For purely inductive components voltage leads current by 90° . Group of answer choicesTrueFalse

Answer

The statement "For purely inductive components voltage leads current by 90°" is True.

Explanation

In an ideal inductor, the relationship between voltage (V) and current (I) is defined by the following equation:

$V(t) = L \frac{di(t)}{dt}$

where $L$ is the inductance. This means that the voltage across an inductor is proportional to the rate of change of current through it.

When the current waveform is a sine wave, the voltage waveform is a cosine wave shifted by 90 degrees, indicating that the voltage leads the current. In terms of their phase relationship in the frequency domain, the current waveform can be represented as:

$I(t) = I_0 \sin(\omega t)$

The corresponding voltage waveform is given by:

$V(t) = L \frac{d}{dt}(I_0 \sin(\omega t)) = L I_0 \omega \cos(\omega t)$

This translates to a phase difference of +90°, confirming that the voltage leads the current in purely inductive components.

Final Answer

True