Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Current Electricity

Question:

An electric bulb rated 220 v and 60 W is connected in series with another electric bulb rated 220 v and 40 W. The combination is connected across 220 volt source of e.m.f. If Power of bulb 1 is $P_1'$ and power of bulb 2 is $P_2'$ then

Options:

$P_1' > P_2'$

$P_1' < P_2'$

$P_1' = P_2'$

None of the above

Correct Answer:

$P_1' < P_2'$

Explanation:

$R = \frac{V^2}{P}$

$\text{Resistance of the first bulb } R_1 = \frac{V^2}{P_1}$

$\text{Resistance of the first bulb } R_2 = \frac{V^2}{P_2}$

$\text{In series same current will pass through each bulb}$

$\text{Ratio of Power is }P_1' = I^2\frac{V^2}{P_1} $

$\text{Ratio of Power is }P_2' = I^2\frac{V^2}{P_2} $

$\Rightarrow \frac{P_1'}{P_2'}= \frac{P_2}{P_1} <1$

$\Rightarrow P_1' < P_2'$