Question

What expression for current is obtained by integrating \[\frac{di}{dt}=\frac{v_m}{L}\sin\omega t\]?

Options

• \[i=\frac{v_m}{\omega L}\sin\omega t+C\]

• \[i=-\frac{v_m}{\omega L}\cos\omega t+C\]

• \[i=\frac{v_m}{\omega L}\cos\omega t+C\]

• \[i=-\frac{v_m\omega}{L}\cos\omega t+C\]

Answer 1

Integrating \[\sin\omega t\] gives \[-\frac{1}{\omega}\cos\omega t\]. The result is \[i=-\frac{v_m}{\omega L}\cos\omega t+C\].