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Alternating Current

155402 In a step-up transformer the voltage in the primary is $220 V$ and the current is $5 A$ . The secondary voltage is found to be $22000 V$ . The current in the secondary (neglect losses) is

1

5 A

2

50 A

3

500 A

4

0.05 A

Explanation:

D Given that,
Primary voltage $(V_{1}) = 220 V$
Primary current $(I_{1}) = 5 A$ ,
Secondary voltage $(V_{2}) = 22000 V$ For Transformer,
$\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
Putting these value, we get -
$\frac{220}{22000} = \frac{I_{2}}{5}$
$\frac{1}{100} = \frac{I_{2}}{5}$
$I_{2} = \frac{1}{20}$
$I_{2} = 0.05 A$

Manipal UGET-2018

Alternating Current

155403 A transformer of $100 %$ efficiency has 200 turns in the primary and 40000 turns in secondary. It is connected to a $220 V$ main supply and secondary feeds to a $100 k Ω$ resistance. The potential difference per turn is

1

1.1 V

2

25 V

3

18 V

4

11 V

Explanation:

A Given,
Number of turns in primary coil $(N_{1}) = 200$
Number of turns in secondary coil $(N_{2}) = 40000$
Primary voltage $(V_{1}) = 220 V$
Secondary resistance $(R_{2}) = 100 k Ω = 10^{5} Ω$
We know that,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}}$
Or $V_{2} = \frac{V_{1} N_{2}}{N_{1}}$
Putting these value, we get -
$V_{2} = \frac{220 \times 40000}{200}$
$∴$ Potential difference per turn
$\frac{V_{2}}{N_{2}} = \frac{220 \times 40, 000}{200 \times 40, 000}$
$\frac{V_{2}}{N_{2}} = \frac{22}{20} = 1.1 V$

Manipal UGET-2018

Alternating Current

155406 An electric motor driven pump fills an overhead tank placed at a height of $20 m$ from the ground- level pump at a rate of 20,000 liter water per hour. The motor has an operating resistance of $22 Ω$ and is connected across a 220 $V$ source. The efficiency of this motor is (Use $g = 9.8 m / s^{2}$ )

1

12.5 %

2

73.5 %

3

49.5 %

4

22.5 %

Explanation:

C Given,
Rate of volume of water,
$V = 20000$ liter $/ hr = \frac{20000 \times 10^{- 3}}{60 \times 60} m^{3} / s = \frac{20}{3600} m^{3} / s$
Mass of the water pumped per second,
$m = \frac{1000 \times 20}{3600} kg$
We know that, input power of motor -
$P_{I / p} = \frac{V^{2}}{R}$
$P_{I / p} = \frac{(220)^{2}}{22} = \frac{220 \times 220}{22} = 10 \times 220$
$P_{I / p} = 2200 W$
And, output power,
$P_{o / p} = mgh = 1000 \times \frac{20}{3600} \times 9.8 \times 20$
$P_{o / p} = 1088.98 W$
$∴$ Efficiency $(η) = \frac{P_{o / p}}{P_{i / p}} \times 100$
$η = \frac{1088.89}{2200} \times 100$
$η = 49.5 %$

TS EAMCET(Medical)-2017

Alternating Current

155407 The output of a step down transformer is measured to be $48 V$ when connected to a $12 W$ bulb. The value of peak current is :

1

\frac{1}{\sqrt{2}} A

2

\sqrt{2} A

3

\frac{1}{2 \sqrt{2}} A

4

\frac{1}{4} A

Explanation:

C Given,
Output voltage $= 48 V = V_{rms}$
Power $= 12 W$
We know that,
$P = \frac{V_{rms}^{2}}{R}$
$Or R = \frac{V_{rms}^{2}}{P} = \frac{48 \times 48}{12} = 4 \times 48$
$R = 192$
Current in secondary coil,
$I_{rms} = \frac{V_{rms}}{R} = \frac{48}{192} = \frac{1}{4}$
$I_{rms} = \frac{1}{4} A$
$∴$ Value of peak current-
$I_{peak} = \sqrt{2} I_{rms}$
$I_{peak} = \sqrt{2} \times \frac{1}{4}$
$I_{peak} = \frac{\sqrt{2} \times \sqrt{2}}{4 \sqrt{2}} = \frac{1}{2 \sqrt{2}}$
$I_{peak} = \frac{1}{2 \sqrt{2}} A$

Karnataka CET-2017

Alternating Current

155402 In a step-up transformer the voltage in the primary is $220 V$ and the current is $5 A$ . The secondary voltage is found to be $22000 V$ . The current in the secondary (neglect losses) is

1

5 A

2

50 A

3

500 A

4

0.05 A

Explanation:

D Given that,
Primary voltage $(V_{1}) = 220 V$
Primary current $(I_{1}) = 5 A$ ,
Secondary voltage $(V_{2}) = 22000 V$ For Transformer,
$\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
Putting these value, we get -
$\frac{220}{22000} = \frac{I_{2}}{5}$
$\frac{1}{100} = \frac{I_{2}}{5}$
$I_{2} = \frac{1}{20}$
$I_{2} = 0.05 A$

Manipal UGET-2018

Alternating Current

155403 A transformer of $100 %$ efficiency has 200 turns in the primary and 40000 turns in secondary. It is connected to a $220 V$ main supply and secondary feeds to a $100 k Ω$ resistance. The potential difference per turn is

1

1.1 V

2

25 V

3

18 V

4

11 V

Explanation:

A Given,
Number of turns in primary coil $(N_{1}) = 200$
Number of turns in secondary coil $(N_{2}) = 40000$
Primary voltage $(V_{1}) = 220 V$
Secondary resistance $(R_{2}) = 100 k Ω = 10^{5} Ω$
We know that,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}}$
Or $V_{2} = \frac{V_{1} N_{2}}{N_{1}}$
Putting these value, we get -
$V_{2} = \frac{220 \times 40000}{200}$
$∴$ Potential difference per turn
$\frac{V_{2}}{N_{2}} = \frac{220 \times 40, 000}{200 \times 40, 000}$
$\frac{V_{2}}{N_{2}} = \frac{22}{20} = 1.1 V$

Manipal UGET-2018

Alternating Current

155406 An electric motor driven pump fills an overhead tank placed at a height of $20 m$ from the ground- level pump at a rate of 20,000 liter water per hour. The motor has an operating resistance of $22 Ω$ and is connected across a 220 $V$ source. The efficiency of this motor is (Use $g = 9.8 m / s^{2}$ )

1

12.5 %

2

73.5 %

3

49.5 %

4

22.5 %

Explanation:

C Given,
Rate of volume of water,
$V = 20000$ liter $/ hr = \frac{20000 \times 10^{- 3}}{60 \times 60} m^{3} / s = \frac{20}{3600} m^{3} / s$
Mass of the water pumped per second,
$m = \frac{1000 \times 20}{3600} kg$
We know that, input power of motor -
$P_{I / p} = \frac{V^{2}}{R}$
$P_{I / p} = \frac{(220)^{2}}{22} = \frac{220 \times 220}{22} = 10 \times 220$
$P_{I / p} = 2200 W$
And, output power,
$P_{o / p} = mgh = 1000 \times \frac{20}{3600} \times 9.8 \times 20$
$P_{o / p} = 1088.98 W$
$∴$ Efficiency $(η) = \frac{P_{o / p}}{P_{i / p}} \times 100$
$η = \frac{1088.89}{2200} \times 100$
$η = 49.5 %$

TS EAMCET(Medical)-2017

Alternating Current

155407 The output of a step down transformer is measured to be $48 V$ when connected to a $12 W$ bulb. The value of peak current is :

1

\frac{1}{\sqrt{2}} A

2

\sqrt{2} A

3

\frac{1}{2 \sqrt{2}} A

4

\frac{1}{4} A

Explanation:

C Given,
Output voltage $= 48 V = V_{rms}$
Power $= 12 W$
We know that,
$P = \frac{V_{rms}^{2}}{R}$
$Or R = \frac{V_{rms}^{2}}{P} = \frac{48 \times 48}{12} = 4 \times 48$
$R = 192$
Current in secondary coil,
$I_{rms} = \frac{V_{rms}}{R} = \frac{48}{192} = \frac{1}{4}$
$I_{rms} = \frac{1}{4} A$
$∴$ Value of peak current-
$I_{peak} = \sqrt{2} I_{rms}$
$I_{peak} = \sqrt{2} \times \frac{1}{4}$
$I_{peak} = \frac{\sqrt{2} \times \sqrt{2}}{4 \sqrt{2}} = \frac{1}{2 \sqrt{2}}$
$I_{peak} = \frac{1}{2 \sqrt{2}} A$

Karnataka CET-2017

Alternating Current

155402 In a step-up transformer the voltage in the primary is $220 V$ and the current is $5 A$ . The secondary voltage is found to be $22000 V$ . The current in the secondary (neglect losses) is

1

5 A

2

50 A

3

500 A

4

0.05 A

Explanation:

D Given that,
Primary voltage $(V_{1}) = 220 V$
Primary current $(I_{1}) = 5 A$ ,
Secondary voltage $(V_{2}) = 22000 V$ For Transformer,
$\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
Putting these value, we get -
$\frac{220}{22000} = \frac{I_{2}}{5}$
$\frac{1}{100} = \frac{I_{2}}{5}$
$I_{2} = \frac{1}{20}$
$I_{2} = 0.05 A$

Manipal UGET-2018

Alternating Current

155403 A transformer of $100 %$ efficiency has 200 turns in the primary and 40000 turns in secondary. It is connected to a $220 V$ main supply and secondary feeds to a $100 k Ω$ resistance. The potential difference per turn is

1

1.1 V

2

25 V

3

18 V

4

11 V

Explanation:

A Given,
Number of turns in primary coil $(N_{1}) = 200$
Number of turns in secondary coil $(N_{2}) = 40000$
Primary voltage $(V_{1}) = 220 V$
Secondary resistance $(R_{2}) = 100 k Ω = 10^{5} Ω$
We know that,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}}$
Or $V_{2} = \frac{V_{1} N_{2}}{N_{1}}$
Putting these value, we get -
$V_{2} = \frac{220 \times 40000}{200}$
$∴$ Potential difference per turn
$\frac{V_{2}}{N_{2}} = \frac{220 \times 40, 000}{200 \times 40, 000}$
$\frac{V_{2}}{N_{2}} = \frac{22}{20} = 1.1 V$

Manipal UGET-2018

Alternating Current

155406 An electric motor driven pump fills an overhead tank placed at a height of $20 m$ from the ground- level pump at a rate of 20,000 liter water per hour. The motor has an operating resistance of $22 Ω$ and is connected across a 220 $V$ source. The efficiency of this motor is (Use $g = 9.8 m / s^{2}$ )

1

12.5 %

2

73.5 %

3

49.5 %

4

22.5 %

Explanation:

C Given,
Rate of volume of water,
$V = 20000$ liter $/ hr = \frac{20000 \times 10^{- 3}}{60 \times 60} m^{3} / s = \frac{20}{3600} m^{3} / s$
Mass of the water pumped per second,
$m = \frac{1000 \times 20}{3600} kg$
We know that, input power of motor -
$P_{I / p} = \frac{V^{2}}{R}$
$P_{I / p} = \frac{(220)^{2}}{22} = \frac{220 \times 220}{22} = 10 \times 220$
$P_{I / p} = 2200 W$
And, output power,
$P_{o / p} = mgh = 1000 \times \frac{20}{3600} \times 9.8 \times 20$
$P_{o / p} = 1088.98 W$
$∴$ Efficiency $(η) = \frac{P_{o / p}}{P_{i / p}} \times 100$
$η = \frac{1088.89}{2200} \times 100$
$η = 49.5 %$

TS EAMCET(Medical)-2017

Alternating Current

155407 The output of a step down transformer is measured to be $48 V$ when connected to a $12 W$ bulb. The value of peak current is :

1

\frac{1}{\sqrt{2}} A

2

\sqrt{2} A

3

\frac{1}{2 \sqrt{2}} A

4

\frac{1}{4} A

Explanation:

C Given,
Output voltage $= 48 V = V_{rms}$
Power $= 12 W$
We know that,
$P = \frac{V_{rms}^{2}}{R}$
$Or R = \frac{V_{rms}^{2}}{P} = \frac{48 \times 48}{12} = 4 \times 48$
$R = 192$
Current in secondary coil,
$I_{rms} = \frac{V_{rms}}{R} = \frac{48}{192} = \frac{1}{4}$
$I_{rms} = \frac{1}{4} A$
$∴$ Value of peak current-
$I_{peak} = \sqrt{2} I_{rms}$
$I_{peak} = \sqrt{2} \times \frac{1}{4}$
$I_{peak} = \frac{\sqrt{2} \times \sqrt{2}}{4 \sqrt{2}} = \frac{1}{2 \sqrt{2}}$
$I_{peak} = \frac{1}{2 \sqrt{2}} A$

Karnataka CET-2017

Alternating Current

155402 In a step-up transformer the voltage in the primary is $220 V$ and the current is $5 A$ . The secondary voltage is found to be $22000 V$ . The current in the secondary (neglect losses) is

1

5 A

2

50 A

3

500 A

4

0.05 A

Explanation:

D Given that,
Primary voltage $(V_{1}) = 220 V$
Primary current $(I_{1}) = 5 A$ ,
Secondary voltage $(V_{2}) = 22000 V$ For Transformer,
$\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
Putting these value, we get -
$\frac{220}{22000} = \frac{I_{2}}{5}$
$\frac{1}{100} = \frac{I_{2}}{5}$
$I_{2} = \frac{1}{20}$
$I_{2} = 0.05 A$

Manipal UGET-2018

Alternating Current

155403 A transformer of $100 %$ efficiency has 200 turns in the primary and 40000 turns in secondary. It is connected to a $220 V$ main supply and secondary feeds to a $100 k Ω$ resistance. The potential difference per turn is

1

1.1 V

2

25 V

3

18 V

4

11 V

Explanation:

A Given,
Number of turns in primary coil $(N_{1}) = 200$
Number of turns in secondary coil $(N_{2}) = 40000$
Primary voltage $(V_{1}) = 220 V$
Secondary resistance $(R_{2}) = 100 k Ω = 10^{5} Ω$
We know that,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}}$
Or $V_{2} = \frac{V_{1} N_{2}}{N_{1}}$
Putting these value, we get -
$V_{2} = \frac{220 \times 40000}{200}$
$∴$ Potential difference per turn
$\frac{V_{2}}{N_{2}} = \frac{220 \times 40, 000}{200 \times 40, 000}$
$\frac{V_{2}}{N_{2}} = \frac{22}{20} = 1.1 V$

Manipal UGET-2018

Alternating Current

155406 An electric motor driven pump fills an overhead tank placed at a height of $20 m$ from the ground- level pump at a rate of 20,000 liter water per hour. The motor has an operating resistance of $22 Ω$ and is connected across a 220 $V$ source. The efficiency of this motor is (Use $g = 9.8 m / s^{2}$ )

1

12.5 %

2

73.5 %

3

49.5 %

4

22.5 %

Explanation:

C Given,
Rate of volume of water,
$V = 20000$ liter $/ hr = \frac{20000 \times 10^{- 3}}{60 \times 60} m^{3} / s = \frac{20}{3600} m^{3} / s$
Mass of the water pumped per second,
$m = \frac{1000 \times 20}{3600} kg$
We know that, input power of motor -
$P_{I / p} = \frac{V^{2}}{R}$
$P_{I / p} = \frac{(220)^{2}}{22} = \frac{220 \times 220}{22} = 10 \times 220$
$P_{I / p} = 2200 W$
And, output power,
$P_{o / p} = mgh = 1000 \times \frac{20}{3600} \times 9.8 \times 20$
$P_{o / p} = 1088.98 W$
$∴$ Efficiency $(η) = \frac{P_{o / p}}{P_{i / p}} \times 100$
$η = \frac{1088.89}{2200} \times 100$
$η = 49.5 %$

TS EAMCET(Medical)-2017

Alternating Current

155407 The output of a step down transformer is measured to be $48 V$ when connected to a $12 W$ bulb. The value of peak current is :

1

\frac{1}{\sqrt{2}} A

2

\sqrt{2} A

3

\frac{1}{2 \sqrt{2}} A

4

\frac{1}{4} A

Explanation:

C Given,
Output voltage $= 48 V = V_{rms}$
Power $= 12 W$
We know that,
$P = \frac{V_{rms}^{2}}{R}$
$Or R = \frac{V_{rms}^{2}}{P} = \frac{48 \times 48}{12} = 4 \times 48$
$R = 192$
Current in secondary coil,
$I_{rms} = \frac{V_{rms}}{R} = \frac{48}{192} = \frac{1}{4}$
$I_{rms} = \frac{1}{4} A$
$∴$ Value of peak current-
$I_{peak} = \sqrt{2} I_{rms}$
$I_{peak} = \sqrt{2} \times \frac{1}{4}$
$I_{peak} = \frac{\sqrt{2} \times \sqrt{2}}{4 \sqrt{2}} = \frac{1}{2 \sqrt{2}}$
$I_{peak} = \frac{1}{2 \sqrt{2}} A$

Karnataka CET-2017

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