QBManager

Alternating Current

155382 A step-up transformer has 300 turns of primary winding and 450 turns of secondary winding. A primary is connected to 150 volt and the current flowing through it is $9 A$ . The current and voltage in the secondary are

1

4.5 A, 100 V

2

13.5 A, 225 V

3

13.5 A, 100 V

4

6.0 A, 225 V

Explanation:

D Given that,
$N_{S} = 450 turns$
$N_{P} = 300 turns$
$V_{P} = 150 V$
$I_{P} = 9 A$
From transformer ratio,
$\frac{N_{S}}{N_{P}} = \frac{V_{S}}{V_{P}} = \frac{I_{P}}{I_{S}}$
$So, \frac{N_{S}}{N_{P}} = \frac{V_{S}}{V_{P}}$
$\frac{450}{300} = \frac{V_{S}}{150}$
$\frac{150 \times 450}{300} = V_{S}$
$Now, \frac{V_{S}}{V_{P}} = \frac{I_{P}}{I_{S}}$
$I_{S} = I_{P} \times \frac{V_{P}}{V_{S}}$
$I_{S} = \frac{9 \times 150}{225 V}$
$I_{S} = 6.0 A$

MHT-CET 2020

Alternating Current

155385 The primary of a transformer has 100 turns and operates at $100 V - 200 W$ . The number of turns in the secondary, if the output voltage is $2000 V$ , then

1 2000

2 200

3 100

4 500

Explanation:

A Given that,
Primary turns $(N_{P}) = 100$
Primary voltage $(V_{P}) = 100 V$
Secondary voltage $(V_{S}) = 2000 V$
From equation of transformer,
$\frac{N_{S}}{N_{P}} = \frac{V_{S}}{V_{P}}$
$N_{S} = \frac{N_{P} \times V_{S}}{V_{P}} = \frac{100 \times 2000}{100}$
$N_{S} = 2000$

AP EAMCET (21.09.2020) Shift-II

Alternating Current

155386 Find the current through the primary coil $(P)$ of the transformer shown below.

1

0.08 A

2

0.04 A

3

0.02 A

4

0.01 A

Explanation:

C Given that,
Primary voltage $(V_{1}) = 230 V$ ,
Secondary voltage $(V_{2}) = 23 V$ ,
Resistance $(R) = 115 Ω$
Current in secondary $(I_{2}) = \frac{V_{2}}{R} = \frac{23}{115}$
From voltage current relation in transformer
$\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$I_{1} = \frac{I_{2} V_{2}}{V_{1}} = \frac{23}{115} \times \frac{23}{230}$
$I_{1} = 0.02 A$

AP EAMCET (21.09.2020) Shift-I

Alternating Current

155387 An ideal transformer has 500 turns in the primary and 2500 in the secondary. The meters of the secondary are indicating $200 V, 8$ $A$ , under these conditions. What would the meters of the primary read?

1

100 V, 16 A

2

40 V, 40 A

3

160 V, 10 A

4

80 V, 20 A

Explanation:

B Given that,
Primary turns $(N_{1}) = 500$
Secondary turns $(N_{2}) = 2500$
Secondary voltage $(V_{2}) = 200 V$
Current in secondary $(I_{2}) = 8 A$
We know that,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}}$
$\frac{500}{2500} = \frac{V_{1}}{200}$
$V_{1} = \frac{500 \times 200}{2500} = 40 V$
And $\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$\frac{40}{200} = \frac{8}{I_{1}}$
$I_{1} = \frac{8 \times 200}{40} = 40 A$
$I_{1} = 40 A$
Hence, the reading on primary meter is $40 V, 40 A$ .

CG PET- 2006

Alternating Current

155382 A step-up transformer has 300 turns of primary winding and 450 turns of secondary winding. A primary is connected to 150 volt and the current flowing through it is $9 A$ . The current and voltage in the secondary are

1

4.5 A, 100 V

2

13.5 A, 225 V

3

13.5 A, 100 V

4

6.0 A, 225 V

Explanation:

D Given that,
$N_{S} = 450 turns$
$N_{P} = 300 turns$
$V_{P} = 150 V$
$I_{P} = 9 A$
From transformer ratio,
$\frac{N_{S}}{N_{P}} = \frac{V_{S}}{V_{P}} = \frac{I_{P}}{I_{S}}$
$So, \frac{N_{S}}{N_{P}} = \frac{V_{S}}{V_{P}}$
$\frac{450}{300} = \frac{V_{S}}{150}$
$\frac{150 \times 450}{300} = V_{S}$
$Now, \frac{V_{S}}{V_{P}} = \frac{I_{P}}{I_{S}}$
$I_{S} = I_{P} \times \frac{V_{P}}{V_{S}}$
$I_{S} = \frac{9 \times 150}{225 V}$
$I_{S} = 6.0 A$

MHT-CET 2020

Alternating Current

155385 The primary of a transformer has 100 turns and operates at $100 V - 200 W$ . The number of turns in the secondary, if the output voltage is $2000 V$ , then

1 2000

2 200

3 100

4 500

Explanation:

A Given that,
Primary turns $(N_{P}) = 100$
Primary voltage $(V_{P}) = 100 V$
Secondary voltage $(V_{S}) = 2000 V$
From equation of transformer,
$\frac{N_{S}}{N_{P}} = \frac{V_{S}}{V_{P}}$
$N_{S} = \frac{N_{P} \times V_{S}}{V_{P}} = \frac{100 \times 2000}{100}$
$N_{S} = 2000$

AP EAMCET (21.09.2020) Shift-II

Alternating Current

155386 Find the current through the primary coil $(P)$ of the transformer shown below.

1

0.08 A

2

0.04 A

3

0.02 A

4

0.01 A

Explanation:

C Given that,
Primary voltage $(V_{1}) = 230 V$ ,
Secondary voltage $(V_{2}) = 23 V$ ,
Resistance $(R) = 115 Ω$
Current in secondary $(I_{2}) = \frac{V_{2}}{R} = \frac{23}{115}$
From voltage current relation in transformer
$\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$I_{1} = \frac{I_{2} V_{2}}{V_{1}} = \frac{23}{115} \times \frac{23}{230}$
$I_{1} = 0.02 A$

AP EAMCET (21.09.2020) Shift-I

Alternating Current

155387 An ideal transformer has 500 turns in the primary and 2500 in the secondary. The meters of the secondary are indicating $200 V, 8$ $A$ , under these conditions. What would the meters of the primary read?

1

100 V, 16 A

2

40 V, 40 A

3

160 V, 10 A

4

80 V, 20 A

Explanation:

B Given that,
Primary turns $(N_{1}) = 500$
Secondary turns $(N_{2}) = 2500$
Secondary voltage $(V_{2}) = 200 V$
Current in secondary $(I_{2}) = 8 A$
We know that,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}}$
$\frac{500}{2500} = \frac{V_{1}}{200}$
$V_{1} = \frac{500 \times 200}{2500} = 40 V$
And $\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$\frac{40}{200} = \frac{8}{I_{1}}$
$I_{1} = \frac{8 \times 200}{40} = 40 A$
$I_{1} = 40 A$
Hence, the reading on primary meter is $40 V, 40 A$ .

CG PET- 2006

Alternating Current

155382 A step-up transformer has 300 turns of primary winding and 450 turns of secondary winding. A primary is connected to 150 volt and the current flowing through it is $9 A$ . The current and voltage in the secondary are

1

4.5 A, 100 V

2

13.5 A, 225 V

3

13.5 A, 100 V

4

6.0 A, 225 V

Explanation:

D Given that,
$N_{S} = 450 turns$
$N_{P} = 300 turns$
$V_{P} = 150 V$
$I_{P} = 9 A$
From transformer ratio,
$\frac{N_{S}}{N_{P}} = \frac{V_{S}}{V_{P}} = \frac{I_{P}}{I_{S}}$
$So, \frac{N_{S}}{N_{P}} = \frac{V_{S}}{V_{P}}$
$\frac{450}{300} = \frac{V_{S}}{150}$
$\frac{150 \times 450}{300} = V_{S}$
$Now, \frac{V_{S}}{V_{P}} = \frac{I_{P}}{I_{S}}$
$I_{S} = I_{P} \times \frac{V_{P}}{V_{S}}$
$I_{S} = \frac{9 \times 150}{225 V}$
$I_{S} = 6.0 A$

MHT-CET 2020

Alternating Current

155385 The primary of a transformer has 100 turns and operates at $100 V - 200 W$ . The number of turns in the secondary, if the output voltage is $2000 V$ , then

1 2000

2 200

3 100

4 500

Explanation:

A Given that,
Primary turns $(N_{P}) = 100$
Primary voltage $(V_{P}) = 100 V$
Secondary voltage $(V_{S}) = 2000 V$
From equation of transformer,
$\frac{N_{S}}{N_{P}} = \frac{V_{S}}{V_{P}}$
$N_{S} = \frac{N_{P} \times V_{S}}{V_{P}} = \frac{100 \times 2000}{100}$
$N_{S} = 2000$

AP EAMCET (21.09.2020) Shift-II

Alternating Current

155386 Find the current through the primary coil $(P)$ of the transformer shown below.

1

0.08 A

2

0.04 A

3

0.02 A

4

0.01 A

Explanation:

C Given that,
Primary voltage $(V_{1}) = 230 V$ ,
Secondary voltage $(V_{2}) = 23 V$ ,
Resistance $(R) = 115 Ω$
Current in secondary $(I_{2}) = \frac{V_{2}}{R} = \frac{23}{115}$
From voltage current relation in transformer
$\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$I_{1} = \frac{I_{2} V_{2}}{V_{1}} = \frac{23}{115} \times \frac{23}{230}$
$I_{1} = 0.02 A$

AP EAMCET (21.09.2020) Shift-I

Alternating Current

155387 An ideal transformer has 500 turns in the primary and 2500 in the secondary. The meters of the secondary are indicating $200 V, 8$ $A$ , under these conditions. What would the meters of the primary read?

1

100 V, 16 A

2

40 V, 40 A

3

160 V, 10 A

4

80 V, 20 A

Explanation:

B Given that,
Primary turns $(N_{1}) = 500$
Secondary turns $(N_{2}) = 2500$
Secondary voltage $(V_{2}) = 200 V$
Current in secondary $(I_{2}) = 8 A$
We know that,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}}$
$\frac{500}{2500} = \frac{V_{1}}{200}$
$V_{1} = \frac{500 \times 200}{2500} = 40 V$
And $\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$\frac{40}{200} = \frac{8}{I_{1}}$
$I_{1} = \frac{8 \times 200}{40} = 40 A$
$I_{1} = 40 A$
Hence, the reading on primary meter is $40 V, 40 A$ .

CG PET- 2006

Alternating Current

155382 A step-up transformer has 300 turns of primary winding and 450 turns of secondary winding. A primary is connected to 150 volt and the current flowing through it is $9 A$ . The current and voltage in the secondary are

1

4.5 A, 100 V

2

13.5 A, 225 V

3

13.5 A, 100 V

4

6.0 A, 225 V

Explanation:

D Given that,
$N_{S} = 450 turns$
$N_{P} = 300 turns$
$V_{P} = 150 V$
$I_{P} = 9 A$
From transformer ratio,
$\frac{N_{S}}{N_{P}} = \frac{V_{S}}{V_{P}} = \frac{I_{P}}{I_{S}}$
$So, \frac{N_{S}}{N_{P}} = \frac{V_{S}}{V_{P}}$
$\frac{450}{300} = \frac{V_{S}}{150}$
$\frac{150 \times 450}{300} = V_{S}$
$Now, \frac{V_{S}}{V_{P}} = \frac{I_{P}}{I_{S}}$
$I_{S} = I_{P} \times \frac{V_{P}}{V_{S}}$
$I_{S} = \frac{9 \times 150}{225 V}$
$I_{S} = 6.0 A$

MHT-CET 2020

Alternating Current

155385 The primary of a transformer has 100 turns and operates at $100 V - 200 W$ . The number of turns in the secondary, if the output voltage is $2000 V$ , then

1 2000

2 200

3 100

4 500

Explanation:

A Given that,
Primary turns $(N_{P}) = 100$
Primary voltage $(V_{P}) = 100 V$
Secondary voltage $(V_{S}) = 2000 V$
From equation of transformer,
$\frac{N_{S}}{N_{P}} = \frac{V_{S}}{V_{P}}$
$N_{S} = \frac{N_{P} \times V_{S}}{V_{P}} = \frac{100 \times 2000}{100}$
$N_{S} = 2000$

AP EAMCET (21.09.2020) Shift-II

Alternating Current

155386 Find the current through the primary coil $(P)$ of the transformer shown below.

1

0.08 A

2

0.04 A

3

0.02 A

4

0.01 A

Explanation:

C Given that,
Primary voltage $(V_{1}) = 230 V$ ,
Secondary voltage $(V_{2}) = 23 V$ ,
Resistance $(R) = 115 Ω$
Current in secondary $(I_{2}) = \frac{V_{2}}{R} = \frac{23}{115}$
From voltage current relation in transformer
$\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$I_{1} = \frac{I_{2} V_{2}}{V_{1}} = \frac{23}{115} \times \frac{23}{230}$
$I_{1} = 0.02 A$

AP EAMCET (21.09.2020) Shift-I

Alternating Current

155387 An ideal transformer has 500 turns in the primary and 2500 in the secondary. The meters of the secondary are indicating $200 V, 8$ $A$ , under these conditions. What would the meters of the primary read?

1

100 V, 16 A

2

40 V, 40 A

3

160 V, 10 A

4

80 V, 20 A

Explanation:

B Given that,
Primary turns $(N_{1}) = 500$
Secondary turns $(N_{2}) = 2500$
Secondary voltage $(V_{2}) = 200 V$
Current in secondary $(I_{2}) = 8 A$
We know that,
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}}$
$\frac{500}{2500} = \frac{V_{1}}{200}$
$V_{1} = \frac{500 \times 200}{2500} = 40 V$
And $\frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}$
$\frac{40}{200} = \frac{8}{I_{1}}$
$I_{1} = \frac{8 \times 200}{40} = 40 A$
$I_{1} = 40 A$
Hence, the reading on primary meter is $40 V, 40 A$ .

CG PET- 2006

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: