Question

You have a special lightbulb with a very delicate wire filament. The wire will break if the current in it ever exceeds 1.30 A , even for an instant.

What is the largest root-mean-square current you can run through this bulb?

Answer 1

about 0.919 amp

Step-by-step explanation:

`I_r_m_s=\frac{I_m_a_x} {√(2)}\\ I_r_m_s=(1.3)/(√(2))\\I_r_m_s\approx 0.919 amp`