An alternating voltage is given by E=100sin⁡ω+(π)/6V . The voltage will be maximum for the first time when is (T = periodic time)

Correct answer: (C) T/6. Given, alternating voltage is E=100sin⁡ωt+(π)/6V The voltage will be maximum, when sin⁡ωt+(π)/6=1 ⇒sin⁡ωt+(π)/6=sin(π)/2 ⇒ωt+(π)/6=(π)/2 ⇒ωt=(π)/3 ⇒t=(π)/(3ω)=(π×T)/(3×2π)∵ω=(2π)/T =T/6 Thus, the voltage will be maximum for the first time when t = T/6 .

Q146 / 150 · MHT CET 2019 (Shift 2) · Physics

An alternating voltage is given by $E = 100 \sin (ω + \frac{π}{6}) V$ . The voltage will be maximum for the first time when is (T = periodic time)

A
$\frac{T}{12}$
B
$\frac{T}{2}$
Correct answer
C
$\frac{T}{6}$
D
$\frac{T}{3}$

Solution

Given, alternating voltage is

E = 100 \sin (ω t + \frac{π}{6}) V

The voltage will be maximum, when

\sin (ω t + \frac{π}{6}) = 1

\Rightarrow \sin (ω t + \frac{π}{6}) = s i n \frac{π}{2}

\Rightarrow ω t + \frac{π}{6} = \frac{π}{2}

\Rightarrow ω t = \frac{π}{3}

\Rightarrow t = \frac{π}{3 ω} = \frac{π \times T}{3 \times 2 π} [∵ ω = \frac{2 π}{T}]

= \frac{T}{6}

Thus, the voltage will be maximum for the first time when t =

\frac{T}{6}

Hence, the correct answer is option: 'C'.

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An alternating voltage is given by E=100sin⁡ω+π6V . The voltage will be maximum for the first time when is (T = periodic time)

Solution

Relevant Questions

An alternating voltage is given by $E = 100 \sin (ω + \frac{π}{6}) V$ . The voltage will be maximum for the first time when is (T = periodic time)