Question

If instantaneous current in a circuit is given by$l=(3+4sin\omega t)A$ os then the maximum value of current in the circuit is

• A. 4 A
• B. 5 A
• C. 7 A
• D. 3 A

Answer 1

The correct option is C 7 A

$i=A(3+4sinwt)A$

maximum when the term

$i=3+4sinwt$ is maximum.

so it maximum for

$wt$= $\pi /2$

so $i=3+4sin\pi /2$ $=\sqrt{3^2+4^2}=5Amp$

$i=7Ampere$