Question

What results from substituting \[v=v_m\sin\omega t\] into \[v=iR\]?

Options

• \[v_mi_m\sin^{2}\omega t=iR\]

• \[v_m\sin\omega t=\frac{i}{R}\]

• \[v_m\sin\omega t=iR\]

• \[i_m\sin\omega t=iR\]

Answer 1

Replace the instantaneous voltage \[v\] in \[v=iR\] with \[v_m\sin\omega t\]. The result is \[v_m\sin\omega t=iR\].