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PHXII07:ALTERNATING CURRENT

356291 A transformer with efficiency 80% works at 4 $k W$ and 100 $V$ . If the secondary voltage is 200 $V$ then the primary and secondary currents are respectively:

1

40 A, 16 A

2

16 A, 40 A

3

20 A, 40 A

4

40 A, 10 A

Explanation:

Efficiency, $η = \frac{o u t p u t p o w e r}{I n p u t p o w e r}$
$\Rightarrow η = \frac{V_{s} I_{s}}{V_{p} I_{p}} \Rightarrow \frac{80}{100} = \frac{200 \times I_{s}}{4000} \Rightarrow I_{s} = 16 A$
Also, $V_{p} I_{p} = 4000 \Rightarrow I_{p} = \frac{4000}{100} = 40 A$

PHXII07:ALTERNATING CURRENT

356292 A power transmission line feeds input power at $2300 V$ to a step down transformer with its primary windings having 4000 turns giving the output power at $230 V$ . If the current in the primary of the transformer is $5 A$ and its efficiency is $90 %$ the output current would be:

1

45 A

2

50 A

3

20 A

4

25 A

Explanation:

Efficiency $η = 0.9 = \frac{P_{S}}{P_{P}}$
$V_{S} I_{S} = 0.9 \times V_{P} I_{P}$
$I_{S} = \frac{0.9 \times 2300 \times 5}{230} = 45 A$

JEE - 2018

PHXII07:ALTERNATING CURRENT

356293 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 $ϕ$ is a constant, then the output voltage across the secondary coil is

1

90 V

2

120 V

3

220 V

4

30 V

Explanation:

The magnetic flux linked with the primary coil is given by
$ϕ = ϕ_{0} + 4 t$
So voltage across primary coil
$V_{p} = \frac{d ϕ}{d t} = \frac{d}{d t} (ϕ_{0} + 4 t) = 4 V$ (as $ϕ_{0} =$ constant)
Also, we have
$N_{p} = 50 a n d N_{s} = 1500$
From relation
$\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}} \Rightarrow V_{s} = V_{p} \frac{N_{s}}{N_{p}} = 4 (\frac{1500}{50}) = 120 V$

PHXII07:ALTERNATING CURRENT

356294 The turns ratio of a transformer is given as 8 : 3. If the current through the primary coil is 3 $A$ , then calculate the current through load resistance.

1

2.5 A

2

1.125 A

3

2 A

4

1.5 A

Explanation:

$\frac{N_{s}}{N_{p}} = \frac{i_{p}}{i_{s}} o r \frac{8}{3} = \frac{3}{i_{s}} \Rightarrow i_{s} = 1.125 A$

PHXII07:ALTERNATING CURRENT

356291 A transformer with efficiency 80% works at 4 $k W$ and 100 $V$ . If the secondary voltage is 200 $V$ then the primary and secondary currents are respectively:

1

40 A, 16 A

2

16 A, 40 A

3

20 A, 40 A

4

40 A, 10 A

Explanation:

Efficiency, $η = \frac{o u t p u t p o w e r}{I n p u t p o w e r}$
$\Rightarrow η = \frac{V_{s} I_{s}}{V_{p} I_{p}} \Rightarrow \frac{80}{100} = \frac{200 \times I_{s}}{4000} \Rightarrow I_{s} = 16 A$
Also, $V_{p} I_{p} = 4000 \Rightarrow I_{p} = \frac{4000}{100} = 40 A$

PHXII07:ALTERNATING CURRENT

356292 A power transmission line feeds input power at $2300 V$ to a step down transformer with its primary windings having 4000 turns giving the output power at $230 V$ . If the current in the primary of the transformer is $5 A$ and its efficiency is $90 %$ the output current would be:

1

45 A

2

50 A

3

20 A

4

25 A

Explanation:

Efficiency $η = 0.9 = \frac{P_{S}}{P_{P}}$
$V_{S} I_{S} = 0.9 \times V_{P} I_{P}$
$I_{S} = \frac{0.9 \times 2300 \times 5}{230} = 45 A$

JEE - 2018

PHXII07:ALTERNATING CURRENT

356293 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 $ϕ$ is a constant, then the output voltage across the secondary coil is

1

90 V

2

120 V

3

220 V

4

30 V

Explanation:

The magnetic flux linked with the primary coil is given by
$ϕ = ϕ_{0} + 4 t$
So voltage across primary coil
$V_{p} = \frac{d ϕ}{d t} = \frac{d}{d t} (ϕ_{0} + 4 t) = 4 V$ (as $ϕ_{0} =$ constant)
Also, we have
$N_{p} = 50 a n d N_{s} = 1500$
From relation
$\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}} \Rightarrow V_{s} = V_{p} \frac{N_{s}}{N_{p}} = 4 (\frac{1500}{50}) = 120 V$

PHXII07:ALTERNATING CURRENT

356294 The turns ratio of a transformer is given as 8 : 3. If the current through the primary coil is 3 $A$ , then calculate the current through load resistance.

1

2.5 A

2

1.125 A

3

2 A

4

1.5 A

Explanation:

$\frac{N_{s}}{N_{p}} = \frac{i_{p}}{i_{s}} o r \frac{8}{3} = \frac{3}{i_{s}} \Rightarrow i_{s} = 1.125 A$

PHXII07:ALTERNATING CURRENT

356291 A transformer with efficiency 80% works at 4 $k W$ and 100 $V$ . If the secondary voltage is 200 $V$ then the primary and secondary currents are respectively:

1

40 A, 16 A

2

16 A, 40 A

3

20 A, 40 A

4

40 A, 10 A

Explanation:

Efficiency, $η = \frac{o u t p u t p o w e r}{I n p u t p o w e r}$
$\Rightarrow η = \frac{V_{s} I_{s}}{V_{p} I_{p}} \Rightarrow \frac{80}{100} = \frac{200 \times I_{s}}{4000} \Rightarrow I_{s} = 16 A$
Also, $V_{p} I_{p} = 4000 \Rightarrow I_{p} = \frac{4000}{100} = 40 A$

PHXII07:ALTERNATING CURRENT

356292 A power transmission line feeds input power at $2300 V$ to a step down transformer with its primary windings having 4000 turns giving the output power at $230 V$ . If the current in the primary of the transformer is $5 A$ and its efficiency is $90 %$ the output current would be:

1

45 A

2

50 A

3

20 A

4

25 A

Explanation:

Efficiency $η = 0.9 = \frac{P_{S}}{P_{P}}$
$V_{S} I_{S} = 0.9 \times V_{P} I_{P}$
$I_{S} = \frac{0.9 \times 2300 \times 5}{230} = 45 A$

JEE - 2018

PHXII07:ALTERNATING CURRENT

356293 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 $ϕ$ is a constant, then the output voltage across the secondary coil is

1

90 V

2

120 V

3

220 V

4

30 V

Explanation:

The magnetic flux linked with the primary coil is given by
$ϕ = ϕ_{0} + 4 t$
So voltage across primary coil
$V_{p} = \frac{d ϕ}{d t} = \frac{d}{d t} (ϕ_{0} + 4 t) = 4 V$ (as $ϕ_{0} =$ constant)
Also, we have
$N_{p} = 50 a n d N_{s} = 1500$
From relation
$\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}} \Rightarrow V_{s} = V_{p} \frac{N_{s}}{N_{p}} = 4 (\frac{1500}{50}) = 120 V$

PHXII07:ALTERNATING CURRENT

356294 The turns ratio of a transformer is given as 8 : 3. If the current through the primary coil is 3 $A$ , then calculate the current through load resistance.

1

2.5 A

2

1.125 A

3

2 A

4

1.5 A

Explanation:

$\frac{N_{s}}{N_{p}} = \frac{i_{p}}{i_{s}} o r \frac{8}{3} = \frac{3}{i_{s}} \Rightarrow i_{s} = 1.125 A$

PHXII07:ALTERNATING CURRENT

356291 A transformer with efficiency 80% works at 4 $k W$ and 100 $V$ . If the secondary voltage is 200 $V$ then the primary and secondary currents are respectively:

1

40 A, 16 A

2

16 A, 40 A

3

20 A, 40 A

4

40 A, 10 A

Explanation:

Efficiency, $η = \frac{o u t p u t p o w e r}{I n p u t p o w e r}$
$\Rightarrow η = \frac{V_{s} I_{s}}{V_{p} I_{p}} \Rightarrow \frac{80}{100} = \frac{200 \times I_{s}}{4000} \Rightarrow I_{s} = 16 A$
Also, $V_{p} I_{p} = 4000 \Rightarrow I_{p} = \frac{4000}{100} = 40 A$

PHXII07:ALTERNATING CURRENT

356292 A power transmission line feeds input power at $2300 V$ to a step down transformer with its primary windings having 4000 turns giving the output power at $230 V$ . If the current in the primary of the transformer is $5 A$ and its efficiency is $90 %$ the output current would be:

1

45 A

2

50 A

3

20 A

4

25 A

Explanation:

Efficiency $η = 0.9 = \frac{P_{S}}{P_{P}}$
$V_{S} I_{S} = 0.9 \times V_{P} I_{P}$
$I_{S} = \frac{0.9 \times 2300 \times 5}{230} = 45 A$

JEE - 2018

PHXII07:ALTERNATING CURRENT

356293 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 $ϕ$ is a constant, then the output voltage across the secondary coil is

1

90 V

2

120 V

3

220 V

4

30 V

Explanation:

The magnetic flux linked with the primary coil is given by
$ϕ = ϕ_{0} + 4 t$
So voltage across primary coil
$V_{p} = \frac{d ϕ}{d t} = \frac{d}{d t} (ϕ_{0} + 4 t) = 4 V$ (as $ϕ_{0} =$ constant)
Also, we have
$N_{p} = 50 a n d N_{s} = 1500$
From relation
$\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}} \Rightarrow V_{s} = V_{p} \frac{N_{s}}{N_{p}} = 4 (\frac{1500}{50}) = 120 V$

PHXII07:ALTERNATING CURRENT

356294 The turns ratio of a transformer is given as 8 : 3. If the current through the primary coil is 3 $A$ , then calculate the current through load resistance.

1

2.5 A

2

1.125 A

3

2 A

4

1.5 A

Explanation:

$\frac{N_{s}}{N_{p}} = \frac{i_{p}}{i_{s}} o r \frac{8}{3} = \frac{3}{i_{s}} \Rightarrow i_{s} = 1.125 A$