Q. A $280\ ohm$ electric bulb is connected to $200V$ electric line. The peak value of current in the bulb will be
A) About one ampere
B) Zero
C) About two ampere
D) About four ampere
Explanation
(a) $i_{rms} = \frac{200}{280} = \frac{5}{7}A$. So $i_{0} = i_{rms} \times \sqrt{2} = \frac{5}{7} \times \sqrt{2} \approx 1A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.