesser wattage will give more bright light and p.d. appeared across it will be more.

Bulbs in Parallel

(i) Total power consumed

            P_total = P₁ + P₂ + P₃ +...+ P_n

(ii) If ‘n’ identical bulbs are in parallel P_total = nP

(iii) P_consumed (Brightness) ∝ P_R ∝ i ∝ 1/R i.e. in parallel combination, bulb of greater wattage will give more bright light and more current will pass through it.

Solved example 1: An electric bulb is rated 220 volt and 100 watt. Power consumed by it when operated on 110 volt is

(A) 50 watt      (B) 75 watt      (C) 90 watt      (D) 25 watt

Solution: (D) Resistance of the bulb V₂/P_Rotate = 220×220/100 = 484 Ω

When connected with 110 V, the power consumed

            P_consumed = V₂/R = 110×110/484 = 25W

Solved example 2: Two bulbs are working in parallel order. Bulb A is brighter than bulb B. If RAB are their resistance respectively then and R

(A) R_A > R_B                             (B) R_A < R_B

(C) R_A = R_B                             (D) None of these

Solution: (B) In parallel P_consumed ∝ Brightness ∝ 1/R

P_A > P_B (given). R_A < R_B

Posted in: Electric Current