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Alternating Current

155398 The number of turns in primary and secondary coils of a transformer is 50 and 200 respectively. If the current in the primary coil is $4 A$ , then the current in the secondary coil is

1

1 A

2

2 A

3

4 A

4

5 A

Explanation:

A Given that,
$N_{P} = 50, N_{S} = 200, I_{P} = 4 A, I_{S} =$ ?
We know that, in a transformer
$\frac{N_{P}}{N_{S}} = \frac{I_{S}}{I_{P}}$
$\frac{50}{200} = \frac{I_{S}}{4}$
$I_{S} = 1 A$

EAMCET-2018

Alternating Current

155399 A transformer has 100 turns in the primary coil and carries $8 A$ current. If input power is $1 kW$ , the number of turns in secondary coil to have $500 V$ of output will be

1 100

2 200

3 400

4 300

Explanation:

C Given that,
Primary number of turns $= 100$
Primary current $= 8 A$
Input power $= 1 kW$
$Power = V.I$
$1000 = 500 . I$
$I = 2 A$
Now, from the transformers equation -
$\frac{I_{1}}{I_{2}} = \frac{N_{2}}{N_{1}}$
$\frac{8}{2} = \frac{N_{2}}{100}$
$N_{2} = 400$

JCECE-2018

Alternating Current

155400 A transformer is used to glow a $140 W - 24 V$ bulb at $240 V$ AC. If current in the primary coil is $0.7 A$ , then the efficiency of the transformer is

1

16.7 %

2

50 %

3

83.3 %

4

25 %

Explanation:

C Given that,
Input voltage $= 240 V$
Primary coil current $= 0.7 A$
Output power $= 140 W$
Power input $= V \times I$
$P = 240 \times 0.7$
$P = 168.0 W$
$∴$ Efficiency of transformer,
$η % = \frac{P_{output}}{P_{input}} \times 100 = \frac{140}{168} \times 100$
$η = 83.33 %$

CG PET -2018

Alternating Current

155401 A distribution transformer with an efficiency of $90 %$ supplies to a colony of 10 homes. All the 10 homes have electrical oven running at the same time, that draw 20 A current from $220 V$ lines. The power dissipated as heat in the transformer is

1

12.2 kW

2

4.9 kW

3

8.4 kW

4

9.9 kW

Explanation:

B Given that,
Efficiency of transformer $(η) = 90 %$
$V = 220$ volt, $I = 20 A$
We know that,
$P_{1} = 10 VI = 10 \times 220 \times 20 = 44 \times 10^{3} W$
So, power delivered by transformer
$P_{2} = \frac{100}{90} \times P_{1} = \frac{10}{9} \times 44 \times 10^{3}$
$P_{2} = 48.9 \times 10^{3} W$
$∴$ Power dissipated as heat
$P_{H} = P_{2} - P_{1}$
$P_{H} = (48.9 - 44) \times 10^{3}$
$P_{H} = 4.9 \times 10^{3}$
$P_{H} = 4.9 kW$

TS- EAMCET-07.05.2018

Alternating Current

155398 The number of turns in primary and secondary coils of a transformer is 50 and 200 respectively. If the current in the primary coil is $4 A$ , then the current in the secondary coil is

1

1 A

2

2 A

3

4 A

4

5 A

Explanation:

A Given that,
$N_{P} = 50, N_{S} = 200, I_{P} = 4 A, I_{S} =$ ?
We know that, in a transformer
$\frac{N_{P}}{N_{S}} = \frac{I_{S}}{I_{P}}$
$\frac{50}{200} = \frac{I_{S}}{4}$
$I_{S} = 1 A$

EAMCET-2018

Alternating Current

155399 A transformer has 100 turns in the primary coil and carries $8 A$ current. If input power is $1 kW$ , the number of turns in secondary coil to have $500 V$ of output will be

1 100

2 200

3 400

4 300

Explanation:

C Given that,
Primary number of turns $= 100$
Primary current $= 8 A$
Input power $= 1 kW$
$Power = V.I$
$1000 = 500 . I$
$I = 2 A$
Now, from the transformers equation -
$\frac{I_{1}}{I_{2}} = \frac{N_{2}}{N_{1}}$
$\frac{8}{2} = \frac{N_{2}}{100}$
$N_{2} = 400$

JCECE-2018

Alternating Current

155400 A transformer is used to glow a $140 W - 24 V$ bulb at $240 V$ AC. If current in the primary coil is $0.7 A$ , then the efficiency of the transformer is

1

16.7 %

2

50 %

3

83.3 %

4

25 %

Explanation:

C Given that,
Input voltage $= 240 V$
Primary coil current $= 0.7 A$
Output power $= 140 W$
Power input $= V \times I$
$P = 240 \times 0.7$
$P = 168.0 W$
$∴$ Efficiency of transformer,
$η % = \frac{P_{output}}{P_{input}} \times 100 = \frac{140}{168} \times 100$
$η = 83.33 %$

CG PET -2018

Alternating Current

155401 A distribution transformer with an efficiency of $90 %$ supplies to a colony of 10 homes. All the 10 homes have electrical oven running at the same time, that draw 20 A current from $220 V$ lines. The power dissipated as heat in the transformer is

1

12.2 kW

2

4.9 kW

3

8.4 kW

4

9.9 kW

Explanation:

B Given that,
Efficiency of transformer $(η) = 90 %$
$V = 220$ volt, $I = 20 A$
We know that,
$P_{1} = 10 VI = 10 \times 220 \times 20 = 44 \times 10^{3} W$
So, power delivered by transformer
$P_{2} = \frac{100}{90} \times P_{1} = \frac{10}{9} \times 44 \times 10^{3}$
$P_{2} = 48.9 \times 10^{3} W$
$∴$ Power dissipated as heat
$P_{H} = P_{2} - P_{1}$
$P_{H} = (48.9 - 44) \times 10^{3}$
$P_{H} = 4.9 \times 10^{3}$
$P_{H} = 4.9 kW$

TS- EAMCET-07.05.2018

Alternating Current

155398 The number of turns in primary and secondary coils of a transformer is 50 and 200 respectively. If the current in the primary coil is $4 A$ , then the current in the secondary coil is

1

1 A

2

2 A

3

4 A

4

5 A

Explanation:

A Given that,
$N_{P} = 50, N_{S} = 200, I_{P} = 4 A, I_{S} =$ ?
We know that, in a transformer
$\frac{N_{P}}{N_{S}} = \frac{I_{S}}{I_{P}}$
$\frac{50}{200} = \frac{I_{S}}{4}$
$I_{S} = 1 A$

EAMCET-2018

Alternating Current

155399 A transformer has 100 turns in the primary coil and carries $8 A$ current. If input power is $1 kW$ , the number of turns in secondary coil to have $500 V$ of output will be

1 100

2 200

3 400

4 300

Explanation:

C Given that,
Primary number of turns $= 100$
Primary current $= 8 A$
Input power $= 1 kW$
$Power = V.I$
$1000 = 500 . I$
$I = 2 A$
Now, from the transformers equation -
$\frac{I_{1}}{I_{2}} = \frac{N_{2}}{N_{1}}$
$\frac{8}{2} = \frac{N_{2}}{100}$
$N_{2} = 400$

JCECE-2018

Alternating Current

155400 A transformer is used to glow a $140 W - 24 V$ bulb at $240 V$ AC. If current in the primary coil is $0.7 A$ , then the efficiency of the transformer is

1

16.7 %

2

50 %

3

83.3 %

4

25 %

Explanation:

C Given that,
Input voltage $= 240 V$
Primary coil current $= 0.7 A$
Output power $= 140 W$
Power input $= V \times I$
$P = 240 \times 0.7$
$P = 168.0 W$
$∴$ Efficiency of transformer,
$η % = \frac{P_{output}}{P_{input}} \times 100 = \frac{140}{168} \times 100$
$η = 83.33 %$

CG PET -2018

Alternating Current

155401 A distribution transformer with an efficiency of $90 %$ supplies to a colony of 10 homes. All the 10 homes have electrical oven running at the same time, that draw 20 A current from $220 V$ lines. The power dissipated as heat in the transformer is

1

12.2 kW

2

4.9 kW

3

8.4 kW

4

9.9 kW

Explanation:

B Given that,
Efficiency of transformer $(η) = 90 %$
$V = 220$ volt, $I = 20 A$
We know that,
$P_{1} = 10 VI = 10 \times 220 \times 20 = 44 \times 10^{3} W$
So, power delivered by transformer
$P_{2} = \frac{100}{90} \times P_{1} = \frac{10}{9} \times 44 \times 10^{3}$
$P_{2} = 48.9 \times 10^{3} W$
$∴$ Power dissipated as heat
$P_{H} = P_{2} - P_{1}$
$P_{H} = (48.9 - 44) \times 10^{3}$
$P_{H} = 4.9 \times 10^{3}$
$P_{H} = 4.9 kW$

TS- EAMCET-07.05.2018

Alternating Current

155398 The number of turns in primary and secondary coils of a transformer is 50 and 200 respectively. If the current in the primary coil is $4 A$ , then the current in the secondary coil is

1

1 A

2

2 A

3

4 A

4

5 A

Explanation:

A Given that,
$N_{P} = 50, N_{S} = 200, I_{P} = 4 A, I_{S} =$ ?
We know that, in a transformer
$\frac{N_{P}}{N_{S}} = \frac{I_{S}}{I_{P}}$
$\frac{50}{200} = \frac{I_{S}}{4}$
$I_{S} = 1 A$

EAMCET-2018

Alternating Current

155399 A transformer has 100 turns in the primary coil and carries $8 A$ current. If input power is $1 kW$ , the number of turns in secondary coil to have $500 V$ of output will be

1 100

2 200

3 400

4 300

Explanation:

C Given that,
Primary number of turns $= 100$
Primary current $= 8 A$
Input power $= 1 kW$
$Power = V.I$
$1000 = 500 . I$
$I = 2 A$
Now, from the transformers equation -
$\frac{I_{1}}{I_{2}} = \frac{N_{2}}{N_{1}}$
$\frac{8}{2} = \frac{N_{2}}{100}$
$N_{2} = 400$

JCECE-2018

Alternating Current

155400 A transformer is used to glow a $140 W - 24 V$ bulb at $240 V$ AC. If current in the primary coil is $0.7 A$ , then the efficiency of the transformer is

1

16.7 %

2

50 %

3

83.3 %

4

25 %

Explanation:

C Given that,
Input voltage $= 240 V$
Primary coil current $= 0.7 A$
Output power $= 140 W$
Power input $= V \times I$
$P = 240 \times 0.7$
$P = 168.0 W$
$∴$ Efficiency of transformer,
$η % = \frac{P_{output}}{P_{input}} \times 100 = \frac{140}{168} \times 100$
$η = 83.33 %$

CG PET -2018

Alternating Current

155401 A distribution transformer with an efficiency of $90 %$ supplies to a colony of 10 homes. All the 10 homes have electrical oven running at the same time, that draw 20 A current from $220 V$ lines. The power dissipated as heat in the transformer is

1

12.2 kW

2

4.9 kW

3

8.4 kW

4

9.9 kW

Explanation:

B Given that,
Efficiency of transformer $(η) = 90 %$
$V = 220$ volt, $I = 20 A$
We know that,
$P_{1} = 10 VI = 10 \times 220 \times 20 = 44 \times 10^{3} W$
So, power delivered by transformer
$P_{2} = \frac{100}{90} \times P_{1} = \frac{10}{9} \times 44 \times 10^{3}$
$P_{2} = 48.9 \times 10^{3} W$
$∴$ Power dissipated as heat
$P_{H} = P_{2} - P_{1}$
$P_{H} = (48.9 - 44) \times 10^{3}$
$P_{H} = 4.9 \times 10^{3}$
$P_{H} = 4.9 kW$

TS- EAMCET-07.05.2018