3 identical bulbs are connected in series and these together dissipate a power P. If now the bulbs are connected in parallel, then the power dissipated will be :

Question

3 identical bulbs are connected in series and these together dissipate a power P. If now the bulbs are connected in parallel, then the power dissipated will be : (A) P/3 (B) 3P (C) 9P (D) P/9

Tardigrade Admin · Accepted Answer

Let the resistance of each bulb is R. Then, total resistance in series =3R ∴ P=(V2/3R) In parallel, total resistance =(R/3) ∴ Power dissipated P'=(V2/R/3) =(3V2/R) =9( (V2/3R) ) =9P

Q. 3 identical bulbs are connected in series and these together dissipate a power P. If now the bulbs are connected in parallel, then the power dissipated will be :

P/3

3P

9P

P/9

Let the resistance of each bulb is R. Then, total resistance in series $= 3 R$ $∴$ $P = \frac{V ^{2}}{3 R}$ In parallel, total resistance $= \frac{R}{3}$ $∴$ Power dissipated $P^{'} = \frac{V ^{2}}{R /3}$ $= \frac{3 V ^{2}}{R}$ $= 9 (\frac{V ^{2}}{3 R})$ $= 9 P$

Q. 3 identical bulbs are connected in series and these together dissipate a power P. If now the bulbs are connected in parallel, then the power dissipated will be :

P/3

3P

9P

P/9

Let the resistance of each bulb is R. Then, total resistance in series =3R ∴ P=3RV2​ In parallel, total resistance =3R​ ∴ Power dissipated P′=R/3V2​ =R3V2​ =9(3RV2​) =9P

Let the resistance of each bulb is R. Then, total resistance in series $= 3 R$ $∴$ $P = \frac{V ^{2}}{3 R}$ In parallel, total resistance $= \frac{R}{3}$ $∴$ Power dissipated $P^{'} = \frac{V ^{2}}{R /3}$ $= \frac{3 V ^{2}}{R}$ $= 9 (\frac{V ^{2}}{3 R})$ $= 9 P$