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Alternating Current

155464 The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux $ϕ$ linked with the primary coil is given by $ϕ = ϕ_{0} + 4 t$ , where $ϕ$ is in weber, $t$ is time in second and $ϕ_{0}$ is a constant, the output voltage across the secondary coil is

1

90 V

2

120 V

3

220 V

4

30 V

Explanation:

B Given that,
$n_{p} = 50, n_{s} = 1500$
$ϕ_{P} = ϕ_{0} + 4 t$
Induced emf on a coil
$ε_{p} = \frac{d ϕ_{p}}{dt} = \frac{d}{dt} (ϕ_{0} + 4 t) = 4 V$
Therefore,
$\frac{ε_{s}}{ε_{p}} = \frac{n_{s}}{n_{p}}$
$ε_{s} = \frac{n_{s}}{n_{p}} \times ε_{p}$
$= \frac{1500 \times 4}{50}$
$ε_{s} = 120 V$

JIPMER-2009

Alternating Current

155367 A $12 V, 60 W$ lamp is connected to the secondary of a step down transformer, whose primary is connected to ac mains of $220 V$ . Assuming the transformer to be ideal, what is the current in the primary winding?

1

2.7 A

2

3.7 A

3

0.37 A

4

0.27 A

Explanation:

D For ideal Transformer-
$V_{S} I_{S} = V_{P} I_{P}$
$\Rightarrow P_{out} = P_{in}$
$\Rightarrow 60 = 220 \times I_{P}$
$I_{P} = \frac{60}{220} = 0.27 A$

NEET (UG)-07.05.2023

Alternating Current

155369 A transformer is used to light a $100 W, 110 V$ lamp from a $220 V$ supply. If the supply current is $0.6 A$ , the efficiency of the transformer is

1

66 %

2

76 %

3

86 %

4

96 %

Explanation:

B Given, $P = 100 W, V_{p} = 220 V, I_{p} = 0.6 A$ The efficiency of transformer
$η = \frac{Output}{Input} \times 100$
$η % = \frac{100}{220 \times 0.6} \times 100 = 75.75 % ◻ 76 %$

COMEDK 2015

Alternating Current

155383 In a step-up transformer, relation between number of turns in primary $(N_{P})$ and number of turns in secondary $(N_{S})$ coils is

1

N_{S}

is greater than

N_{P}

2

N_{P}

is greater than

N_{S}

3

N_{S}

is equal to

N_{P}

4

N_{P} = 2 Ns

Explanation:

A In a step-up transformer, the number of turns in primary coil is less than the number turns in secondary coil .
$∴ \frac{N_{s}}{N_{p}} > 1$
$N_{s} > N_{p} .$

COMEDK 2020

Alternating Current

155464 The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux $ϕ$ linked with the primary coil is given by $ϕ = ϕ_{0} + 4 t$ , where $ϕ$ is in weber, $t$ is time in second and $ϕ_{0}$ is a constant, the output voltage across the secondary coil is

1

90 V

2

120 V

3

220 V

4

30 V

Explanation:

B Given that,
$n_{p} = 50, n_{s} = 1500$
$ϕ_{P} = ϕ_{0} + 4 t$
Induced emf on a coil
$ε_{p} = \frac{d ϕ_{p}}{dt} = \frac{d}{dt} (ϕ_{0} + 4 t) = 4 V$
Therefore,
$\frac{ε_{s}}{ε_{p}} = \frac{n_{s}}{n_{p}}$
$ε_{s} = \frac{n_{s}}{n_{p}} \times ε_{p}$
$= \frac{1500 \times 4}{50}$
$ε_{s} = 120 V$

JIPMER-2009

Alternating Current

155367 A $12 V, 60 W$ lamp is connected to the secondary of a step down transformer, whose primary is connected to ac mains of $220 V$ . Assuming the transformer to be ideal, what is the current in the primary winding?

1

2.7 A

2

3.7 A

3

0.37 A

4

0.27 A

Explanation:

D For ideal Transformer-
$V_{S} I_{S} = V_{P} I_{P}$
$\Rightarrow P_{out} = P_{in}$
$\Rightarrow 60 = 220 \times I_{P}$
$I_{P} = \frac{60}{220} = 0.27 A$

NEET (UG)-07.05.2023

Alternating Current

155369 A transformer is used to light a $100 W, 110 V$ lamp from a $220 V$ supply. If the supply current is $0.6 A$ , the efficiency of the transformer is

1

66 %

2

76 %

3

86 %

4

96 %

Explanation:

B Given, $P = 100 W, V_{p} = 220 V, I_{p} = 0.6 A$ The efficiency of transformer
$η = \frac{Output}{Input} \times 100$
$η % = \frac{100}{220 \times 0.6} \times 100 = 75.75 % ◻ 76 %$

COMEDK 2015

Alternating Current

155383 In a step-up transformer, relation between number of turns in primary $(N_{P})$ and number of turns in secondary $(N_{S})$ coils is

1

N_{S}

is greater than

N_{P}

2

N_{P}

is greater than

N_{S}

3

N_{S}

is equal to

N_{P}

4

N_{P} = 2 Ns

Explanation:

A In a step-up transformer, the number of turns in primary coil is less than the number turns in secondary coil .
$∴ \frac{N_{s}}{N_{p}} > 1$
$N_{s} > N_{p} .$

COMEDK 2020

Alternating Current

155464 The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux $ϕ$ linked with the primary coil is given by $ϕ = ϕ_{0} + 4 t$ , where $ϕ$ is in weber, $t$ is time in second and $ϕ_{0}$ is a constant, the output voltage across the secondary coil is

1

90 V

2

120 V

3

220 V

4

30 V

Explanation:

B Given that,
$n_{p} = 50, n_{s} = 1500$
$ϕ_{P} = ϕ_{0} + 4 t$
Induced emf on a coil
$ε_{p} = \frac{d ϕ_{p}}{dt} = \frac{d}{dt} (ϕ_{0} + 4 t) = 4 V$
Therefore,
$\frac{ε_{s}}{ε_{p}} = \frac{n_{s}}{n_{p}}$
$ε_{s} = \frac{n_{s}}{n_{p}} \times ε_{p}$
$= \frac{1500 \times 4}{50}$
$ε_{s} = 120 V$

JIPMER-2009

Alternating Current

155367 A $12 V, 60 W$ lamp is connected to the secondary of a step down transformer, whose primary is connected to ac mains of $220 V$ . Assuming the transformer to be ideal, what is the current in the primary winding?

1

2.7 A

2

3.7 A

3

0.37 A

4

0.27 A

Explanation:

D For ideal Transformer-
$V_{S} I_{S} = V_{P} I_{P}$
$\Rightarrow P_{out} = P_{in}$
$\Rightarrow 60 = 220 \times I_{P}$
$I_{P} = \frac{60}{220} = 0.27 A$

NEET (UG)-07.05.2023

Alternating Current

155369 A transformer is used to light a $100 W, 110 V$ lamp from a $220 V$ supply. If the supply current is $0.6 A$ , the efficiency of the transformer is

1

66 %

2

76 %

3

86 %

4

96 %

Explanation:

B Given, $P = 100 W, V_{p} = 220 V, I_{p} = 0.6 A$ The efficiency of transformer
$η = \frac{Output}{Input} \times 100$
$η % = \frac{100}{220 \times 0.6} \times 100 = 75.75 % ◻ 76 %$

COMEDK 2015

Alternating Current

155383 In a step-up transformer, relation between number of turns in primary $(N_{P})$ and number of turns in secondary $(N_{S})$ coils is

1

N_{S}

is greater than

N_{P}

2

N_{P}

is greater than

N_{S}

3

N_{S}

is equal to

N_{P}

4

N_{P} = 2 Ns

Explanation:

A In a step-up transformer, the number of turns in primary coil is less than the number turns in secondary coil .
$∴ \frac{N_{s}}{N_{p}} > 1$
$N_{s} > N_{p} .$

COMEDK 2020

Alternating Current

155464 The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux $ϕ$ linked with the primary coil is given by $ϕ = ϕ_{0} + 4 t$ , where $ϕ$ is in weber, $t$ is time in second and $ϕ_{0}$ is a constant, the output voltage across the secondary coil is

1

90 V

2

120 V

3

220 V

4

30 V

Explanation:

B Given that,
$n_{p} = 50, n_{s} = 1500$
$ϕ_{P} = ϕ_{0} + 4 t$
Induced emf on a coil
$ε_{p} = \frac{d ϕ_{p}}{dt} = \frac{d}{dt} (ϕ_{0} + 4 t) = 4 V$
Therefore,
$\frac{ε_{s}}{ε_{p}} = \frac{n_{s}}{n_{p}}$
$ε_{s} = \frac{n_{s}}{n_{p}} \times ε_{p}$
$= \frac{1500 \times 4}{50}$
$ε_{s} = 120 V$

JIPMER-2009

Alternating Current

155367 A $12 V, 60 W$ lamp is connected to the secondary of a step down transformer, whose primary is connected to ac mains of $220 V$ . Assuming the transformer to be ideal, what is the current in the primary winding?

1

2.7 A

2

3.7 A

3

0.37 A

4

0.27 A

Explanation:

D For ideal Transformer-
$V_{S} I_{S} = V_{P} I_{P}$
$\Rightarrow P_{out} = P_{in}$
$\Rightarrow 60 = 220 \times I_{P}$
$I_{P} = \frac{60}{220} = 0.27 A$

NEET (UG)-07.05.2023

Alternating Current

155369 A transformer is used to light a $100 W, 110 V$ lamp from a $220 V$ supply. If the supply current is $0.6 A$ , the efficiency of the transformer is

1

66 %

2

76 %

3

86 %

4

96 %

Explanation:

B Given, $P = 100 W, V_{p} = 220 V, I_{p} = 0.6 A$ The efficiency of transformer
$η = \frac{Output}{Input} \times 100$
$η % = \frac{100}{220 \times 0.6} \times 100 = 75.75 % ◻ 76 %$

COMEDK 2015

Alternating Current

155383 In a step-up transformer, relation between number of turns in primary $(N_{P})$ and number of turns in secondary $(N_{S})$ coils is

1

N_{S}

is greater than

N_{P}

2

N_{P}

is greater than

N_{S}

3

N_{S}

is equal to

N_{P}

4

N_{P} = 2 Ns

Explanation:

A In a step-up transformer, the number of turns in primary coil is less than the number turns in secondary coil .
$∴ \frac{N_{s}}{N_{p}} > 1$
$N_{s} > N_{p} .$

COMEDK 2020