Question

A bulb of resistance \(280\,\Omega\) is supplied with a voltage \(V = 400\sin \pi t\). The peak current is

• A. \(2.22\,A\)
• B. \(2.02\,A\)
• C. \(1.11\,A\)
• D. \(1.43\,A\)

Hint:

Peak current in AC circuit is found using \(I_0 = \frac{V_0}{R}\) when only resistance is present.

Answer 1

The Correct Option is D

Step 1: Identify peak voltage.
Given voltage is:
\[ V = 400 \sin \pi t \] So peak voltage \(V_0 = 400\,V\).

Step 2: Use Ohm’s law for AC peak values.
\[ I_0 = \frac{V_0}{R} \]

Step 3: Substitute values.
\[ I_0 = \frac{400}{280} \]

Step 4: Calculate current.
\[ I_0 = 1.43\,A \]

Step 5: Final conclusion.
\[ \boxed{1.43\,A} \] Hence, correct answer is option (D).