A voltage phasor of length 200 V rotates anticlockwise with an angular speed of \(100\pi\ \mathrm{rad\,s^{-1}}\). At \(t=0\), the phasor lies along the positive horizontal axis and the voltage is zero. The instantaneous voltage at \(t=\frac{1}{600}\ \mathrm{s}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 100 V
The length of a phasor is the peak value \( v_m \). The value at any instant is the projection of the phasor on the vertical axis. The phasor starts on the horizontal axis, where the voltage is zero. So after a time \( t \) it has turned through \( \omega t \) and \( v=v_m\sin\omega t \).
\( \begin{aligned} \omega t &= 100\pi\times\frac{1}{600} \\ &= \frac{\pi}{6} \end{aligned} \)
\( \begin{aligned} v &= v_m\sin\omega t \\ &= 200\times\sin30^\circ \\ &= 100\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Representation of AC Current and Voltage by Rotating Vectors — Phasors – “… The projection of voltage and current phasors on vertical axis, i.e., vm sinωt and im sinωt, respectively represent the value of voltage and current at that instant. …”