155443
A transformer rated at $10 \mathrm{~kW}$ is used to connect a $5 \mathrm{kV}$ transmission line to a $240 \mathrm{~V}$ 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 $\left(\mathrm{V}_{1}\right)=5 \mathrm{kV}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=240 \mathrm{~V}$ The ratio of the turns in the windings, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=$ ? From the transformer ratio, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{5 \mathrm{kV}}{240 \mathrm{~V}}=\frac{5 \times 1000}{240}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=20.8$
VITEEE-2008
Alternating Current
155444
A 400 turns primary coil of an ideal transformer is connected to an alternating current power line of $120 \mathrm{~V}$. A secondary coil of 100 turns is connected to a light bulb of $60 \Omega$ resistance. The maximum current in the secondary would be
1 $2 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $0.5 \mathrm{~A}$
4 $0.25 \mathrm{~A}$
Explanation:
C Given that, Primary turns $\left(\mathrm{N}_{1}\right)=400$ Secondary turns $\left(\mathrm{N}_{2}\right)=100$ Primary voltage $\left(\mathrm{V}_{1}\right)=120 \mathrm{~V}$ From the transformer ratio, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\mathrm{~V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}$ $\mathrm{~V}_{2}=\frac{120 \times 100}{400}=30 \mathrm{~V}$ $\because$ Resistance of bulb connected in secondary coil $=60 \Omega$ $\therefore$ Current in secondary coil $\left(\mathrm{I}_{2}\right)=\frac{\mathrm{V}_{2}}{\mathrm{R}}=\frac{30}{60}=0.5 \mathrm{~A}$ $\mathrm{I}_{2}=0.5 \mathrm{~A}$
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 $220 \mathrm{~V}, 1 \mathrm{~A}$ A.C. source, what is output current of the transformer?
1 $\frac{1}{20} \mathrm{~A}$
2 $20 \mathrm{~A}$
3 $100 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
B Given that, Secondary turns $\left(\mathrm{N}_{2}\right)=50$ Primary turns $\left(\mathrm{N}_{1}\right)=1000$ Primary voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Primary current $\left(\mathrm{I}_{1}\right)=1 \mathrm{~A}$ Secondary current $\left(\mathrm{I}_{2}\right)=$ ? From the transformer ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{1000}{50}=\frac{\mathrm{I}_{2}}{1}$ $\mathrm{I}_{2}=20 \mathrm{~A}$
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 $2 \mathrm{~A}$ and the voltage applied to it is $100 \mathrm{~V}$. Then the voltage and the current flowing in the secondary coil are respectively
1 $150 \mathrm{~V}, 1.5 \mathrm{~A}$
2 $300 \mathrm{~V}, 0.5 \mathrm{~A}$
3 $300 \mathrm{~V}, 1.5 \mathrm{~A}$
4 $150 \mathrm{~V}, 0.5 \mathrm{~A}$
Explanation:
B Given, Current in the primary $\left(\mathrm{I}_{1}\right)=2 \mathrm{~A}$ Current in the secondary $\left(\mathrm{I}_{2}\right)=$ ? Primary voltage $\left(\mathrm{V}_{1}\right)=100 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=$ ? Turn Ratio, $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\frac{3}{1}$ From the transformer ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{3}$ $\mathrm{~V}_{2}=3 \mathrm{~V}_{1}=3 \times 100$ $\mathrm{~V}_{2}=300 \mathrm{~V}$ $\because \quad \text { Efficiency }=\frac{\text { Output power }}{\text { Input power }}$ $\therefore \quad 0.75=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{\mathrm{~V}_{1} \mathrm{I}_{1}}$ $\mathrm{I}_{2}=\frac{0.75 \times \mathrm{V}_{1} \mathrm{I}_{1}}{\mathrm{~V}_{2}}=\frac{0.75 \times 100 \times 2}{300}$ $\mathrm{I}_{2}=0.25 \times 2$ $\mathrm{I}_{2}=0.50 \mathrm{~A}$ $\mathrm{~V}_{2}=300 \mathrm{~V}$ Hence, option (b) is correct.
155443
A transformer rated at $10 \mathrm{~kW}$ is used to connect a $5 \mathrm{kV}$ transmission line to a $240 \mathrm{~V}$ 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 $\left(\mathrm{V}_{1}\right)=5 \mathrm{kV}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=240 \mathrm{~V}$ The ratio of the turns in the windings, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=$ ? From the transformer ratio, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{5 \mathrm{kV}}{240 \mathrm{~V}}=\frac{5 \times 1000}{240}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=20.8$
VITEEE-2008
Alternating Current
155444
A 400 turns primary coil of an ideal transformer is connected to an alternating current power line of $120 \mathrm{~V}$. A secondary coil of 100 turns is connected to a light bulb of $60 \Omega$ resistance. The maximum current in the secondary would be
1 $2 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $0.5 \mathrm{~A}$
4 $0.25 \mathrm{~A}$
Explanation:
C Given that, Primary turns $\left(\mathrm{N}_{1}\right)=400$ Secondary turns $\left(\mathrm{N}_{2}\right)=100$ Primary voltage $\left(\mathrm{V}_{1}\right)=120 \mathrm{~V}$ From the transformer ratio, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\mathrm{~V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}$ $\mathrm{~V}_{2}=\frac{120 \times 100}{400}=30 \mathrm{~V}$ $\because$ Resistance of bulb connected in secondary coil $=60 \Omega$ $\therefore$ Current in secondary coil $\left(\mathrm{I}_{2}\right)=\frac{\mathrm{V}_{2}}{\mathrm{R}}=\frac{30}{60}=0.5 \mathrm{~A}$ $\mathrm{I}_{2}=0.5 \mathrm{~A}$
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 $220 \mathrm{~V}, 1 \mathrm{~A}$ A.C. source, what is output current of the transformer?
1 $\frac{1}{20} \mathrm{~A}$
2 $20 \mathrm{~A}$
3 $100 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
B Given that, Secondary turns $\left(\mathrm{N}_{2}\right)=50$ Primary turns $\left(\mathrm{N}_{1}\right)=1000$ Primary voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Primary current $\left(\mathrm{I}_{1}\right)=1 \mathrm{~A}$ Secondary current $\left(\mathrm{I}_{2}\right)=$ ? From the transformer ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{1000}{50}=\frac{\mathrm{I}_{2}}{1}$ $\mathrm{I}_{2}=20 \mathrm{~A}$
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 $2 \mathrm{~A}$ and the voltage applied to it is $100 \mathrm{~V}$. Then the voltage and the current flowing in the secondary coil are respectively
1 $150 \mathrm{~V}, 1.5 \mathrm{~A}$
2 $300 \mathrm{~V}, 0.5 \mathrm{~A}$
3 $300 \mathrm{~V}, 1.5 \mathrm{~A}$
4 $150 \mathrm{~V}, 0.5 \mathrm{~A}$
Explanation:
B Given, Current in the primary $\left(\mathrm{I}_{1}\right)=2 \mathrm{~A}$ Current in the secondary $\left(\mathrm{I}_{2}\right)=$ ? Primary voltage $\left(\mathrm{V}_{1}\right)=100 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=$ ? Turn Ratio, $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\frac{3}{1}$ From the transformer ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{3}$ $\mathrm{~V}_{2}=3 \mathrm{~V}_{1}=3 \times 100$ $\mathrm{~V}_{2}=300 \mathrm{~V}$ $\because \quad \text { Efficiency }=\frac{\text { Output power }}{\text { Input power }}$ $\therefore \quad 0.75=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{\mathrm{~V}_{1} \mathrm{I}_{1}}$ $\mathrm{I}_{2}=\frac{0.75 \times \mathrm{V}_{1} \mathrm{I}_{1}}{\mathrm{~V}_{2}}=\frac{0.75 \times 100 \times 2}{300}$ $\mathrm{I}_{2}=0.25 \times 2$ $\mathrm{I}_{2}=0.50 \mathrm{~A}$ $\mathrm{~V}_{2}=300 \mathrm{~V}$ Hence, option (b) is correct.
155443
A transformer rated at $10 \mathrm{~kW}$ is used to connect a $5 \mathrm{kV}$ transmission line to a $240 \mathrm{~V}$ 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 $\left(\mathrm{V}_{1}\right)=5 \mathrm{kV}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=240 \mathrm{~V}$ The ratio of the turns in the windings, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=$ ? From the transformer ratio, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{5 \mathrm{kV}}{240 \mathrm{~V}}=\frac{5 \times 1000}{240}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=20.8$
VITEEE-2008
Alternating Current
155444
A 400 turns primary coil of an ideal transformer is connected to an alternating current power line of $120 \mathrm{~V}$. A secondary coil of 100 turns is connected to a light bulb of $60 \Omega$ resistance. The maximum current in the secondary would be
1 $2 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $0.5 \mathrm{~A}$
4 $0.25 \mathrm{~A}$
Explanation:
C Given that, Primary turns $\left(\mathrm{N}_{1}\right)=400$ Secondary turns $\left(\mathrm{N}_{2}\right)=100$ Primary voltage $\left(\mathrm{V}_{1}\right)=120 \mathrm{~V}$ From the transformer ratio, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\mathrm{~V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}$ $\mathrm{~V}_{2}=\frac{120 \times 100}{400}=30 \mathrm{~V}$ $\because$ Resistance of bulb connected in secondary coil $=60 \Omega$ $\therefore$ Current in secondary coil $\left(\mathrm{I}_{2}\right)=\frac{\mathrm{V}_{2}}{\mathrm{R}}=\frac{30}{60}=0.5 \mathrm{~A}$ $\mathrm{I}_{2}=0.5 \mathrm{~A}$
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 $220 \mathrm{~V}, 1 \mathrm{~A}$ A.C. source, what is output current of the transformer?
1 $\frac{1}{20} \mathrm{~A}$
2 $20 \mathrm{~A}$
3 $100 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
B Given that, Secondary turns $\left(\mathrm{N}_{2}\right)=50$ Primary turns $\left(\mathrm{N}_{1}\right)=1000$ Primary voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Primary current $\left(\mathrm{I}_{1}\right)=1 \mathrm{~A}$ Secondary current $\left(\mathrm{I}_{2}\right)=$ ? From the transformer ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{1000}{50}=\frac{\mathrm{I}_{2}}{1}$ $\mathrm{I}_{2}=20 \mathrm{~A}$
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 $2 \mathrm{~A}$ and the voltage applied to it is $100 \mathrm{~V}$. Then the voltage and the current flowing in the secondary coil are respectively
1 $150 \mathrm{~V}, 1.5 \mathrm{~A}$
2 $300 \mathrm{~V}, 0.5 \mathrm{~A}$
3 $300 \mathrm{~V}, 1.5 \mathrm{~A}$
4 $150 \mathrm{~V}, 0.5 \mathrm{~A}$
Explanation:
B Given, Current in the primary $\left(\mathrm{I}_{1}\right)=2 \mathrm{~A}$ Current in the secondary $\left(\mathrm{I}_{2}\right)=$ ? Primary voltage $\left(\mathrm{V}_{1}\right)=100 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=$ ? Turn Ratio, $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\frac{3}{1}$ From the transformer ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{3}$ $\mathrm{~V}_{2}=3 \mathrm{~V}_{1}=3 \times 100$ $\mathrm{~V}_{2}=300 \mathrm{~V}$ $\because \quad \text { Efficiency }=\frac{\text { Output power }}{\text { Input power }}$ $\therefore \quad 0.75=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{\mathrm{~V}_{1} \mathrm{I}_{1}}$ $\mathrm{I}_{2}=\frac{0.75 \times \mathrm{V}_{1} \mathrm{I}_{1}}{\mathrm{~V}_{2}}=\frac{0.75 \times 100 \times 2}{300}$ $\mathrm{I}_{2}=0.25 \times 2$ $\mathrm{I}_{2}=0.50 \mathrm{~A}$ $\mathrm{~V}_{2}=300 \mathrm{~V}$ Hence, option (b) is correct.
155443
A transformer rated at $10 \mathrm{~kW}$ is used to connect a $5 \mathrm{kV}$ transmission line to a $240 \mathrm{~V}$ 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 $\left(\mathrm{V}_{1}\right)=5 \mathrm{kV}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=240 \mathrm{~V}$ The ratio of the turns in the windings, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=$ ? From the transformer ratio, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{5 \mathrm{kV}}{240 \mathrm{~V}}=\frac{5 \times 1000}{240}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=20.8$
VITEEE-2008
Alternating Current
155444
A 400 turns primary coil of an ideal transformer is connected to an alternating current power line of $120 \mathrm{~V}$. A secondary coil of 100 turns is connected to a light bulb of $60 \Omega$ resistance. The maximum current in the secondary would be
1 $2 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $0.5 \mathrm{~A}$
4 $0.25 \mathrm{~A}$
Explanation:
C Given that, Primary turns $\left(\mathrm{N}_{1}\right)=400$ Secondary turns $\left(\mathrm{N}_{2}\right)=100$ Primary voltage $\left(\mathrm{V}_{1}\right)=120 \mathrm{~V}$ From the transformer ratio, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\mathrm{~V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~N}_{2}}{\mathrm{~N}_{1}}$ $\mathrm{~V}_{2}=\frac{120 \times 100}{400}=30 \mathrm{~V}$ $\because$ Resistance of bulb connected in secondary coil $=60 \Omega$ $\therefore$ Current in secondary coil $\left(\mathrm{I}_{2}\right)=\frac{\mathrm{V}_{2}}{\mathrm{R}}=\frac{30}{60}=0.5 \mathrm{~A}$ $\mathrm{I}_{2}=0.5 \mathrm{~A}$
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 $220 \mathrm{~V}, 1 \mathrm{~A}$ A.C. source, what is output current of the transformer?
1 $\frac{1}{20} \mathrm{~A}$
2 $20 \mathrm{~A}$
3 $100 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
B Given that, Secondary turns $\left(\mathrm{N}_{2}\right)=50$ Primary turns $\left(\mathrm{N}_{1}\right)=1000$ Primary voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Primary current $\left(\mathrm{I}_{1}\right)=1 \mathrm{~A}$ Secondary current $\left(\mathrm{I}_{2}\right)=$ ? From the transformer ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{1000}{50}=\frac{\mathrm{I}_{2}}{1}$ $\mathrm{I}_{2}=20 \mathrm{~A}$
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 $2 \mathrm{~A}$ and the voltage applied to it is $100 \mathrm{~V}$. Then the voltage and the current flowing in the secondary coil are respectively
1 $150 \mathrm{~V}, 1.5 \mathrm{~A}$
2 $300 \mathrm{~V}, 0.5 \mathrm{~A}$
3 $300 \mathrm{~V}, 1.5 \mathrm{~A}$
4 $150 \mathrm{~V}, 0.5 \mathrm{~A}$
Explanation:
B Given, Current in the primary $\left(\mathrm{I}_{1}\right)=2 \mathrm{~A}$ Current in the secondary $\left(\mathrm{I}_{2}\right)=$ ? Primary voltage $\left(\mathrm{V}_{1}\right)=100 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=$ ? Turn Ratio, $\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\frac{3}{1}$ From the transformer ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{3}$ $\mathrm{~V}_{2}=3 \mathrm{~V}_{1}=3 \times 100$ $\mathrm{~V}_{2}=300 \mathrm{~V}$ $\because \quad \text { Efficiency }=\frac{\text { Output power }}{\text { Input power }}$ $\therefore \quad 0.75=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{\mathrm{~V}_{1} \mathrm{I}_{1}}$ $\mathrm{I}_{2}=\frac{0.75 \times \mathrm{V}_{1} \mathrm{I}_{1}}{\mathrm{~V}_{2}}=\frac{0.75 \times 100 \times 2}{300}$ $\mathrm{I}_{2}=0.25 \times 2$ $\mathrm{I}_{2}=0.50 \mathrm{~A}$ $\mathrm{~V}_{2}=300 \mathrm{~V}$ Hence, option (b) is correct.
