QBManager

AC (NCERT)

274682 A transformer is employed to

1 convert A.C. into D.C.

2 convert D.C. into A.C.

3 obtain a suitable A.C. voltage

4 obtain a suitable D.C. voltage

NCERT Page-260 / N-194

AC (NCERT)

274692 A transformer has an efficiency of $80$ . It works at $4 kW$ and $100 V$ . If secondary voltage is $240 V$ , the current in primary coil is

1

0.4 A

2

4 A

3

10 A

4

40 A

Explanation:

(d) As $E_{p} I_{p} = P_{i} ∴ I_{p} = \frac{P_{i}}{E_{p}} = \frac{4000}{100} = 40 A$ .

NCERT Page-261 / N-195

AC (NCERT)

274693 The primary of a transformer has 400 turns while the secondary has 2000 turns. If the power output from the secondary at $1000 V$ is $12 kW$ , what is the primary voltage?

1

200 V

2

300 V

3

400 V

4

500 V

Explanation:

(a) $N_{P} = 400, N_{s} = 2000$ and $V_{s} = 1000 V$ .
$\frac{V_{P}}{V_{S}} = \frac{N_{P}}{N_{S}}$ of, $V_{P} = \frac{V_{S} \times N_{P}}{N_{S}} = \frac{1000 \times 400}{2000} = 200 V$ .

NCERT Page-261 / N-195

AC (NCERT)

274689 A transformer is used to light a $140 W, 24 V$ bulb from a $240 V$ a.c. mains. The current in the main cable is $0.7 A$ . The efficiency of the transformer is

1

63.8

2

83.3

3

16.7

4

36.2

Explanation:

(b) Power of source $= EI = 240 \times 0.7 = 166$
$\Rightarrow$ Efficiency $= \frac{140}{166} \Rightarrow η = 83.3$

NCERT Page-261 / N-195

AC (NCERT)

274690 Eddy currents in the core of transformer can't be developed by

1 increasing the number of turns in secondary coil

2 taking laminated transformer

3 making step down transformer

4 using a weak a.c. at high potential

NCERT Page-261 / N-195

AC (NCERT)

274682 A transformer is employed to

1 convert A.C. into D.C.

2 convert D.C. into A.C.

3 obtain a suitable A.C. voltage

4 obtain a suitable D.C. voltage

NCERT Page-260 / N-194

AC (NCERT)

274692 A transformer has an efficiency of $80$ . It works at $4 kW$ and $100 V$ . If secondary voltage is $240 V$ , the current in primary coil is

1

0.4 A

2

4 A

3

10 A

4

40 A

Explanation:

(d) As $E_{p} I_{p} = P_{i} ∴ I_{p} = \frac{P_{i}}{E_{p}} = \frac{4000}{100} = 40 A$ .

NCERT Page-261 / N-195

AC (NCERT)

274693 The primary of a transformer has 400 turns while the secondary has 2000 turns. If the power output from the secondary at $1000 V$ is $12 kW$ , what is the primary voltage?

1

200 V

2

300 V

3

400 V

4

500 V

Explanation:

(a) $N_{P} = 400, N_{s} = 2000$ and $V_{s} = 1000 V$ .
$\frac{V_{P}}{V_{S}} = \frac{N_{P}}{N_{S}}$ of, $V_{P} = \frac{V_{S} \times N_{P}}{N_{S}} = \frac{1000 \times 400}{2000} = 200 V$ .

NCERT Page-261 / N-195

AC (NCERT)

274689 A transformer is used to light a $140 W, 24 V$ bulb from a $240 V$ a.c. mains. The current in the main cable is $0.7 A$ . The efficiency of the transformer is

1

63.8

2

83.3

3

16.7

4

36.2

Explanation:

(b) Power of source $= EI = 240 \times 0.7 = 166$
$\Rightarrow$ Efficiency $= \frac{140}{166} \Rightarrow η = 83.3$

NCERT Page-261 / N-195

AC (NCERT)

274690 Eddy currents in the core of transformer can't be developed by

1 increasing the number of turns in secondary coil

2 taking laminated transformer

3 making step down transformer

4 using a weak a.c. at high potential

NCERT Page-261 / N-195

AC (NCERT)

274682 A transformer is employed to

1 convert A.C. into D.C.

2 convert D.C. into A.C.

3 obtain a suitable A.C. voltage

4 obtain a suitable D.C. voltage

NCERT Page-260 / N-194

AC (NCERT)

274692 A transformer has an efficiency of $80$ . It works at $4 kW$ and $100 V$ . If secondary voltage is $240 V$ , the current in primary coil is

1

0.4 A

2

4 A

3

10 A

4

40 A

Explanation:

(d) As $E_{p} I_{p} = P_{i} ∴ I_{p} = \frac{P_{i}}{E_{p}} = \frac{4000}{100} = 40 A$ .

NCERT Page-261 / N-195

AC (NCERT)

274693 The primary of a transformer has 400 turns while the secondary has 2000 turns. If the power output from the secondary at $1000 V$ is $12 kW$ , what is the primary voltage?

1

200 V

2

300 V

3

400 V

4

500 V

Explanation:

(a) $N_{P} = 400, N_{s} = 2000$ and $V_{s} = 1000 V$ .
$\frac{V_{P}}{V_{S}} = \frac{N_{P}}{N_{S}}$ of, $V_{P} = \frac{V_{S} \times N_{P}}{N_{S}} = \frac{1000 \times 400}{2000} = 200 V$ .

NCERT Page-261 / N-195

AC (NCERT)

274689 A transformer is used to light a $140 W, 24 V$ bulb from a $240 V$ a.c. mains. The current in the main cable is $0.7 A$ . The efficiency of the transformer is

1

63.8

2

83.3

3

16.7

4

36.2

Explanation:

(b) Power of source $= EI = 240 \times 0.7 = 166$
$\Rightarrow$ Efficiency $= \frac{140}{166} \Rightarrow η = 83.3$

NCERT Page-261 / N-195

AC (NCERT)

274690 Eddy currents in the core of transformer can't be developed by

1 increasing the number of turns in secondary coil

2 taking laminated transformer

3 making step down transformer

4 using a weak a.c. at high potential

NCERT Page-261 / N-195

AC (NCERT)

274682 A transformer is employed to

1 convert A.C. into D.C.

2 convert D.C. into A.C.

3 obtain a suitable A.C. voltage

4 obtain a suitable D.C. voltage

NCERT Page-260 / N-194

AC (NCERT)

274692 A transformer has an efficiency of $80$ . It works at $4 kW$ and $100 V$ . If secondary voltage is $240 V$ , the current in primary coil is

1

0.4 A

2

4 A

3

10 A

4

40 A

Explanation:

(d) As $E_{p} I_{p} = P_{i} ∴ I_{p} = \frac{P_{i}}{E_{p}} = \frac{4000}{100} = 40 A$ .

NCERT Page-261 / N-195

AC (NCERT)

274693 The primary of a transformer has 400 turns while the secondary has 2000 turns. If the power output from the secondary at $1000 V$ is $12 kW$ , what is the primary voltage?

1

200 V

2

300 V

3

400 V

4

500 V

Explanation:

(a) $N_{P} = 400, N_{s} = 2000$ and $V_{s} = 1000 V$ .
$\frac{V_{P}}{V_{S}} = \frac{N_{P}}{N_{S}}$ of, $V_{P} = \frac{V_{S} \times N_{P}}{N_{S}} = \frac{1000 \times 400}{2000} = 200 V$ .

NCERT Page-261 / N-195

AC (NCERT)

274689 A transformer is used to light a $140 W, 24 V$ bulb from a $240 V$ a.c. mains. The current in the main cable is $0.7 A$ . The efficiency of the transformer is

1

63.8

2

83.3

3

16.7

4

36.2

Explanation:

(b) Power of source $= EI = 240 \times 0.7 = 166$
$\Rightarrow$ Efficiency $= \frac{140}{166} \Rightarrow η = 83.3$

NCERT Page-261 / N-195

AC (NCERT)

274690 Eddy currents in the core of transformer can't be developed by

1 increasing the number of turns in secondary coil

2 taking laminated transformer

3 making step down transformer

4 using a weak a.c. at high potential

NCERT Page-261 / N-195

AC (NCERT)

274682 A transformer is employed to

1 convert A.C. into D.C.

2 convert D.C. into A.C.

3 obtain a suitable A.C. voltage

4 obtain a suitable D.C. voltage

NCERT Page-260 / N-194

AC (NCERT)

274692 A transformer has an efficiency of $80$ . It works at $4 kW$ and $100 V$ . If secondary voltage is $240 V$ , the current in primary coil is

1

0.4 A

2

4 A

3

10 A

4

40 A

Explanation:

(d) As $E_{p} I_{p} = P_{i} ∴ I_{p} = \frac{P_{i}}{E_{p}} = \frac{4000}{100} = 40 A$ .

NCERT Page-261 / N-195

AC (NCERT)

274693 The primary of a transformer has 400 turns while the secondary has 2000 turns. If the power output from the secondary at $1000 V$ is $12 kW$ , what is the primary voltage?

1

200 V

2

300 V

3

400 V

4

500 V

Explanation:

(a) $N_{P} = 400, N_{s} = 2000$ and $V_{s} = 1000 V$ .
$\frac{V_{P}}{V_{S}} = \frac{N_{P}}{N_{S}}$ of, $V_{P} = \frac{V_{S} \times N_{P}}{N_{S}} = \frac{1000 \times 400}{2000} = 200 V$ .

NCERT Page-261 / N-195

AC (NCERT)

274689 A transformer is used to light a $140 W, 24 V$ bulb from a $240 V$ a.c. mains. The current in the main cable is $0.7 A$ . The efficiency of the transformer is

1

63.8

2

83.3

3

16.7

4

36.2

Explanation:

(b) Power of source $= EI = 240 \times 0.7 = 166$
$\Rightarrow$ Efficiency $= \frac{140}{166} \Rightarrow η = 83.3$

NCERT Page-261 / N-195

AC (NCERT)

274690 Eddy currents in the core of transformer can't be developed by

1 increasing the number of turns in secondary coil

2 taking laminated transformer

3 making step down transformer

4 using a weak a.c. at high potential

NCERT Page-261 / N-195

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