Question

A current is made of two components, a DC component of 3A and AC component of $4\sqrt{2}sin\omega tA$. The reading of the hot wire ammeter is

• A. $3A$
• B. $4\sqrt{2}A$
• C. $(3+4\sqrt{2})A$
• D. $5A$

Answer 1

The correct option is D

$5A$

$l_{rms}=\sqrt{\frac{{\int }_0^T(9+4\sqrt{2}sin\omega t{)}^{2}dt}{T}}=\sqrt{\frac{{\int }_0^T(9+32si{n}^2\omega t+8\sqrt{2}sin\omega t)dt}{T}}=\sqrt{9+\frac{32}{2}}=5A$