The peak value of an alternating emf ' $\mathrm{e}$ ' given by $\mathrm{e}=\mathrm{e}_0 \cos \omega…

The peak value of an alternating emf ' $\mathrm{e}$ ' given by $\mathrm{e}=\mathrm{e}_0 \cos \omega \mathrm{t}$ is 10 volt and its frequency is $50 \mathrm{~Hz}$. At time $\mathrm{t}=\frac{1}{600} \mathrm{~s}$, the instantaneous e.m.f is
  1. 10 V
  2. $\frac{10}{\sqrt{3}} \mathrm{~V}$
  3. 5 V
  4. $5 \sqrt{3} \mathrm{~V}$

Solution

$\begin{aligned} & \mathrm{e}=\mathrm{e}_0 \cos \omega \mathrm{t}=10 \cos 2 \pi \mathrm{ft} \\ & \mathrm{f}=50 \mathrm{~Hz}, \mathrm{t}=\frac{1}{600} \mathrm{~s} \\ & \therefore \mathrm{e}=10 \cos 100 \pi \times \frac{1}{600}=10 \cos \frac{\pi}{6}=5 \sqrt{3} \mathrm{~V}\end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 1)

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