155451
The number of turns in the primary and the secondary turns of a transformer are 1000 and 3000 respectively. If $80 \mathrm{~V}$ A.C. is applied to the primary coil of the transformer, then the potential difference per turn of secondary coil is
1 0.24 volt
2 0.08 volt
3 240 volt
4 2400 volt
Explanation:
B Given Number of primary turns $\mathrm{N}_{1}=1000$ Number of secondary turns $\mathrm{N}_{2}=3000$ Primary voltage $\mathrm{V}_{1}=80 \mathrm{~V}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\therefore \quad \mathrm{V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}=\frac{80 \times 3000}{1000}=240 \mathrm{~V}$ $\mathrm{V}_{2}=240$ $\therefore \quad$ Voltage/turn $=\frac{\mathrm{V}_{2}}{\mathrm{~N}_{2}}=\frac{240}{3000}$ Voltage/turns $=0.08$
J and K CET- 2001
Alternating Current
155454
The number of turns in the primary coil of a transformer is 200 and the number of turns in secondary coil is 10 .If $240 \mathrm{~V}$ A.C is applied to the primary, the output from secondary will be
1 $48 \mathrm{~V}$
2 $24 \mathrm{~V}$
3 $12 \mathrm{~V}$
4 $6 \mathrm{~V}$
Explanation:
C Given, Primary turns $\left(\mathrm{N}_{1}\right)=200$ Secondary turns $\left(\mathrm{N}_{2}\right)=10$ Primary voltage $\left(\mathrm{V}_{1}\right)=240 \mathrm{~V}$ We know that, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ Or $\quad \mathrm{V}_{2} =\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}=\frac{240 \times 10}{200}$ $\mathrm{~V}_{2} =12 \mathrm{~V}$
J and K CET- 1999
Alternating Current
155455
A transformer has 200 primary turns and 150 secondary turns. If $400 \mathrm{~V}$ are applied in primary, the voltage in secondary will be
1 $6 \mathrm{~V}$
2 $300 \mathrm{~V}$
3 $80,000 \mathrm{~V}$
4 $2 \mathrm{~V}$
Explanation:
B Given, Primary voltage $\left(\mathrm{V}_{1}\right)=400 \mathrm{~V}$ Primary turn $\left(\mathrm{N}_{1}\right)=200$ Secondary turns $\left(\mathrm{N}_{2}\right)=150$ We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\mathrm{~V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}=\frac{400 \times 150}{200}$ $\mathrm{~V}_{2}=300 \mathrm{~V}$
J and K CET- 1998
Alternating Current
155456
A transformer is used to reduce the voltage from $230 \mathrm{~V}$ to $6 \mathrm{~V}$. The number of turns on the secondary is 48 . Then the number of turns on the primary is
1 700
2 1840
3 3400
4 340
Explanation:
B Given, Primary voltage $\left(\mathrm{V}_{1}\right)=230 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=6 \mathrm{~V}$ Number of turns in the secondary coil $\left(\mathrm{N}_{2}\right)=48 \mathrm{~V}$ $\because \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\mathrm{N}_{1}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~V}_{2}}$ $\mathrm{N}_{1}=\frac{230 \times 48}{6}$ $\mathrm{N}_{1}=1840$ turns
J and K CET- 1997
Alternating Current
155458
In a transformer the number of primary turns is four times that of the secondary turns. If primary is connected to an a.c. source of voltage $V$. Then
1 Current through its secondary is about four times that of the current through its primary
2 voltage across its secondary is about four times that of the voltage across its primary
3 voltage across its secondary is about two times that of the voltage across its primary
4 voltage across its secondary is about $\frac{1}{2 \sqrt{2}}$ times that of the voltage across its primary
Explanation:
A We know that, $\because \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ Given, $\therefore \quad \mathrm{N}_{1}=4 \mathrm{~N}_{2}$ $\quad \frac{4 \mathrm{~N}_{2}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=4$ $\mathrm{~V}_{1}=4 \mathrm{~V}_{2}$ Current through secondary, $\frac{V_{1}}{V_{2}}=\frac{I_{2}}{I_{1}}$ $\frac{4 V_{2}}{V_{2}}=\frac{I_{2}}{I_{1}}$ $4=\frac{I_{2}}{I_{1}}$ $\therefore \quad \mathrm{I}_{2}=4 \mathrm{I}_{1}$
155451
The number of turns in the primary and the secondary turns of a transformer are 1000 and 3000 respectively. If $80 \mathrm{~V}$ A.C. is applied to the primary coil of the transformer, then the potential difference per turn of secondary coil is
1 0.24 volt
2 0.08 volt
3 240 volt
4 2400 volt
Explanation:
B Given Number of primary turns $\mathrm{N}_{1}=1000$ Number of secondary turns $\mathrm{N}_{2}=3000$ Primary voltage $\mathrm{V}_{1}=80 \mathrm{~V}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\therefore \quad \mathrm{V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}=\frac{80 \times 3000}{1000}=240 \mathrm{~V}$ $\mathrm{V}_{2}=240$ $\therefore \quad$ Voltage/turn $=\frac{\mathrm{V}_{2}}{\mathrm{~N}_{2}}=\frac{240}{3000}$ Voltage/turns $=0.08$
J and K CET- 2001
Alternating Current
155454
The number of turns in the primary coil of a transformer is 200 and the number of turns in secondary coil is 10 .If $240 \mathrm{~V}$ A.C is applied to the primary, the output from secondary will be
1 $48 \mathrm{~V}$
2 $24 \mathrm{~V}$
3 $12 \mathrm{~V}$
4 $6 \mathrm{~V}$
Explanation:
C Given, Primary turns $\left(\mathrm{N}_{1}\right)=200$ Secondary turns $\left(\mathrm{N}_{2}\right)=10$ Primary voltage $\left(\mathrm{V}_{1}\right)=240 \mathrm{~V}$ We know that, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ Or $\quad \mathrm{V}_{2} =\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}=\frac{240 \times 10}{200}$ $\mathrm{~V}_{2} =12 \mathrm{~V}$
J and K CET- 1999
Alternating Current
155455
A transformer has 200 primary turns and 150 secondary turns. If $400 \mathrm{~V}$ are applied in primary, the voltage in secondary will be
1 $6 \mathrm{~V}$
2 $300 \mathrm{~V}$
3 $80,000 \mathrm{~V}$
4 $2 \mathrm{~V}$
Explanation:
B Given, Primary voltage $\left(\mathrm{V}_{1}\right)=400 \mathrm{~V}$ Primary turn $\left(\mathrm{N}_{1}\right)=200$ Secondary turns $\left(\mathrm{N}_{2}\right)=150$ We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\mathrm{~V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}=\frac{400 \times 150}{200}$ $\mathrm{~V}_{2}=300 \mathrm{~V}$
J and K CET- 1998
Alternating Current
155456
A transformer is used to reduce the voltage from $230 \mathrm{~V}$ to $6 \mathrm{~V}$. The number of turns on the secondary is 48 . Then the number of turns on the primary is
1 700
2 1840
3 3400
4 340
Explanation:
B Given, Primary voltage $\left(\mathrm{V}_{1}\right)=230 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=6 \mathrm{~V}$ Number of turns in the secondary coil $\left(\mathrm{N}_{2}\right)=48 \mathrm{~V}$ $\because \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\mathrm{N}_{1}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~V}_{2}}$ $\mathrm{N}_{1}=\frac{230 \times 48}{6}$ $\mathrm{N}_{1}=1840$ turns
J and K CET- 1997
Alternating Current
155458
In a transformer the number of primary turns is four times that of the secondary turns. If primary is connected to an a.c. source of voltage $V$. Then
1 Current through its secondary is about four times that of the current through its primary
2 voltage across its secondary is about four times that of the voltage across its primary
3 voltage across its secondary is about two times that of the voltage across its primary
4 voltage across its secondary is about $\frac{1}{2 \sqrt{2}}$ times that of the voltage across its primary
Explanation:
A We know that, $\because \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ Given, $\therefore \quad \mathrm{N}_{1}=4 \mathrm{~N}_{2}$ $\quad \frac{4 \mathrm{~N}_{2}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=4$ $\mathrm{~V}_{1}=4 \mathrm{~V}_{2}$ Current through secondary, $\frac{V_{1}}{V_{2}}=\frac{I_{2}}{I_{1}}$ $\frac{4 V_{2}}{V_{2}}=\frac{I_{2}}{I_{1}}$ $4=\frac{I_{2}}{I_{1}}$ $\therefore \quad \mathrm{I}_{2}=4 \mathrm{I}_{1}$
155451
The number of turns in the primary and the secondary turns of a transformer are 1000 and 3000 respectively. If $80 \mathrm{~V}$ A.C. is applied to the primary coil of the transformer, then the potential difference per turn of secondary coil is
1 0.24 volt
2 0.08 volt
3 240 volt
4 2400 volt
Explanation:
B Given Number of primary turns $\mathrm{N}_{1}=1000$ Number of secondary turns $\mathrm{N}_{2}=3000$ Primary voltage $\mathrm{V}_{1}=80 \mathrm{~V}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\therefore \quad \mathrm{V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}=\frac{80 \times 3000}{1000}=240 \mathrm{~V}$ $\mathrm{V}_{2}=240$ $\therefore \quad$ Voltage/turn $=\frac{\mathrm{V}_{2}}{\mathrm{~N}_{2}}=\frac{240}{3000}$ Voltage/turns $=0.08$
J and K CET- 2001
Alternating Current
155454
The number of turns in the primary coil of a transformer is 200 and the number of turns in secondary coil is 10 .If $240 \mathrm{~V}$ A.C is applied to the primary, the output from secondary will be
1 $48 \mathrm{~V}$
2 $24 \mathrm{~V}$
3 $12 \mathrm{~V}$
4 $6 \mathrm{~V}$
Explanation:
C Given, Primary turns $\left(\mathrm{N}_{1}\right)=200$ Secondary turns $\left(\mathrm{N}_{2}\right)=10$ Primary voltage $\left(\mathrm{V}_{1}\right)=240 \mathrm{~V}$ We know that, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ Or $\quad \mathrm{V}_{2} =\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}=\frac{240 \times 10}{200}$ $\mathrm{~V}_{2} =12 \mathrm{~V}$
J and K CET- 1999
Alternating Current
155455
A transformer has 200 primary turns and 150 secondary turns. If $400 \mathrm{~V}$ are applied in primary, the voltage in secondary will be
1 $6 \mathrm{~V}$
2 $300 \mathrm{~V}$
3 $80,000 \mathrm{~V}$
4 $2 \mathrm{~V}$
Explanation:
B Given, Primary voltage $\left(\mathrm{V}_{1}\right)=400 \mathrm{~V}$ Primary turn $\left(\mathrm{N}_{1}\right)=200$ Secondary turns $\left(\mathrm{N}_{2}\right)=150$ We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\mathrm{~V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}=\frac{400 \times 150}{200}$ $\mathrm{~V}_{2}=300 \mathrm{~V}$
J and K CET- 1998
Alternating Current
155456
A transformer is used to reduce the voltage from $230 \mathrm{~V}$ to $6 \mathrm{~V}$. The number of turns on the secondary is 48 . Then the number of turns on the primary is
1 700
2 1840
3 3400
4 340
Explanation:
B Given, Primary voltage $\left(\mathrm{V}_{1}\right)=230 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=6 \mathrm{~V}$ Number of turns in the secondary coil $\left(\mathrm{N}_{2}\right)=48 \mathrm{~V}$ $\because \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\mathrm{N}_{1}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~V}_{2}}$ $\mathrm{N}_{1}=\frac{230 \times 48}{6}$ $\mathrm{N}_{1}=1840$ turns
J and K CET- 1997
Alternating Current
155458
In a transformer the number of primary turns is four times that of the secondary turns. If primary is connected to an a.c. source of voltage $V$. Then
1 Current through its secondary is about four times that of the current through its primary
2 voltage across its secondary is about four times that of the voltage across its primary
3 voltage across its secondary is about two times that of the voltage across its primary
4 voltage across its secondary is about $\frac{1}{2 \sqrt{2}}$ times that of the voltage across its primary
Explanation:
A We know that, $\because \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ Given, $\therefore \quad \mathrm{N}_{1}=4 \mathrm{~N}_{2}$ $\quad \frac{4 \mathrm{~N}_{2}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=4$ $\mathrm{~V}_{1}=4 \mathrm{~V}_{2}$ Current through secondary, $\frac{V_{1}}{V_{2}}=\frac{I_{2}}{I_{1}}$ $\frac{4 V_{2}}{V_{2}}=\frac{I_{2}}{I_{1}}$ $4=\frac{I_{2}}{I_{1}}$ $\therefore \quad \mathrm{I}_{2}=4 \mathrm{I}_{1}$
155451
The number of turns in the primary and the secondary turns of a transformer are 1000 and 3000 respectively. If $80 \mathrm{~V}$ A.C. is applied to the primary coil of the transformer, then the potential difference per turn of secondary coil is
1 0.24 volt
2 0.08 volt
3 240 volt
4 2400 volt
Explanation:
B Given Number of primary turns $\mathrm{N}_{1}=1000$ Number of secondary turns $\mathrm{N}_{2}=3000$ Primary voltage $\mathrm{V}_{1}=80 \mathrm{~V}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\therefore \quad \mathrm{V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}=\frac{80 \times 3000}{1000}=240 \mathrm{~V}$ $\mathrm{V}_{2}=240$ $\therefore \quad$ Voltage/turn $=\frac{\mathrm{V}_{2}}{\mathrm{~N}_{2}}=\frac{240}{3000}$ Voltage/turns $=0.08$
J and K CET- 2001
Alternating Current
155454
The number of turns in the primary coil of a transformer is 200 and the number of turns in secondary coil is 10 .If $240 \mathrm{~V}$ A.C is applied to the primary, the output from secondary will be
1 $48 \mathrm{~V}$
2 $24 \mathrm{~V}$
3 $12 \mathrm{~V}$
4 $6 \mathrm{~V}$
Explanation:
C Given, Primary turns $\left(\mathrm{N}_{1}\right)=200$ Secondary turns $\left(\mathrm{N}_{2}\right)=10$ Primary voltage $\left(\mathrm{V}_{1}\right)=240 \mathrm{~V}$ We know that, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ Or $\quad \mathrm{V}_{2} =\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}=\frac{240 \times 10}{200}$ $\mathrm{~V}_{2} =12 \mathrm{~V}$
J and K CET- 1999
Alternating Current
155455
A transformer has 200 primary turns and 150 secondary turns. If $400 \mathrm{~V}$ are applied in primary, the voltage in secondary will be
1 $6 \mathrm{~V}$
2 $300 \mathrm{~V}$
3 $80,000 \mathrm{~V}$
4 $2 \mathrm{~V}$
Explanation:
B Given, Primary voltage $\left(\mathrm{V}_{1}\right)=400 \mathrm{~V}$ Primary turn $\left(\mathrm{N}_{1}\right)=200$ Secondary turns $\left(\mathrm{N}_{2}\right)=150$ We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\mathrm{~V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}=\frac{400 \times 150}{200}$ $\mathrm{~V}_{2}=300 \mathrm{~V}$
J and K CET- 1998
Alternating Current
155456
A transformer is used to reduce the voltage from $230 \mathrm{~V}$ to $6 \mathrm{~V}$. The number of turns on the secondary is 48 . Then the number of turns on the primary is
1 700
2 1840
3 3400
4 340
Explanation:
B Given, Primary voltage $\left(\mathrm{V}_{1}\right)=230 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=6 \mathrm{~V}$ Number of turns in the secondary coil $\left(\mathrm{N}_{2}\right)=48 \mathrm{~V}$ $\because \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\mathrm{N}_{1}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~V}_{2}}$ $\mathrm{N}_{1}=\frac{230 \times 48}{6}$ $\mathrm{N}_{1}=1840$ turns
J and K CET- 1997
Alternating Current
155458
In a transformer the number of primary turns is four times that of the secondary turns. If primary is connected to an a.c. source of voltage $V$. Then
1 Current through its secondary is about four times that of the current through its primary
2 voltage across its secondary is about four times that of the voltage across its primary
3 voltage across its secondary is about two times that of the voltage across its primary
4 voltage across its secondary is about $\frac{1}{2 \sqrt{2}}$ times that of the voltage across its primary
Explanation:
A We know that, $\because \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ Given, $\therefore \quad \mathrm{N}_{1}=4 \mathrm{~N}_{2}$ $\quad \frac{4 \mathrm{~N}_{2}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=4$ $\mathrm{~V}_{1}=4 \mathrm{~V}_{2}$ Current through secondary, $\frac{V_{1}}{V_{2}}=\frac{I_{2}}{I_{1}}$ $\frac{4 V_{2}}{V_{2}}=\frac{I_{2}}{I_{1}}$ $4=\frac{I_{2}}{I_{1}}$ $\therefore \quad \mathrm{I}_{2}=4 \mathrm{I}_{1}$
155451
The number of turns in the primary and the secondary turns of a transformer are 1000 and 3000 respectively. If $80 \mathrm{~V}$ A.C. is applied to the primary coil of the transformer, then the potential difference per turn of secondary coil is
1 0.24 volt
2 0.08 volt
3 240 volt
4 2400 volt
Explanation:
B Given Number of primary turns $\mathrm{N}_{1}=1000$ Number of secondary turns $\mathrm{N}_{2}=3000$ Primary voltage $\mathrm{V}_{1}=80 \mathrm{~V}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\therefore \quad \mathrm{V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}=\frac{80 \times 3000}{1000}=240 \mathrm{~V}$ $\mathrm{V}_{2}=240$ $\therefore \quad$ Voltage/turn $=\frac{\mathrm{V}_{2}}{\mathrm{~N}_{2}}=\frac{240}{3000}$ Voltage/turns $=0.08$
J and K CET- 2001
Alternating Current
155454
The number of turns in the primary coil of a transformer is 200 and the number of turns in secondary coil is 10 .If $240 \mathrm{~V}$ A.C is applied to the primary, the output from secondary will be
1 $48 \mathrm{~V}$
2 $24 \mathrm{~V}$
3 $12 \mathrm{~V}$
4 $6 \mathrm{~V}$
Explanation:
C Given, Primary turns $\left(\mathrm{N}_{1}\right)=200$ Secondary turns $\left(\mathrm{N}_{2}\right)=10$ Primary voltage $\left(\mathrm{V}_{1}\right)=240 \mathrm{~V}$ We know that, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ Or $\quad \mathrm{V}_{2} =\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}=\frac{240 \times 10}{200}$ $\mathrm{~V}_{2} =12 \mathrm{~V}$
J and K CET- 1999
Alternating Current
155455
A transformer has 200 primary turns and 150 secondary turns. If $400 \mathrm{~V}$ are applied in primary, the voltage in secondary will be
1 $6 \mathrm{~V}$
2 $300 \mathrm{~V}$
3 $80,000 \mathrm{~V}$
4 $2 \mathrm{~V}$
Explanation:
B Given, Primary voltage $\left(\mathrm{V}_{1}\right)=400 \mathrm{~V}$ Primary turn $\left(\mathrm{N}_{1}\right)=200$ Secondary turns $\left(\mathrm{N}_{2}\right)=150$ We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\mathrm{~V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}=\frac{400 \times 150}{200}$ $\mathrm{~V}_{2}=300 \mathrm{~V}$
J and K CET- 1998
Alternating Current
155456
A transformer is used to reduce the voltage from $230 \mathrm{~V}$ to $6 \mathrm{~V}$. The number of turns on the secondary is 48 . Then the number of turns on the primary is
1 700
2 1840
3 3400
4 340
Explanation:
B Given, Primary voltage $\left(\mathrm{V}_{1}\right)=230 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=6 \mathrm{~V}$ Number of turns in the secondary coil $\left(\mathrm{N}_{2}\right)=48 \mathrm{~V}$ $\because \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\mathrm{N}_{1}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~V}_{2}}$ $\mathrm{N}_{1}=\frac{230 \times 48}{6}$ $\mathrm{N}_{1}=1840$ turns
J and K CET- 1997
Alternating Current
155458
In a transformer the number of primary turns is four times that of the secondary turns. If primary is connected to an a.c. source of voltage $V$. Then
1 Current through its secondary is about four times that of the current through its primary
2 voltage across its secondary is about four times that of the voltage across its primary
3 voltage across its secondary is about two times that of the voltage across its primary
4 voltage across its secondary is about $\frac{1}{2 \sqrt{2}}$ times that of the voltage across its primary
Explanation:
A We know that, $\because \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ Given, $\therefore \quad \mathrm{N}_{1}=4 \mathrm{~N}_{2}$ $\quad \frac{4 \mathrm{~N}_{2}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=4$ $\mathrm{~V}_{1}=4 \mathrm{~V}_{2}$ Current through secondary, $\frac{V_{1}}{V_{2}}=\frac{I_{2}}{I_{1}}$ $\frac{4 V_{2}}{V_{2}}=\frac{I_{2}}{I_{1}}$ $4=\frac{I_{2}}{I_{1}}$ $\therefore \quad \mathrm{I}_{2}=4 \mathrm{I}_{1}$
