Question

How is peak current \[i_m\] related to peak voltage \[v_m\] in a pure resistor AC circuit?

Options

• \[i_m=\frac{R}{v_m}\]

• \[i_m=\frac{v_m}{\sqrt{2}}\]

• \[i_m=v_mR\]

• \[i_m=\frac{v_m}{R}\]

Answer 1

The coefficient of \[\sin\omega t\] in the current equation is the peak current. Therefore, \[i_m=\frac{v_m}{R}\].