Question

In the figure shown ${\text{B}}_1,{\text{B}}_{2}and{\text{B}}_3$are the three bulbs rated as $(200\text{V},25\text{W}$ ) respectively.

Which bulb will give more light?

• A. ${\text{B}}_1$
• B. ${\text{B}}_2$
• C. ${\text{B}}_3$
• D. All three will give equal light

Answer 1

The correct option is C ${\text{B}}_3$

Using the relation $P=\frac{v^2}{R}$for power rating of each bulb.

$R_1=\frac{(V{)}^2}{P_1}=\frac{(200{)}^2}{50}=800\Omega$
$R_2=\frac{(V{)}^2}{P_2}=\frac{(200{)}^2}{100}=200\Omega$
$R_3=\frac{(V{)}^2}{P_3}=\frac{(200{)}^2}{25}=1600\Omega$

All the three bulbs are in series thus current supplied by battery,

$i=\frac{V}{R_{eq}}=\frac{200}{R_1+R_2+R_3}$

Also same current will pass through each bulb. Using relation for power consumed by bulb,

$P=i^{2}R$.......(i)

$\because R_3>R_1>R_2$ and $i_1=i_2=i_3=i$

From Eq.($i$)we get,

$P\propto R$

$\Rightarrow P_3>P_1>P_2$

Power consumed by bulb ${\text{B}}_3$ is maximum hence it will glow the brightest.

$\therefore {\text{B}}_3$ will give more light.