Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Current Electricity

Question:

Options:

a

b

c

d

Correct Answer:

c

Explanation:

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

$ R_1 = \frac{220^2}{25}= 1936\Omega , R_2 = \frac{220^2}{100} = 484\Omega$

$ R_{eq} = R_1 + R_2 = 2420$

$ I = \frac{440}{2420} = 0.18A$

$ P_1 = I^2R_1=62.72W , P_2 = I^2R_2 = 15.68W$

$\text{ 25W bulb will fuse first}$