Question

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 amperes
• D. About four amperes

Answer 1

Answer: (A)

$i_{ms}=\frac{200}{280}=\frac{5}{7}A.$
So $i_0=i_{rms}^{\prime }\times \sqrt{2}$
$=\frac{5}{7}\times \sqrt{2}\approx 1A$