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PHXII07:ALTERNATING CURRENT

356054 The average current of a sinusoidally varying alternating current of peak value $5 A$ with initial phase zero, between the instants $t = T / 8$ to $t = T / 4$ is (where ' $T$ ' is time period)

1

\frac{10}{π} \sqrt{2} A

2

\frac{5}{π} \sqrt{2} A

3

\frac{20 \sqrt{2}}{π} A

4

\frac{10}{π} A

Explanation:

$= \frac{\int_{T / 8}^{T / 4} i d t}{\int_{T / 8}^{T / 4} d t}$
Then, $i = 5 \sin ω t$ .
So, correct option is (1).

PHXII07:ALTERNATING CURRENT

356055 Alternating current in circuit is given by $I = I_{0} \sin 2 π n t$ . Then the time taken by the current to rise from zero to r.m.s. value is equal to

1

1 / 2 n

2

1 / n

3

1 / 4 n

4

1 / 8 n

Explanation:

$I = I_{0} \sin (2 π n t)$
$\frac{I_{0}}{\sqrt{2}} = I_{0} \sin (2 π n t)$
$\Rightarrow \sin (2 π n t) = \frac{1}{\sqrt{2}} = \sin \frac{π}{4}$
$\Rightarrow 2 π n t = \frac{π}{4}$
$\Rightarrow t = \frac{1}{8 n}$
So, correct option is (4).

PHXII07:ALTERNATING CURRENT

356056 An alternating voltage is given by $E = 100 \sin (ω t + \frac{π}{6}) V$ .The voltage will be maximum for the first time when $t$ is [ $T =$ periodic time]

1

\frac{T}{12}

2

\frac{T}{2}

3

\frac{T}{6}

4

\frac{T}{3}

Explanation:

Given, alternating voltage is
$E = 100 \sin (ω t + \frac{π}{6}) V$
The voltage will be maximum at time $t$ , when
$\sin (ω t + \frac{π}{6}) = 1$
$\Rightarrow ω t + \frac{π}{6} = \sin^{- 1} (1) = \frac{π}{2}$
$\Rightarrow ω t = \frac{π}{3}$
$\Rightarrow t = \frac{π}{3 ω} = \frac{π \times T}{3 \times 2 π} [∵ ω = \frac{2 π}{T}]$
$\Rightarrow t = \frac{T}{6}$

MHTCET - 2019

PHXII07:ALTERNATING CURRENT

356057 An $A C$ source is $240 - V - 60 H z$ . The value of voltage after $\frac{1}{720} s$ starting from zero will be:

1

20.2 V

2

42.4 V

3

169.7 V

4

106.8 V

Explanation:

$V = V_{0} \sin ω t \Rightarrow V = V_{r m s} \sqrt{2} \sin ω t$
After, $t = \frac{1}{720} s,$
$V = 240 \sqrt{2} \sin 2 π v t = 240 \sqrt{2} \sin 2 π \times 60 \times \frac{1}{720}$
$= 120 \sqrt{2} = 169.7 V$

PHXII07:ALTERNATING CURRENT

356058 In an $a . c$ circuit, peak value of voltage is $423 v o l t s,$ its effective voltage is:

1

400 V

2

323 V

3

300 V

4

340 V

Explanation:

We know,
The $r . m . s .$ or effective voltage is $V_{m s} = \frac{V_{0}}{\sqrt{2}}$
$\Rightarrow V_{r m s} = \frac{423}{\sqrt{2}} = 300 V$
So, correct option is (3).

PHXII07:ALTERNATING CURRENT

356054 The average current of a sinusoidally varying alternating current of peak value $5 A$ with initial phase zero, between the instants $t = T / 8$ to $t = T / 4$ is (where ' $T$ ' is time period)

1

\frac{10}{π} \sqrt{2} A

2

\frac{5}{π} \sqrt{2} A

3

\frac{20 \sqrt{2}}{π} A

4

\frac{10}{π} A

Explanation:

$= \frac{\int_{T / 8}^{T / 4} i d t}{\int_{T / 8}^{T / 4} d t}$
Then, $i = 5 \sin ω t$ .
So, correct option is (1).

PHXII07:ALTERNATING CURRENT

356055 Alternating current in circuit is given by $I = I_{0} \sin 2 π n t$ . Then the time taken by the current to rise from zero to r.m.s. value is equal to

1

1 / 2 n

2

1 / n

3

1 / 4 n

4

1 / 8 n

Explanation:

$I = I_{0} \sin (2 π n t)$
$\frac{I_{0}}{\sqrt{2}} = I_{0} \sin (2 π n t)$
$\Rightarrow \sin (2 π n t) = \frac{1}{\sqrt{2}} = \sin \frac{π}{4}$
$\Rightarrow 2 π n t = \frac{π}{4}$
$\Rightarrow t = \frac{1}{8 n}$
So, correct option is (4).

PHXII07:ALTERNATING CURRENT

356056 An alternating voltage is given by $E = 100 \sin (ω t + \frac{π}{6}) V$ .The voltage will be maximum for the first time when $t$ is [ $T =$ periodic time]

1

\frac{T}{12}

2

\frac{T}{2}

3

\frac{T}{6}

4

\frac{T}{3}

Explanation:

Given, alternating voltage is
$E = 100 \sin (ω t + \frac{π}{6}) V$
The voltage will be maximum at time $t$ , when
$\sin (ω t + \frac{π}{6}) = 1$
$\Rightarrow ω t + \frac{π}{6} = \sin^{- 1} (1) = \frac{π}{2}$
$\Rightarrow ω t = \frac{π}{3}$
$\Rightarrow t = \frac{π}{3 ω} = \frac{π \times T}{3 \times 2 π} [∵ ω = \frac{2 π}{T}]$
$\Rightarrow t = \frac{T}{6}$

MHTCET - 2019

PHXII07:ALTERNATING CURRENT

356057 An $A C$ source is $240 - V - 60 H z$ . The value of voltage after $\frac{1}{720} s$ starting from zero will be:

1

20.2 V

2

42.4 V

3

169.7 V

4

106.8 V

Explanation:

$V = V_{0} \sin ω t \Rightarrow V = V_{r m s} \sqrt{2} \sin ω t$
After, $t = \frac{1}{720} s,$
$V = 240 \sqrt{2} \sin 2 π v t = 240 \sqrt{2} \sin 2 π \times 60 \times \frac{1}{720}$
$= 120 \sqrt{2} = 169.7 V$

PHXII07:ALTERNATING CURRENT

356058 In an $a . c$ circuit, peak value of voltage is $423 v o l t s,$ its effective voltage is:

1

400 V

2

323 V

3

300 V

4

340 V

Explanation:

We know,
The $r . m . s .$ or effective voltage is $V_{m s} = \frac{V_{0}}{\sqrt{2}}$
$\Rightarrow V_{r m s} = \frac{423}{\sqrt{2}} = 300 V$
So, correct option is (3).

PHXII07:ALTERNATING CURRENT

356054 The average current of a sinusoidally varying alternating current of peak value $5 A$ with initial phase zero, between the instants $t = T / 8$ to $t = T / 4$ is (where ' $T$ ' is time period)

1

\frac{10}{π} \sqrt{2} A

2

\frac{5}{π} \sqrt{2} A

3

\frac{20 \sqrt{2}}{π} A

4

\frac{10}{π} A

Explanation:

$= \frac{\int_{T / 8}^{T / 4} i d t}{\int_{T / 8}^{T / 4} d t}$
Then, $i = 5 \sin ω t$ .
So, correct option is (1).

PHXII07:ALTERNATING CURRENT

356055 Alternating current in circuit is given by $I = I_{0} \sin 2 π n t$ . Then the time taken by the current to rise from zero to r.m.s. value is equal to

1

1 / 2 n

2

1 / n

3

1 / 4 n

4

1 / 8 n

Explanation:

$I = I_{0} \sin (2 π n t)$
$\frac{I_{0}}{\sqrt{2}} = I_{0} \sin (2 π n t)$
$\Rightarrow \sin (2 π n t) = \frac{1}{\sqrt{2}} = \sin \frac{π}{4}$
$\Rightarrow 2 π n t = \frac{π}{4}$
$\Rightarrow t = \frac{1}{8 n}$
So, correct option is (4).

PHXII07:ALTERNATING CURRENT

356056 An alternating voltage is given by $E = 100 \sin (ω t + \frac{π}{6}) V$ .The voltage will be maximum for the first time when $t$ is [ $T =$ periodic time]

1

\frac{T}{12}

2

\frac{T}{2}

3

\frac{T}{6}

4

\frac{T}{3}

Explanation:

Given, alternating voltage is
$E = 100 \sin (ω t + \frac{π}{6}) V$
The voltage will be maximum at time $t$ , when
$\sin (ω t + \frac{π}{6}) = 1$
$\Rightarrow ω t + \frac{π}{6} = \sin^{- 1} (1) = \frac{π}{2}$
$\Rightarrow ω t = \frac{π}{3}$
$\Rightarrow t = \frac{π}{3 ω} = \frac{π \times T}{3 \times 2 π} [∵ ω = \frac{2 π}{T}]$
$\Rightarrow t = \frac{T}{6}$

MHTCET - 2019

PHXII07:ALTERNATING CURRENT

356057 An $A C$ source is $240 - V - 60 H z$ . The value of voltage after $\frac{1}{720} s$ starting from zero will be:

1

20.2 V

2

42.4 V

3

169.7 V

4

106.8 V

Explanation:

$V = V_{0} \sin ω t \Rightarrow V = V_{r m s} \sqrt{2} \sin ω t$
After, $t = \frac{1}{720} s,$
$V = 240 \sqrt{2} \sin 2 π v t = 240 \sqrt{2} \sin 2 π \times 60 \times \frac{1}{720}$
$= 120 \sqrt{2} = 169.7 V$

PHXII07:ALTERNATING CURRENT

356058 In an $a . c$ circuit, peak value of voltage is $423 v o l t s,$ its effective voltage is:

1

400 V

2

323 V

3

300 V

4

340 V

Explanation:

We know,
The $r . m . s .$ or effective voltage is $V_{m s} = \frac{V_{0}}{\sqrt{2}}$
$\Rightarrow V_{r m s} = \frac{423}{\sqrt{2}} = 300 V$
So, correct option is (3).

PHXII07:ALTERNATING CURRENT

356054 The average current of a sinusoidally varying alternating current of peak value $5 A$ with initial phase zero, between the instants $t = T / 8$ to $t = T / 4$ is (where ' $T$ ' is time period)

1

\frac{10}{π} \sqrt{2} A

2

\frac{5}{π} \sqrt{2} A

3

\frac{20 \sqrt{2}}{π} A

4

\frac{10}{π} A

Explanation:

$= \frac{\int_{T / 8}^{T / 4} i d t}{\int_{T / 8}^{T / 4} d t}$
Then, $i = 5 \sin ω t$ .
So, correct option is (1).

PHXII07:ALTERNATING CURRENT

356055 Alternating current in circuit is given by $I = I_{0} \sin 2 π n t$ . Then the time taken by the current to rise from zero to r.m.s. value is equal to

1

1 / 2 n

2

1 / n

3

1 / 4 n

4

1 / 8 n

Explanation:

$I = I_{0} \sin (2 π n t)$
$\frac{I_{0}}{\sqrt{2}} = I_{0} \sin (2 π n t)$
$\Rightarrow \sin (2 π n t) = \frac{1}{\sqrt{2}} = \sin \frac{π}{4}$
$\Rightarrow 2 π n t = \frac{π}{4}$
$\Rightarrow t = \frac{1}{8 n}$
So, correct option is (4).

PHXII07:ALTERNATING CURRENT

356056 An alternating voltage is given by $E = 100 \sin (ω t + \frac{π}{6}) V$ .The voltage will be maximum for the first time when $t$ is [ $T =$ periodic time]

1

\frac{T}{12}

2

\frac{T}{2}

3

\frac{T}{6}

4

\frac{T}{3}

Explanation:

Given, alternating voltage is
$E = 100 \sin (ω t + \frac{π}{6}) V$
The voltage will be maximum at time $t$ , when
$\sin (ω t + \frac{π}{6}) = 1$
$\Rightarrow ω t + \frac{π}{6} = \sin^{- 1} (1) = \frac{π}{2}$
$\Rightarrow ω t = \frac{π}{3}$
$\Rightarrow t = \frac{π}{3 ω} = \frac{π \times T}{3 \times 2 π} [∵ ω = \frac{2 π}{T}]$
$\Rightarrow t = \frac{T}{6}$

MHTCET - 2019

PHXII07:ALTERNATING CURRENT

356057 An $A C$ source is $240 - V - 60 H z$ . The value of voltage after $\frac{1}{720} s$ starting from zero will be:

1

20.2 V

2

42.4 V

3

169.7 V

4

106.8 V

Explanation:

$V = V_{0} \sin ω t \Rightarrow V = V_{r m s} \sqrt{2} \sin ω t$
After, $t = \frac{1}{720} s,$
$V = 240 \sqrt{2} \sin 2 π v t = 240 \sqrt{2} \sin 2 π \times 60 \times \frac{1}{720}$
$= 120 \sqrt{2} = 169.7 V$

PHXII07:ALTERNATING CURRENT

356058 In an $a . c$ circuit, peak value of voltage is $423 v o l t s,$ its effective voltage is:

1

400 V

2

323 V

3

300 V

4

340 V

Explanation:

We know,
The $r . m . s .$ or effective voltage is $V_{m s} = \frac{V_{0}}{\sqrt{2}}$
$\Rightarrow V_{r m s} = \frac{423}{\sqrt{2}} = 300 V$
So, correct option is (3).

PHXII07:ALTERNATING CURRENT

356054 The average current of a sinusoidally varying alternating current of peak value $5 A$ with initial phase zero, between the instants $t = T / 8$ to $t = T / 4$ is (where ' $T$ ' is time period)

1

\frac{10}{π} \sqrt{2} A

2

\frac{5}{π} \sqrt{2} A

3

\frac{20 \sqrt{2}}{π} A

4

\frac{10}{π} A

Explanation:

$= \frac{\int_{T / 8}^{T / 4} i d t}{\int_{T / 8}^{T / 4} d t}$
Then, $i = 5 \sin ω t$ .
So, correct option is (1).

PHXII07:ALTERNATING CURRENT

356055 Alternating current in circuit is given by $I = I_{0} \sin 2 π n t$ . Then the time taken by the current to rise from zero to r.m.s. value is equal to

1

1 / 2 n

2

1 / n

3

1 / 4 n

4

1 / 8 n

Explanation:

$I = I_{0} \sin (2 π n t)$
$\frac{I_{0}}{\sqrt{2}} = I_{0} \sin (2 π n t)$
$\Rightarrow \sin (2 π n t) = \frac{1}{\sqrt{2}} = \sin \frac{π}{4}$
$\Rightarrow 2 π n t = \frac{π}{4}$
$\Rightarrow t = \frac{1}{8 n}$
So, correct option is (4).

PHXII07:ALTERNATING CURRENT

356056 An alternating voltage is given by $E = 100 \sin (ω t + \frac{π}{6}) V$ .The voltage will be maximum for the first time when $t$ is [ $T =$ periodic time]

1

\frac{T}{12}

2

\frac{T}{2}

3

\frac{T}{6}

4

\frac{T}{3}

Explanation:

Given, alternating voltage is
$E = 100 \sin (ω t + \frac{π}{6}) V$
The voltage will be maximum at time $t$ , when
$\sin (ω t + \frac{π}{6}) = 1$
$\Rightarrow ω t + \frac{π}{6} = \sin^{- 1} (1) = \frac{π}{2}$
$\Rightarrow ω t = \frac{π}{3}$
$\Rightarrow t = \frac{π}{3 ω} = \frac{π \times T}{3 \times 2 π} [∵ ω = \frac{2 π}{T}]$
$\Rightarrow t = \frac{T}{6}$

MHTCET - 2019

PHXII07:ALTERNATING CURRENT

356057 An $A C$ source is $240 - V - 60 H z$ . The value of voltage after $\frac{1}{720} s$ starting from zero will be:

1

20.2 V

2

42.4 V

3

169.7 V

4

106.8 V

Explanation:

$V = V_{0} \sin ω t \Rightarrow V = V_{r m s} \sqrt{2} \sin ω t$
After, $t = \frac{1}{720} s,$
$V = 240 \sqrt{2} \sin 2 π v t = 240 \sqrt{2} \sin 2 π \times 60 \times \frac{1}{720}$
$= 120 \sqrt{2} = 169.7 V$

PHXII07:ALTERNATING CURRENT

356058 In an $a . c$ circuit, peak value of voltage is $423 v o l t s,$ its effective voltage is:

1

400 V

2

323 V

3

300 V

4

340 V

Explanation:

We know,
The $r . m . s .$ or effective voltage is $V_{m s} = \frac{V_{0}}{\sqrt{2}}$
$\Rightarrow V_{r m s} = \frac{423}{\sqrt{2}} = 300 V$
So, correct option is (3).

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