QBManager

Alternating Current

155370 A transformer is used to light a $100 W$ and 110 $V$ lamp from a $220 V$ mains. If the main current is $0.5 A$ , the efficiency of the transformer is approximately

1

30 %

2

50 %

3

90 %

4

10 %

Explanation:

C Given,
Output power $(P_{output}) = 100 W$
Voltage in primary coil $(V_{P}) = 220 V$
Current in primary coil $(I_{P}) = 0.5 A$
We know that,
$η = \frac{P_{o / p}}{P_{I / p}} \times 100$
$= \frac{100}{V_{P} I_{P}} \times 100$
$η = \frac{100}{220 \times 0.5} \times 100$
$η = 90.9 %$
$η ◻ 90 %$

AIPMT- 2007

Alternating Current

155371 In a transformer 220 AC voltage is increased to $2200 V$ . If the number of turns in the secondary are 2000 , then find the number of turns in the primary:

1 200

2 100

3 50

4 20

Explanation:

A Given,
Voltage in primary coil $(V_{1}) = 220 V$
Voltage in secondary coil $(V_{2}) = 2200 V$
Number of turns in secondary coil $(N_{2}) = 2000$
Number of turns in primary coil $(N_{1}) =$ ?
$We know that,$
$∵ \frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$
$\frac{220}{2200} = \frac{N_{1}}{2000}$
$N_{1} = 200$

AP EAPCET-12.07.2022

Alternating Current

155372 A transformer operating at primary voltage 8 $kV$ and secondary voltage $160 V$ serves a load of $80 kW$ . Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be

1

800 Ω

and

1.06 Ω

2

10 Ω

and

500 Ω

3

800 Ω

and

0.32 Ω

4

1.06 Ω

and

500 Ω

Explanation:

C Given,
Primary voltage $(V_{1}) = 8 kV = 8000 V$
Secondary voltage $(V_{2}) = 160 V$
Let resistance on secondary Load is $R_{s}$
$∵ P_{o / P} = \frac{V_{2}^{2}}{R_{s}}$
$80 KW = \frac{(160)^{2}}{R_{s}}$
$∴ R_{s} = \frac{160 \times 160}{80 \times 1000}$
$R_{s} = 0.32 Ω$
$∵$ Transformer is ideal,
Input power $=$ output power
$\frac{V_{1}^{2}}{R_{p}} = 80 \times 1000$
$R_{p} = \frac{V_{1}^{2}}{80 \times 1000} = \frac{(8000)^{2}}{80 \times 1000}$
$= \frac{8000 \times 8000}{80 \times 1000}$
$R_{p} = 800 Ω$
$R_{s} = 0.32 Ω$

JEE Main-28.07.2022

Alternating Current

155373 A cell phone charger consists of a stepdown transformer to convert $A C$ voltage of $120 V$ to AC voltage of $5.0 V$ . If the secondary coil contains 30 turns and charger supplies $720 mA$ , calculate the current in the primary coil.

1

12 mA

2

30 mA

3

85 mA

4

160 mA

Explanation:

B Given,
Primary voltage $(V_{1}) = 120 V$
Secondary voltage $(V_{2}) = 5 V$
Turns in secondary coil $= 30$
Output current or secondary current $= 720 mA$ We know that,
$\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$Or I_{1} = \frac{I_{2} V_{2}}{V_{1}}$
$I_{1} = \frac{720 \times 5 \times 10^{- 3}}{120} = 30 \times 10^{- 3} A$
$I_{1} = 30 mA$

TS EAMCET 31.07.2022

Alternating Current

155375 A step down transformer connected to an AC mains supply of $220 V$ is made to operate at $11 V, 44 W$ lamp. Ignoring power losses in the transformer, what is the current in the primary circuit ?

1

0.2 A

2

0.4 A

3

2 A

4

4 A

Explanation:

A Given,
Primary voltage $(V_{1}) = 220 V$
Secondary voltage $(V_{2}) = 11 V$
Power output $= 44 W$
$∵$ There is no power loss,
Input power $=$ Output power
$V_{1} I_{1} = 44$
$I_{1} = \frac{44}{220}$
$I_{1} = 0.2 A$

NEET- 2021

Alternating Current

155370 A transformer is used to light a $100 W$ and 110 $V$ lamp from a $220 V$ mains. If the main current is $0.5 A$ , the efficiency of the transformer is approximately

1

30 %

2

50 %

3

90 %

4

10 %

Explanation:

C Given,
Output power $(P_{output}) = 100 W$
Voltage in primary coil $(V_{P}) = 220 V$
Current in primary coil $(I_{P}) = 0.5 A$
We know that,
$η = \frac{P_{o / p}}{P_{I / p}} \times 100$
$= \frac{100}{V_{P} I_{P}} \times 100$
$η = \frac{100}{220 \times 0.5} \times 100$
$η = 90.9 %$
$η ◻ 90 %$

AIPMT- 2007

Alternating Current

155371 In a transformer 220 AC voltage is increased to $2200 V$ . If the number of turns in the secondary are 2000 , then find the number of turns in the primary:

1 200

2 100

3 50

4 20

Explanation:

A Given,
Voltage in primary coil $(V_{1}) = 220 V$
Voltage in secondary coil $(V_{2}) = 2200 V$
Number of turns in secondary coil $(N_{2}) = 2000$
Number of turns in primary coil $(N_{1}) =$ ?
$We know that,$
$∵ \frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$
$\frac{220}{2200} = \frac{N_{1}}{2000}$
$N_{1} = 200$

AP EAPCET-12.07.2022

Alternating Current

155372 A transformer operating at primary voltage 8 $kV$ and secondary voltage $160 V$ serves a load of $80 kW$ . Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be

1

800 Ω

and

1.06 Ω

2

10 Ω

and

500 Ω

3

800 Ω

and

0.32 Ω

4

1.06 Ω

and

500 Ω

Explanation:

C Given,
Primary voltage $(V_{1}) = 8 kV = 8000 V$
Secondary voltage $(V_{2}) = 160 V$
Let resistance on secondary Load is $R_{s}$
$∵ P_{o / P} = \frac{V_{2}^{2}}{R_{s}}$
$80 KW = \frac{(160)^{2}}{R_{s}}$
$∴ R_{s} = \frac{160 \times 160}{80 \times 1000}$
$R_{s} = 0.32 Ω$
$∵$ Transformer is ideal,
Input power $=$ output power
$\frac{V_{1}^{2}}{R_{p}} = 80 \times 1000$
$R_{p} = \frac{V_{1}^{2}}{80 \times 1000} = \frac{(8000)^{2}}{80 \times 1000}$
$= \frac{8000 \times 8000}{80 \times 1000}$
$R_{p} = 800 Ω$
$R_{s} = 0.32 Ω$

JEE Main-28.07.2022

Alternating Current

155373 A cell phone charger consists of a stepdown transformer to convert $A C$ voltage of $120 V$ to AC voltage of $5.0 V$ . If the secondary coil contains 30 turns and charger supplies $720 mA$ , calculate the current in the primary coil.

1

12 mA

2

30 mA

3

85 mA

4

160 mA

Explanation:

B Given,
Primary voltage $(V_{1}) = 120 V$
Secondary voltage $(V_{2}) = 5 V$
Turns in secondary coil $= 30$
Output current or secondary current $= 720 mA$ We know that,
$\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$Or I_{1} = \frac{I_{2} V_{2}}{V_{1}}$
$I_{1} = \frac{720 \times 5 \times 10^{- 3}}{120} = 30 \times 10^{- 3} A$
$I_{1} = 30 mA$

TS EAMCET 31.07.2022

Alternating Current

155375 A step down transformer connected to an AC mains supply of $220 V$ is made to operate at $11 V, 44 W$ lamp. Ignoring power losses in the transformer, what is the current in the primary circuit ?

1

0.2 A

2

0.4 A

3

2 A

4

4 A

Explanation:

A Given,
Primary voltage $(V_{1}) = 220 V$
Secondary voltage $(V_{2}) = 11 V$
Power output $= 44 W$
$∵$ There is no power loss,
Input power $=$ Output power
$V_{1} I_{1} = 44$
$I_{1} = \frac{44}{220}$
$I_{1} = 0.2 A$

NEET- 2021

Alternating Current

155370 A transformer is used to light a $100 W$ and 110 $V$ lamp from a $220 V$ mains. If the main current is $0.5 A$ , the efficiency of the transformer is approximately

1

30 %

2

50 %

3

90 %

4

10 %

Explanation:

C Given,
Output power $(P_{output}) = 100 W$
Voltage in primary coil $(V_{P}) = 220 V$
Current in primary coil $(I_{P}) = 0.5 A$
We know that,
$η = \frac{P_{o / p}}{P_{I / p}} \times 100$
$= \frac{100}{V_{P} I_{P}} \times 100$
$η = \frac{100}{220 \times 0.5} \times 100$
$η = 90.9 %$
$η ◻ 90 %$

AIPMT- 2007

Alternating Current

155371 In a transformer 220 AC voltage is increased to $2200 V$ . If the number of turns in the secondary are 2000 , then find the number of turns in the primary:

1 200

2 100

3 50

4 20

Explanation:

A Given,
Voltage in primary coil $(V_{1}) = 220 V$
Voltage in secondary coil $(V_{2}) = 2200 V$
Number of turns in secondary coil $(N_{2}) = 2000$
Number of turns in primary coil $(N_{1}) =$ ?
$We know that,$
$∵ \frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$
$\frac{220}{2200} = \frac{N_{1}}{2000}$
$N_{1} = 200$

AP EAPCET-12.07.2022

Alternating Current

155372 A transformer operating at primary voltage 8 $kV$ and secondary voltage $160 V$ serves a load of $80 kW$ . Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be

1

800 Ω

and

1.06 Ω

2

10 Ω

and

500 Ω

3

800 Ω

and

0.32 Ω

4

1.06 Ω

and

500 Ω

Explanation:

C Given,
Primary voltage $(V_{1}) = 8 kV = 8000 V$
Secondary voltage $(V_{2}) = 160 V$
Let resistance on secondary Load is $R_{s}$
$∵ P_{o / P} = \frac{V_{2}^{2}}{R_{s}}$
$80 KW = \frac{(160)^{2}}{R_{s}}$
$∴ R_{s} = \frac{160 \times 160}{80 \times 1000}$
$R_{s} = 0.32 Ω$
$∵$ Transformer is ideal,
Input power $=$ output power
$\frac{V_{1}^{2}}{R_{p}} = 80 \times 1000$
$R_{p} = \frac{V_{1}^{2}}{80 \times 1000} = \frac{(8000)^{2}}{80 \times 1000}$
$= \frac{8000 \times 8000}{80 \times 1000}$
$R_{p} = 800 Ω$
$R_{s} = 0.32 Ω$

JEE Main-28.07.2022

Alternating Current

155373 A cell phone charger consists of a stepdown transformer to convert $A C$ voltage of $120 V$ to AC voltage of $5.0 V$ . If the secondary coil contains 30 turns and charger supplies $720 mA$ , calculate the current in the primary coil.

1

12 mA

2

30 mA

3

85 mA

4

160 mA

Explanation:

B Given,
Primary voltage $(V_{1}) = 120 V$
Secondary voltage $(V_{2}) = 5 V$
Turns in secondary coil $= 30$
Output current or secondary current $= 720 mA$ We know that,
$\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$Or I_{1} = \frac{I_{2} V_{2}}{V_{1}}$
$I_{1} = \frac{720 \times 5 \times 10^{- 3}}{120} = 30 \times 10^{- 3} A$
$I_{1} = 30 mA$

TS EAMCET 31.07.2022

Alternating Current

155375 A step down transformer connected to an AC mains supply of $220 V$ is made to operate at $11 V, 44 W$ lamp. Ignoring power losses in the transformer, what is the current in the primary circuit ?

1

0.2 A

2

0.4 A

3

2 A

4

4 A

Explanation:

A Given,
Primary voltage $(V_{1}) = 220 V$
Secondary voltage $(V_{2}) = 11 V$
Power output $= 44 W$
$∵$ There is no power loss,
Input power $=$ Output power
$V_{1} I_{1} = 44$
$I_{1} = \frac{44}{220}$
$I_{1} = 0.2 A$

NEET- 2021

Alternating Current

155370 A transformer is used to light a $100 W$ and 110 $V$ lamp from a $220 V$ mains. If the main current is $0.5 A$ , the efficiency of the transformer is approximately

1

30 %

2

50 %

3

90 %

4

10 %

Explanation:

C Given,
Output power $(P_{output}) = 100 W$
Voltage in primary coil $(V_{P}) = 220 V$
Current in primary coil $(I_{P}) = 0.5 A$
We know that,
$η = \frac{P_{o / p}}{P_{I / p}} \times 100$
$= \frac{100}{V_{P} I_{P}} \times 100$
$η = \frac{100}{220 \times 0.5} \times 100$
$η = 90.9 %$
$η ◻ 90 %$

AIPMT- 2007

Alternating Current

155371 In a transformer 220 AC voltage is increased to $2200 V$ . If the number of turns in the secondary are 2000 , then find the number of turns in the primary:

1 200

2 100

3 50

4 20

Explanation:

A Given,
Voltage in primary coil $(V_{1}) = 220 V$
Voltage in secondary coil $(V_{2}) = 2200 V$
Number of turns in secondary coil $(N_{2}) = 2000$
Number of turns in primary coil $(N_{1}) =$ ?
$We know that,$
$∵ \frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$
$\frac{220}{2200} = \frac{N_{1}}{2000}$
$N_{1} = 200$

AP EAPCET-12.07.2022

Alternating Current

155372 A transformer operating at primary voltage 8 $kV$ and secondary voltage $160 V$ serves a load of $80 kW$ . Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be

1

800 Ω

and

1.06 Ω

2

10 Ω

and

500 Ω

3

800 Ω

and

0.32 Ω

4

1.06 Ω

and

500 Ω

Explanation:

C Given,
Primary voltage $(V_{1}) = 8 kV = 8000 V$
Secondary voltage $(V_{2}) = 160 V$
Let resistance on secondary Load is $R_{s}$
$∵ P_{o / P} = \frac{V_{2}^{2}}{R_{s}}$
$80 KW = \frac{(160)^{2}}{R_{s}}$
$∴ R_{s} = \frac{160 \times 160}{80 \times 1000}$
$R_{s} = 0.32 Ω$
$∵$ Transformer is ideal,
Input power $=$ output power
$\frac{V_{1}^{2}}{R_{p}} = 80 \times 1000$
$R_{p} = \frac{V_{1}^{2}}{80 \times 1000} = \frac{(8000)^{2}}{80 \times 1000}$
$= \frac{8000 \times 8000}{80 \times 1000}$
$R_{p} = 800 Ω$
$R_{s} = 0.32 Ω$

JEE Main-28.07.2022

Alternating Current

155373 A cell phone charger consists of a stepdown transformer to convert $A C$ voltage of $120 V$ to AC voltage of $5.0 V$ . If the secondary coil contains 30 turns and charger supplies $720 mA$ , calculate the current in the primary coil.

1

12 mA

2

30 mA

3

85 mA

4

160 mA

Explanation:

B Given,
Primary voltage $(V_{1}) = 120 V$
Secondary voltage $(V_{2}) = 5 V$
Turns in secondary coil $= 30$
Output current or secondary current $= 720 mA$ We know that,
$\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$Or I_{1} = \frac{I_{2} V_{2}}{V_{1}}$
$I_{1} = \frac{720 \times 5 \times 10^{- 3}}{120} = 30 \times 10^{- 3} A$
$I_{1} = 30 mA$

TS EAMCET 31.07.2022

Alternating Current

155375 A step down transformer connected to an AC mains supply of $220 V$ is made to operate at $11 V, 44 W$ lamp. Ignoring power losses in the transformer, what is the current in the primary circuit ?

1

0.2 A

2

0.4 A

3

2 A

4

4 A

Explanation:

A Given,
Primary voltage $(V_{1}) = 220 V$
Secondary voltage $(V_{2}) = 11 V$
Power output $= 44 W$
$∵$ There is no power loss,
Input power $=$ Output power
$V_{1} I_{1} = 44$
$I_{1} = \frac{44}{220}$
$I_{1} = 0.2 A$

NEET- 2021

Alternating Current

155370 A transformer is used to light a $100 W$ and 110 $V$ lamp from a $220 V$ mains. If the main current is $0.5 A$ , the efficiency of the transformer is approximately

1

30 %

2

50 %

3

90 %

4

10 %

Explanation:

C Given,
Output power $(P_{output}) = 100 W$
Voltage in primary coil $(V_{P}) = 220 V$
Current in primary coil $(I_{P}) = 0.5 A$
We know that,
$η = \frac{P_{o / p}}{P_{I / p}} \times 100$
$= \frac{100}{V_{P} I_{P}} \times 100$
$η = \frac{100}{220 \times 0.5} \times 100$
$η = 90.9 %$
$η ◻ 90 %$

AIPMT- 2007

Alternating Current

155371 In a transformer 220 AC voltage is increased to $2200 V$ . If the number of turns in the secondary are 2000 , then find the number of turns in the primary:

1 200

2 100

3 50

4 20

Explanation:

A Given,
Voltage in primary coil $(V_{1}) = 220 V$
Voltage in secondary coil $(V_{2}) = 2200 V$
Number of turns in secondary coil $(N_{2}) = 2000$
Number of turns in primary coil $(N_{1}) =$ ?
$We know that,$
$∵ \frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$
$\frac{220}{2200} = \frac{N_{1}}{2000}$
$N_{1} = 200$

AP EAPCET-12.07.2022

Alternating Current

155372 A transformer operating at primary voltage 8 $kV$ and secondary voltage $160 V$ serves a load of $80 kW$ . Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be

1

800 Ω

and

1.06 Ω

2

10 Ω

and

500 Ω

3

800 Ω

and

0.32 Ω

4

1.06 Ω

and

500 Ω

Explanation:

C Given,
Primary voltage $(V_{1}) = 8 kV = 8000 V$
Secondary voltage $(V_{2}) = 160 V$
Let resistance on secondary Load is $R_{s}$
$∵ P_{o / P} = \frac{V_{2}^{2}}{R_{s}}$
$80 KW = \frac{(160)^{2}}{R_{s}}$
$∴ R_{s} = \frac{160 \times 160}{80 \times 1000}$
$R_{s} = 0.32 Ω$
$∵$ Transformer is ideal,
Input power $=$ output power
$\frac{V_{1}^{2}}{R_{p}} = 80 \times 1000$
$R_{p} = \frac{V_{1}^{2}}{80 \times 1000} = \frac{(8000)^{2}}{80 \times 1000}$
$= \frac{8000 \times 8000}{80 \times 1000}$
$R_{p} = 800 Ω$
$R_{s} = 0.32 Ω$

JEE Main-28.07.2022

Alternating Current

155373 A cell phone charger consists of a stepdown transformer to convert $A C$ voltage of $120 V$ to AC voltage of $5.0 V$ . If the secondary coil contains 30 turns and charger supplies $720 mA$ , calculate the current in the primary coil.

1

12 mA

2

30 mA

3

85 mA

4

160 mA

Explanation:

B Given,
Primary voltage $(V_{1}) = 120 V$
Secondary voltage $(V_{2}) = 5 V$
Turns in secondary coil $= 30$
Output current or secondary current $= 720 mA$ We know that,
$\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$Or I_{1} = \frac{I_{2} V_{2}}{V_{1}}$
$I_{1} = \frac{720 \times 5 \times 10^{- 3}}{120} = 30 \times 10^{- 3} A$
$I_{1} = 30 mA$

TS EAMCET 31.07.2022

Alternating Current

155375 A step down transformer connected to an AC mains supply of $220 V$ is made to operate at $11 V, 44 W$ lamp. Ignoring power losses in the transformer, what is the current in the primary circuit ?

1

0.2 A

2

0.4 A

3

2 A

4

4 A

Explanation:

A Given,
Primary voltage $(V_{1}) = 220 V$
Secondary voltage $(V_{2}) = 11 V$
Power output $= 44 W$
$∵$ There is no power loss,
Input power $=$ Output power
$V_{1} I_{1} = 44$
$I_{1} = \frac{44}{220}$
$I_{1} = 0.2 A$

NEET- 2021

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