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Alternating Current

155443 A transformer rated at $10 kW$ is used to connect a $5 kV$ transmission line to a $240 V$ circuit. The ratio of turns in the windings of the transformer is

1 5

2 20.8

3 104

4 40

Explanation:

B Given that,
Primary voltage $(V_{1}) = 5 kV$
Secondary voltage $(V_{2}) = 240 V$
The ratio of the turns in the windings, $\frac{N_{1}}{N_{2}} =$ ?
From the transformer ratio,
$\frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$
$\frac{N_{1}}{N_{2}} = \frac{5 kV}{240 V} = \frac{5 \times 1000}{240}$
$\frac{N_{1}}{N_{2}} = 20.8$

VITEEE-2008

Alternating Current

155444 A 400 turns primary coil of an ideal transformer is connected to an alternating current power line of $120 V$ . A secondary coil of 100 turns is connected to a light bulb of $60 Ω$ resistance. The maximum current in the secondary would be

1

2 A

2

1 A

3

0.5 A

4

0.25 A

Explanation:

C Given that,
Primary turns $(N_{1}) = 400$
Secondary turns $(N_{2}) = 100$
Primary voltage $(V_{1}) = 120 V$
From the transformer ratio,
$\frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$
$V_{2} = \frac{V_{1} N_{2}}{N_{1}}$
$V_{2} = \frac{120 \times 100}{400} = 30 V$
$∵$ Resistance of bulb connected in secondary coil $= 60 Ω$
$∴$ Current in secondary coil $(I_{2}) = \frac{V_{2}}{R} = \frac{30}{60} = 0.5 A$
$I_{2} = 0.5 A$

SCRA-2013

Alternating Current

155445 A step down transformer has 50 turns on secondary and 1000 turns on primary winding. If a transformer is connected to $220 V, 1 A$ A.C. source, what is output current of the transformer?

1

\frac{1}{20} A

2

20 A

3

100 A

4

2 A

Explanation:

B Given that,
Secondary turns $(N_{2}) = 50$
Primary turns $(N_{1}) = 1000$
Primary voltage $(V_{1}) = 220 V$
Primary current $(I_{1}) = 1 A$
Secondary current $(I_{2}) =$ ?
From the transformer ratio,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$\frac{1000}{50} = \frac{I_{2}}{1}$
$I_{2} = 20 A$

Karnataka CET-2014

Alternating Current

155446 For a transformer, the turns ratio is 3 and its efficiency is 0.75 . The current flowing in the primary coil is $2 A$ and the voltage applied to it is $100 V$ . Then the voltage and the current flowing in the secondary coil are respectively

1

150 V, 1.5 A

2

300 V, 0.5 A

3

300 V, 1.5 A

4

150 V, 0.5 A

Explanation:

B Given,
Current in the primary $(I_{1}) = 2 A$
Current in the secondary $(I_{2}) =$ ?
Primary voltage $(V_{1}) = 100 V$
Secondary voltage $(V_{2}) =$ ?
Turn Ratio, $\frac{N_{2}}{N_{1}} = \frac{3}{1}$
From the transformer ratio,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$\frac{V_{1}}{V_{2}} = \frac{1}{3}$
$V_{2} = 3 V_{1} = 3 \times 100$
$V_{2} = 300 V$
$∵ Efficiency = \frac{Output power}{Input power}$
$∴ 0.75 = \frac{V_{2} I_{2}}{V_{1} I_{1}}$
$I_{2} = \frac{0.75 \times V_{1} I_{1}}{V_{2}} = \frac{0.75 \times 100 \times 2}{300}$
$I_{2} = 0.25 \times 2$
$I_{2} = 0.50 A$
$V_{2} = 300 V$
Hence, option (b) is correct.

Karnataka CET-2013

Alternating Current

155443 A transformer rated at $10 kW$ is used to connect a $5 kV$ transmission line to a $240 V$ circuit. The ratio of turns in the windings of the transformer is

1 5

2 20.8

3 104

4 40

Explanation:

B Given that,
Primary voltage $(V_{1}) = 5 kV$
Secondary voltage $(V_{2}) = 240 V$
The ratio of the turns in the windings, $\frac{N_{1}}{N_{2}} =$ ?
From the transformer ratio,
$\frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$
$\frac{N_{1}}{N_{2}} = \frac{5 kV}{240 V} = \frac{5 \times 1000}{240}$
$\frac{N_{1}}{N_{2}} = 20.8$

VITEEE-2008

Alternating Current

155444 A 400 turns primary coil of an ideal transformer is connected to an alternating current power line of $120 V$ . A secondary coil of 100 turns is connected to a light bulb of $60 Ω$ resistance. The maximum current in the secondary would be

1

2 A

2

1 A

3

0.5 A

4

0.25 A

Explanation:

C Given that,
Primary turns $(N_{1}) = 400$
Secondary turns $(N_{2}) = 100$
Primary voltage $(V_{1}) = 120 V$
From the transformer ratio,
$\frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$
$V_{2} = \frac{V_{1} N_{2}}{N_{1}}$
$V_{2} = \frac{120 \times 100}{400} = 30 V$
$∵$ Resistance of bulb connected in secondary coil $= 60 Ω$
$∴$ Current in secondary coil $(I_{2}) = \frac{V_{2}}{R} = \frac{30}{60} = 0.5 A$
$I_{2} = 0.5 A$

SCRA-2013

Alternating Current

155445 A step down transformer has 50 turns on secondary and 1000 turns on primary winding. If a transformer is connected to $220 V, 1 A$ A.C. source, what is output current of the transformer?

1

\frac{1}{20} A

2

20 A

3

100 A

4

2 A

Explanation:

B Given that,
Secondary turns $(N_{2}) = 50$
Primary turns $(N_{1}) = 1000$
Primary voltage $(V_{1}) = 220 V$
Primary current $(I_{1}) = 1 A$
Secondary current $(I_{2}) =$ ?
From the transformer ratio,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$\frac{1000}{50} = \frac{I_{2}}{1}$
$I_{2} = 20 A$

Karnataka CET-2014

Alternating Current

155446 For a transformer, the turns ratio is 3 and its efficiency is 0.75 . The current flowing in the primary coil is $2 A$ and the voltage applied to it is $100 V$ . Then the voltage and the current flowing in the secondary coil are respectively

1

150 V, 1.5 A

2

300 V, 0.5 A

3

300 V, 1.5 A

4

150 V, 0.5 A

Explanation:

B Given,
Current in the primary $(I_{1}) = 2 A$
Current in the secondary $(I_{2}) =$ ?
Primary voltage $(V_{1}) = 100 V$
Secondary voltage $(V_{2}) =$ ?
Turn Ratio, $\frac{N_{2}}{N_{1}} = \frac{3}{1}$
From the transformer ratio,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$\frac{V_{1}}{V_{2}} = \frac{1}{3}$
$V_{2} = 3 V_{1} = 3 \times 100$
$V_{2} = 300 V$
$∵ Efficiency = \frac{Output power}{Input power}$
$∴ 0.75 = \frac{V_{2} I_{2}}{V_{1} I_{1}}$
$I_{2} = \frac{0.75 \times V_{1} I_{1}}{V_{2}} = \frac{0.75 \times 100 \times 2}{300}$
$I_{2} = 0.25 \times 2$
$I_{2} = 0.50 A$
$V_{2} = 300 V$
Hence, option (b) is correct.

Karnataka CET-2013

Alternating Current

155443 A transformer rated at $10 kW$ is used to connect a $5 kV$ transmission line to a $240 V$ circuit. The ratio of turns in the windings of the transformer is

1 5

2 20.8

3 104

4 40

Explanation:

B Given that,
Primary voltage $(V_{1}) = 5 kV$
Secondary voltage $(V_{2}) = 240 V$
The ratio of the turns in the windings, $\frac{N_{1}}{N_{2}} =$ ?
From the transformer ratio,
$\frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$
$\frac{N_{1}}{N_{2}} = \frac{5 kV}{240 V} = \frac{5 \times 1000}{240}$
$\frac{N_{1}}{N_{2}} = 20.8$

VITEEE-2008

Alternating Current

155444 A 400 turns primary coil of an ideal transformer is connected to an alternating current power line of $120 V$ . A secondary coil of 100 turns is connected to a light bulb of $60 Ω$ resistance. The maximum current in the secondary would be

1

2 A

2

1 A

3

0.5 A

4

0.25 A

Explanation:

C Given that,
Primary turns $(N_{1}) = 400$
Secondary turns $(N_{2}) = 100$
Primary voltage $(V_{1}) = 120 V$
From the transformer ratio,
$\frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$
$V_{2} = \frac{V_{1} N_{2}}{N_{1}}$
$V_{2} = \frac{120 \times 100}{400} = 30 V$
$∵$ Resistance of bulb connected in secondary coil $= 60 Ω$
$∴$ Current in secondary coil $(I_{2}) = \frac{V_{2}}{R} = \frac{30}{60} = 0.5 A$
$I_{2} = 0.5 A$

SCRA-2013

Alternating Current

155445 A step down transformer has 50 turns on secondary and 1000 turns on primary winding. If a transformer is connected to $220 V, 1 A$ A.C. source, what is output current of the transformer?

1

\frac{1}{20} A

2

20 A

3

100 A

4

2 A

Explanation:

B Given that,
Secondary turns $(N_{2}) = 50$
Primary turns $(N_{1}) = 1000$
Primary voltage $(V_{1}) = 220 V$
Primary current $(I_{1}) = 1 A$
Secondary current $(I_{2}) =$ ?
From the transformer ratio,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$\frac{1000}{50} = \frac{I_{2}}{1}$
$I_{2} = 20 A$

Karnataka CET-2014

Alternating Current

155446 For a transformer, the turns ratio is 3 and its efficiency is 0.75 . The current flowing in the primary coil is $2 A$ and the voltage applied to it is $100 V$ . Then the voltage and the current flowing in the secondary coil are respectively

1

150 V, 1.5 A

2

300 V, 0.5 A

3

300 V, 1.5 A

4

150 V, 0.5 A

Explanation:

B Given,
Current in the primary $(I_{1}) = 2 A$
Current in the secondary $(I_{2}) =$ ?
Primary voltage $(V_{1}) = 100 V$
Secondary voltage $(V_{2}) =$ ?
Turn Ratio, $\frac{N_{2}}{N_{1}} = \frac{3}{1}$
From the transformer ratio,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$\frac{V_{1}}{V_{2}} = \frac{1}{3}$
$V_{2} = 3 V_{1} = 3 \times 100$
$V_{2} = 300 V$
$∵ Efficiency = \frac{Output power}{Input power}$
$∴ 0.75 = \frac{V_{2} I_{2}}{V_{1} I_{1}}$
$I_{2} = \frac{0.75 \times V_{1} I_{1}}{V_{2}} = \frac{0.75 \times 100 \times 2}{300}$
$I_{2} = 0.25 \times 2$
$I_{2} = 0.50 A$
$V_{2} = 300 V$
Hence, option (b) is correct.

Karnataka CET-2013

Alternating Current

155443 A transformer rated at $10 kW$ is used to connect a $5 kV$ transmission line to a $240 V$ circuit. The ratio of turns in the windings of the transformer is

1 5

2 20.8

3 104

4 40

Explanation:

B Given that,
Primary voltage $(V_{1}) = 5 kV$
Secondary voltage $(V_{2}) = 240 V$
The ratio of the turns in the windings, $\frac{N_{1}}{N_{2}} =$ ?
From the transformer ratio,
$\frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$
$\frac{N_{1}}{N_{2}} = \frac{5 kV}{240 V} = \frac{5 \times 1000}{240}$
$\frac{N_{1}}{N_{2}} = 20.8$

VITEEE-2008

Alternating Current

155444 A 400 turns primary coil of an ideal transformer is connected to an alternating current power line of $120 V$ . A secondary coil of 100 turns is connected to a light bulb of $60 Ω$ resistance. The maximum current in the secondary would be

1

2 A

2

1 A

3

0.5 A

4

0.25 A

Explanation:

C Given that,
Primary turns $(N_{1}) = 400$
Secondary turns $(N_{2}) = 100$
Primary voltage $(V_{1}) = 120 V$
From the transformer ratio,
$\frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$
$V_{2} = \frac{V_{1} N_{2}}{N_{1}}$
$V_{2} = \frac{120 \times 100}{400} = 30 V$
$∵$ Resistance of bulb connected in secondary coil $= 60 Ω$
$∴$ Current in secondary coil $(I_{2}) = \frac{V_{2}}{R} = \frac{30}{60} = 0.5 A$
$I_{2} = 0.5 A$

SCRA-2013

Alternating Current

155445 A step down transformer has 50 turns on secondary and 1000 turns on primary winding. If a transformer is connected to $220 V, 1 A$ A.C. source, what is output current of the transformer?

1

\frac{1}{20} A

2

20 A

3

100 A

4

2 A

Explanation:

B Given that,
Secondary turns $(N_{2}) = 50$
Primary turns $(N_{1}) = 1000$
Primary voltage $(V_{1}) = 220 V$
Primary current $(I_{1}) = 1 A$
Secondary current $(I_{2}) =$ ?
From the transformer ratio,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$\frac{1000}{50} = \frac{I_{2}}{1}$
$I_{2} = 20 A$

Karnataka CET-2014

Alternating Current

155446 For a transformer, the turns ratio is 3 and its efficiency is 0.75 . The current flowing in the primary coil is $2 A$ and the voltage applied to it is $100 V$ . Then the voltage and the current flowing in the secondary coil are respectively

1

150 V, 1.5 A

2

300 V, 0.5 A

3

300 V, 1.5 A

4

150 V, 0.5 A

Explanation:

B Given,
Current in the primary $(I_{1}) = 2 A$
Current in the secondary $(I_{2}) =$ ?
Primary voltage $(V_{1}) = 100 V$
Secondary voltage $(V_{2}) =$ ?
Turn Ratio, $\frac{N_{2}}{N_{1}} = \frac{3}{1}$
From the transformer ratio,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$\frac{V_{1}}{V_{2}} = \frac{1}{3}$
$V_{2} = 3 V_{1} = 3 \times 100$
$V_{2} = 300 V$
$∵ Efficiency = \frac{Output power}{Input power}$
$∴ 0.75 = \frac{V_{2} I_{2}}{V_{1} I_{1}}$
$I_{2} = \frac{0.75 \times V_{1} I_{1}}{V_{2}} = \frac{0.75 \times 100 \times 2}{300}$
$I_{2} = 0.25 \times 2$
$I_{2} = 0.50 A$
$V_{2} = 300 V$
Hence, option (b) is correct.

Karnataka CET-2013