The peak value of an alternating e-m-f- E is given by E - E0 cos - t is 10 volts and its frequency is 50 Hz- At time t - 1-600sec- the instantaneous e-m-f- is

E = E_0 cos ω t = E_0 cos 2 π t/T = 10 cos 2 π× 50 × 1/600 = 10 cos π/6π/6 = 5 √(3) volt. ...

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Standard XII

Physics

Sinusoidal AC

The peak valu...

Question

The peak value of an alternating e.m.f. E is given by $E = E_{0} c o s ω t$ is 10 volts and its frequency is 50 Hz. At time $t = \frac{1}{600} s e c,$ the instantaneous e.m.f. is

10 V

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5 \sqrt{3} V

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5 V

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1 V

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Solution

The correct option is B $5 \sqrt{3} V$
$E = E_{0} c o s ω t = E_{0} c o s \frac{2 π t}{T} = 10 c o s \frac{2 π \times 50 \times 1}{600} = 10 c o s \frac{π}{6} \frac{π}{6} = 5 \sqrt{3} v o l t .$

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