Electric bulb 50 W-100 V glowing at full power is to be used in parallel with battery 120 V, 10 ω. Maximum number of bulbs that can be connected so that they glow in full power is

Question

Electric bulb 50 W-100 V glowing at full power is to be used in parallel with battery 120 V, 10 ω. Maximum number of bulbs that can be connected so that they glow in full power is (A) 2 (B) 8 (C) 4 (D) 6

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/c1b14d2e0724b20d1097a9641f844e72-.png" /> When each bulb is glowing at full power, Current from each bulb =( P / V )=(50/100)=(1/2) A So main current i =( n /2) A Apply KVL E = V + ir ⇒ 120 =100+(( n /2)) × 10 ⇒ n =4

Q. Electric bulb $50 W - 100 V$ glowing at full power is to be used in parallel with battery $120 V, 10 ω$ . Maximum number of bulbs that can be connected so that they glow in full power is

2

8

4

6

When each bulb is glowing at full power, Current from each bulb
$= \frac{P}{V} = \frac{50}{100} = \frac{1}{2} A$
So main current $i = \frac{n}{2} A$
Apply $K V L$
$E = V + i r \Rightarrow 120$
$= 100 + (\frac{n}{2}) \times 10$
$\Rightarrow n = 4$

Q. Electric bulb 50W−100V glowing at full power is to be used in parallel with battery 120V,10ω. Maximum number of bulbs that can be connected so that they glow in full power is

2

8

4

6

When each bulb is glowing at full power, Current from each bulb =VP​=10050​=21​A So main current i=2n​A Apply KVL E=V+ir⇒120 =100+(2n​)×10 ⇒n=4

Q. Electric bulb $50 W - 100 V$ glowing at full power is to be used in parallel with battery $120 V, 10 ω$ . Maximum number of bulbs that can be connected so that they glow in full power is

When each bulb is glowing at full power, Current from each bulb
$= \frac{P}{V} = \frac{50}{100} = \frac{1}{2} A$
So main current $i = \frac{n}{2} A$
Apply $K V L$
$E = V + i r \Rightarrow 120$
$= 100 + (\frac{n}{2}) \times 10$
$\Rightarrow n = 4$