Three identical bulbs B₁, B₂ and B₃ each of power rating 18 W, 12 V are connected to a battery of 12 V.

(a) Calculate :

1. the resistance of each bulb
2. the current drawn from the cell

(b) If the bulb B₃ is removed from the circuit, then will the brightness of the bulb B₁ increase, decrease or remain the same?

Given,

Battery emf (ε) = 12 V

Voltage rating of eac bulb (V) = 12 V

Power rating of eac bulb (P) = 18 W

(a)

1. Let resistance of each bulb be R.

Then,

$\text P = \dfrac {\text V^2}{\text R} \\[1 em] ⇒ \text R = \dfrac {\text V^2}{\text P} = \dfrac {12^2}{18} = \dfrac {144}{18}\\[1 em] ⇒ \text R = 8\ \text Ω$

2. As, B₂ and B₃ are in parallel combination then

$\dfrac{1}{\text R$\text P} = \dfrac{1}{\text R1} + \dfrac{1}{\text R_2} \\[1 em]

Here, R₁ = R₂ = 8 Ω

$⇒ \dfrac{1}{\text R$\text P} = \dfrac{1}{8} + \dfrac{1}{8} = \dfrac{2}{8}\\[1 em] ⇒ \text R\text P = \dfrac{8}{2} = 4\ \text Ω

Now this combination is in series with B₁ then

R_S = R_P + R = 4 + 8 = 12 Ω

So,

$\text {Current drawn} = \dfrac{\text {Battery emf}}{\text {Total resistance}} = \dfrac{\text ε}{\text R_\text S}\[1 em] =\dfrac{12}{12} = 1\ \text A$

Hence, current drawn from the battery is 1 A.

(b) When B₃ is removed, the circuit becomes :

Now, the equivalent resistance = 8 + 8 = 16 Ω.

Since, the resistance increases, current in the circuit decreases.

Hence, brightness of B₁ decreases.