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PHXII07:ALTERNATING CURRENT

356050 The peak value of an alternating $e . m . f .$ $E$ is given by $E = E_{0} \cos ω t$ is $10 v o l t s$ and its frequency is $50 H z$ . At time $t = \frac{1}{600} s$ , the instantaneous emf is

1

10 V

2

5 \sqrt{3} V

3

5 V

4

1 V

Explanation:

$E = E_{o} \cos ω t$
$E_{o} = 10 V$
$f = 50 H_{Z}$
$\Rightarrow ω = 2 π \times 50 = 100 π$
$t = \frac{1}{600} s$
The instantaneous value of emf
$= 10 \cos (100 π \times \frac{1}{600})$
$= 10 \cos (\frac{π}{6}) = 5 \sqrt{3} V$

PHXII07:ALTERNATING CURRENT

356051 An $a c$ voltage is represented by $E = 220 \sqrt{2} \cos (50 π) t$ How many times will the current becomes zero in 1 $s$ ?

1

100 t i m e s

2

50 t i m e s

3

25 t i m e s

4

30 t i m e s

Explanation:

$E = E_{o} \cos ω t$
$∴ ω = 50 π$
$2 π f = 50 π \Rightarrow f = 25 H z$
In one cycle ac current becomes zero twice therefore, 50 times the current becomes zero in $1 s$ .

PHXII07:ALTERNATING CURRENT

356052 An $a c$ generator produced an output voltage $E = 170 \sin (377 t)$ volts, where $t$ is in seconds. The frequency of $a c$ voltage is

1

50 H z

2

110 H z

3

60 H z

4

230 H z

Explanation:

$2 π v = 377 \Rightarrow v = 60.03 H z$

PHXII07:ALTERNATING CURRENT

356053 The peak voltage of $440 V o l t$ $A C$ mains is equal to (in $V o l t$ )

1 155.6

2 220.0

3 622.0

4 440.0

Explanation:

$V_{0} = \sqrt{2} \cdot V_{r.m.s} = \sqrt{2} \times 440 = 622$ $V o l t$
So correct option is (3).

PHXII07:ALTERNATING CURRENT

356050 The peak value of an alternating $e . m . f .$ $E$ is given by $E = E_{0} \cos ω t$ is $10 v o l t s$ and its frequency is $50 H z$ . At time $t = \frac{1}{600} s$ , the instantaneous emf is

1

10 V

2

5 \sqrt{3} V

3

5 V

4

1 V

Explanation:

$E = E_{o} \cos ω t$
$E_{o} = 10 V$
$f = 50 H_{Z}$
$\Rightarrow ω = 2 π \times 50 = 100 π$
$t = \frac{1}{600} s$
The instantaneous value of emf
$= 10 \cos (100 π \times \frac{1}{600})$
$= 10 \cos (\frac{π}{6}) = 5 \sqrt{3} V$

PHXII07:ALTERNATING CURRENT

356051 An $a c$ voltage is represented by $E = 220 \sqrt{2} \cos (50 π) t$ How many times will the current becomes zero in 1 $s$ ?

1

100 t i m e s

2

50 t i m e s

3

25 t i m e s

4

30 t i m e s

Explanation:

$E = E_{o} \cos ω t$
$∴ ω = 50 π$
$2 π f = 50 π \Rightarrow f = 25 H z$
In one cycle ac current becomes zero twice therefore, 50 times the current becomes zero in $1 s$ .

PHXII07:ALTERNATING CURRENT

356052 An $a c$ generator produced an output voltage $E = 170 \sin (377 t)$ volts, where $t$ is in seconds. The frequency of $a c$ voltage is

1

50 H z

2

110 H z

3

60 H z

4

230 H z

Explanation:

$2 π v = 377 \Rightarrow v = 60.03 H z$

PHXII07:ALTERNATING CURRENT

356053 The peak voltage of $440 V o l t$ $A C$ mains is equal to (in $V o l t$ )

1 155.6

2 220.0

3 622.0

4 440.0

Explanation:

$V_{0} = \sqrt{2} \cdot V_{r.m.s} = \sqrt{2} \times 440 = 622$ $V o l t$
So correct option is (3).

PHXII07:ALTERNATING CURRENT

356050 The peak value of an alternating $e . m . f .$ $E$ is given by $E = E_{0} \cos ω t$ is $10 v o l t s$ and its frequency is $50 H z$ . At time $t = \frac{1}{600} s$ , the instantaneous emf is

1

10 V

2

5 \sqrt{3} V

3

5 V

4

1 V

Explanation:

$E = E_{o} \cos ω t$
$E_{o} = 10 V$
$f = 50 H_{Z}$
$\Rightarrow ω = 2 π \times 50 = 100 π$
$t = \frac{1}{600} s$
The instantaneous value of emf
$= 10 \cos (100 π \times \frac{1}{600})$
$= 10 \cos (\frac{π}{6}) = 5 \sqrt{3} V$

PHXII07:ALTERNATING CURRENT

356051 An $a c$ voltage is represented by $E = 220 \sqrt{2} \cos (50 π) t$ How many times will the current becomes zero in 1 $s$ ?

1

100 t i m e s

2

50 t i m e s

3

25 t i m e s

4

30 t i m e s

Explanation:

$E = E_{o} \cos ω t$
$∴ ω = 50 π$
$2 π f = 50 π \Rightarrow f = 25 H z$
In one cycle ac current becomes zero twice therefore, 50 times the current becomes zero in $1 s$ .

PHXII07:ALTERNATING CURRENT

356052 An $a c$ generator produced an output voltage $E = 170 \sin (377 t)$ volts, where $t$ is in seconds. The frequency of $a c$ voltage is

1

50 H z

2

110 H z

3

60 H z

4

230 H z

Explanation:

$2 π v = 377 \Rightarrow v = 60.03 H z$

PHXII07:ALTERNATING CURRENT

356053 The peak voltage of $440 V o l t$ $A C$ mains is equal to (in $V o l t$ )

1 155.6

2 220.0

3 622.0

4 440.0

Explanation:

$V_{0} = \sqrt{2} \cdot V_{r.m.s} = \sqrt{2} \times 440 = 622$ $V o l t$
So correct option is (3).

PHXII07:ALTERNATING CURRENT

356050 The peak value of an alternating $e . m . f .$ $E$ is given by $E = E_{0} \cos ω t$ is $10 v o l t s$ and its frequency is $50 H z$ . At time $t = \frac{1}{600} s$ , the instantaneous emf is

1

10 V

2

5 \sqrt{3} V

3

5 V

4

1 V

Explanation:

$E = E_{o} \cos ω t$
$E_{o} = 10 V$
$f = 50 H_{Z}$
$\Rightarrow ω = 2 π \times 50 = 100 π$
$t = \frac{1}{600} s$
The instantaneous value of emf
$= 10 \cos (100 π \times \frac{1}{600})$
$= 10 \cos (\frac{π}{6}) = 5 \sqrt{3} V$

PHXII07:ALTERNATING CURRENT

356051 An $a c$ voltage is represented by $E = 220 \sqrt{2} \cos (50 π) t$ How many times will the current becomes zero in 1 $s$ ?

1

100 t i m e s

2

50 t i m e s

3

25 t i m e s

4

30 t i m e s

Explanation:

$E = E_{o} \cos ω t$
$∴ ω = 50 π$
$2 π f = 50 π \Rightarrow f = 25 H z$
In one cycle ac current becomes zero twice therefore, 50 times the current becomes zero in $1 s$ .

PHXII07:ALTERNATING CURRENT

356052 An $a c$ generator produced an output voltage $E = 170 \sin (377 t)$ volts, where $t$ is in seconds. The frequency of $a c$ voltage is

1

50 H z

2

110 H z

3

60 H z

4

230 H z

Explanation:

$2 π v = 377 \Rightarrow v = 60.03 H z$

PHXII07:ALTERNATING CURRENT

356053 The peak voltage of $440 V o l t$ $A C$ mains is equal to (in $V o l t$ )

1 155.6

2 220.0

3 622.0

4 440.0

Explanation:

$V_{0} = \sqrt{2} \cdot V_{r.m.s} = \sqrt{2} \times 440 = 622$ $V o l t$
So correct option is (3).

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