155439
When a current of $2 A$ is passed through a coil of 100 turns, flux associated with it is $5 \times 10^{-5}$ Wb. Find the self-inductance of the coil
155440
A step down transformer, transforms a supply line voltage of $2200 \mathrm{~V}$ into $220 \mathrm{~V}$. The primary coil has 5000 turns. The efficiency and power transmitted by the transformer are $90 \%$ and 8 $\mathrm{kW}$, respectively. Then the power supplied is
1 $9.89 \mathrm{~kW}$
2 $8.89 \mathrm{~kW}$
3 $88.9 \mathrm{~kW}$
4 $889 \mathrm{~kW}$
Explanation:
B Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=2200 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=220 \mathrm{~V}$ Primary turns $\left(\mathrm{N}_{1}\right)=5000$ turns Power output $\left(\mathrm{P}_{\text {output }}\right)=8 \mathrm{~kW}$ Efficiency of transformer $=90 \%$ We know that, Efficiency of transformer $(\eta)=\frac{P_{\text {output }}}{P_{\text {input }}} \times 100$ $90=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{\mathrm{~V}_{1} \mathrm{I}_{1}} \times 100$ $\frac{90}{100}=\frac{8 \mathrm{~kW}}{\mathrm{P}_{\text {input }}}$ $\mathrm{P}_{\text {input }}=\frac{8 \times 100}{90}=\frac{800}{90}$ $\mathrm{P}_{\text {input }}=8.89 \mathrm{~kW}$
MHT-CET 2006
Alternating Current
155441
The turn ratio of transformers is given as $2: 3$. If the current through the primary coil is $3 \mathrm{~A}$, thus calculate the current through load resistance
1 $1 \mathrm{~A}$
2 $4.5 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
C Given, Turn ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{2}{3}$ Current in primary, $\mathrm{I}_{1}=3 \mathrm{~A}$ $\because \quad \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ Alternating Current $\therefore \quad \frac{2}{3}=\frac{\mathrm{I}_{2}}{3}$ $\mathrm{I}_{2}=\frac{2 \times 3}{3}=2 \mathrm{~A}$ $\mathrm{I}_{2}=2 \mathrm{~A}$
VITEEE-2012
Alternating Current
155442
The output voltage of a transformer connected to $220 \mathrm{~V}$ line is $1100 \mathrm{~V}$ at $2 \mathrm{~A}$ current. Its efficiency is $100 \%$. The current coming from the line is
1 $20 \mathrm{~A}$
2 $10 \mathrm{~A}$
3 $11 \mathrm{~A}$
4 $22 \mathrm{~A}$
Explanation:
B Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=1100 \mathrm{~V}$ Secondary current $\left(\mathrm{I}_{2}\right)=2 \mathrm{~A}$ Primary current $\left(\mathrm{I}_{1}\right)=$ ? Efficiency of transformer $=100 \%$ For $100 \%$ efficiency, $\mathrm{V}_{1} \mathrm{I}_{1}=\mathrm{V}_{2} \mathrm{I}_{2}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\therefore \quad \frac{220}{1100}=\frac{2}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\frac{2 \times 1100}{220}$ $\mathrm{I}_{1}=10 \mathrm{~A}$
155439
When a current of $2 A$ is passed through a coil of 100 turns, flux associated with it is $5 \times 10^{-5}$ Wb. Find the self-inductance of the coil
155440
A step down transformer, transforms a supply line voltage of $2200 \mathrm{~V}$ into $220 \mathrm{~V}$. The primary coil has 5000 turns. The efficiency and power transmitted by the transformer are $90 \%$ and 8 $\mathrm{kW}$, respectively. Then the power supplied is
1 $9.89 \mathrm{~kW}$
2 $8.89 \mathrm{~kW}$
3 $88.9 \mathrm{~kW}$
4 $889 \mathrm{~kW}$
Explanation:
B Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=2200 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=220 \mathrm{~V}$ Primary turns $\left(\mathrm{N}_{1}\right)=5000$ turns Power output $\left(\mathrm{P}_{\text {output }}\right)=8 \mathrm{~kW}$ Efficiency of transformer $=90 \%$ We know that, Efficiency of transformer $(\eta)=\frac{P_{\text {output }}}{P_{\text {input }}} \times 100$ $90=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{\mathrm{~V}_{1} \mathrm{I}_{1}} \times 100$ $\frac{90}{100}=\frac{8 \mathrm{~kW}}{\mathrm{P}_{\text {input }}}$ $\mathrm{P}_{\text {input }}=\frac{8 \times 100}{90}=\frac{800}{90}$ $\mathrm{P}_{\text {input }}=8.89 \mathrm{~kW}$
MHT-CET 2006
Alternating Current
155441
The turn ratio of transformers is given as $2: 3$. If the current through the primary coil is $3 \mathrm{~A}$, thus calculate the current through load resistance
1 $1 \mathrm{~A}$
2 $4.5 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
C Given, Turn ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{2}{3}$ Current in primary, $\mathrm{I}_{1}=3 \mathrm{~A}$ $\because \quad \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ Alternating Current $\therefore \quad \frac{2}{3}=\frac{\mathrm{I}_{2}}{3}$ $\mathrm{I}_{2}=\frac{2 \times 3}{3}=2 \mathrm{~A}$ $\mathrm{I}_{2}=2 \mathrm{~A}$
VITEEE-2012
Alternating Current
155442
The output voltage of a transformer connected to $220 \mathrm{~V}$ line is $1100 \mathrm{~V}$ at $2 \mathrm{~A}$ current. Its efficiency is $100 \%$. The current coming from the line is
1 $20 \mathrm{~A}$
2 $10 \mathrm{~A}$
3 $11 \mathrm{~A}$
4 $22 \mathrm{~A}$
Explanation:
B Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=1100 \mathrm{~V}$ Secondary current $\left(\mathrm{I}_{2}\right)=2 \mathrm{~A}$ Primary current $\left(\mathrm{I}_{1}\right)=$ ? Efficiency of transformer $=100 \%$ For $100 \%$ efficiency, $\mathrm{V}_{1} \mathrm{I}_{1}=\mathrm{V}_{2} \mathrm{I}_{2}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\therefore \quad \frac{220}{1100}=\frac{2}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\frac{2 \times 1100}{220}$ $\mathrm{I}_{1}=10 \mathrm{~A}$
155439
When a current of $2 A$ is passed through a coil of 100 turns, flux associated with it is $5 \times 10^{-5}$ Wb. Find the self-inductance of the coil
155440
A step down transformer, transforms a supply line voltage of $2200 \mathrm{~V}$ into $220 \mathrm{~V}$. The primary coil has 5000 turns. The efficiency and power transmitted by the transformer are $90 \%$ and 8 $\mathrm{kW}$, respectively. Then the power supplied is
1 $9.89 \mathrm{~kW}$
2 $8.89 \mathrm{~kW}$
3 $88.9 \mathrm{~kW}$
4 $889 \mathrm{~kW}$
Explanation:
B Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=2200 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=220 \mathrm{~V}$ Primary turns $\left(\mathrm{N}_{1}\right)=5000$ turns Power output $\left(\mathrm{P}_{\text {output }}\right)=8 \mathrm{~kW}$ Efficiency of transformer $=90 \%$ We know that, Efficiency of transformer $(\eta)=\frac{P_{\text {output }}}{P_{\text {input }}} \times 100$ $90=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{\mathrm{~V}_{1} \mathrm{I}_{1}} \times 100$ $\frac{90}{100}=\frac{8 \mathrm{~kW}}{\mathrm{P}_{\text {input }}}$ $\mathrm{P}_{\text {input }}=\frac{8 \times 100}{90}=\frac{800}{90}$ $\mathrm{P}_{\text {input }}=8.89 \mathrm{~kW}$
MHT-CET 2006
Alternating Current
155441
The turn ratio of transformers is given as $2: 3$. If the current through the primary coil is $3 \mathrm{~A}$, thus calculate the current through load resistance
1 $1 \mathrm{~A}$
2 $4.5 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
C Given, Turn ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{2}{3}$ Current in primary, $\mathrm{I}_{1}=3 \mathrm{~A}$ $\because \quad \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ Alternating Current $\therefore \quad \frac{2}{3}=\frac{\mathrm{I}_{2}}{3}$ $\mathrm{I}_{2}=\frac{2 \times 3}{3}=2 \mathrm{~A}$ $\mathrm{I}_{2}=2 \mathrm{~A}$
VITEEE-2012
Alternating Current
155442
The output voltage of a transformer connected to $220 \mathrm{~V}$ line is $1100 \mathrm{~V}$ at $2 \mathrm{~A}$ current. Its efficiency is $100 \%$. The current coming from the line is
1 $20 \mathrm{~A}$
2 $10 \mathrm{~A}$
3 $11 \mathrm{~A}$
4 $22 \mathrm{~A}$
Explanation:
B Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=1100 \mathrm{~V}$ Secondary current $\left(\mathrm{I}_{2}\right)=2 \mathrm{~A}$ Primary current $\left(\mathrm{I}_{1}\right)=$ ? Efficiency of transformer $=100 \%$ For $100 \%$ efficiency, $\mathrm{V}_{1} \mathrm{I}_{1}=\mathrm{V}_{2} \mathrm{I}_{2}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\therefore \quad \frac{220}{1100}=\frac{2}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\frac{2 \times 1100}{220}$ $\mathrm{I}_{1}=10 \mathrm{~A}$
155439
When a current of $2 A$ is passed through a coil of 100 turns, flux associated with it is $5 \times 10^{-5}$ Wb. Find the self-inductance of the coil
155440
A step down transformer, transforms a supply line voltage of $2200 \mathrm{~V}$ into $220 \mathrm{~V}$. The primary coil has 5000 turns. The efficiency and power transmitted by the transformer are $90 \%$ and 8 $\mathrm{kW}$, respectively. Then the power supplied is
1 $9.89 \mathrm{~kW}$
2 $8.89 \mathrm{~kW}$
3 $88.9 \mathrm{~kW}$
4 $889 \mathrm{~kW}$
Explanation:
B Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=2200 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=220 \mathrm{~V}$ Primary turns $\left(\mathrm{N}_{1}\right)=5000$ turns Power output $\left(\mathrm{P}_{\text {output }}\right)=8 \mathrm{~kW}$ Efficiency of transformer $=90 \%$ We know that, Efficiency of transformer $(\eta)=\frac{P_{\text {output }}}{P_{\text {input }}} \times 100$ $90=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{\mathrm{~V}_{1} \mathrm{I}_{1}} \times 100$ $\frac{90}{100}=\frac{8 \mathrm{~kW}}{\mathrm{P}_{\text {input }}}$ $\mathrm{P}_{\text {input }}=\frac{8 \times 100}{90}=\frac{800}{90}$ $\mathrm{P}_{\text {input }}=8.89 \mathrm{~kW}$
MHT-CET 2006
Alternating Current
155441
The turn ratio of transformers is given as $2: 3$. If the current through the primary coil is $3 \mathrm{~A}$, thus calculate the current through load resistance
1 $1 \mathrm{~A}$
2 $4.5 \mathrm{~A}$
3 $2 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
C Given, Turn ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{2}{3}$ Current in primary, $\mathrm{I}_{1}=3 \mathrm{~A}$ $\because \quad \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ Alternating Current $\therefore \quad \frac{2}{3}=\frac{\mathrm{I}_{2}}{3}$ $\mathrm{I}_{2}=\frac{2 \times 3}{3}=2 \mathrm{~A}$ $\mathrm{I}_{2}=2 \mathrm{~A}$
VITEEE-2012
Alternating Current
155442
The output voltage of a transformer connected to $220 \mathrm{~V}$ line is $1100 \mathrm{~V}$ at $2 \mathrm{~A}$ current. Its efficiency is $100 \%$. The current coming from the line is
1 $20 \mathrm{~A}$
2 $10 \mathrm{~A}$
3 $11 \mathrm{~A}$
4 $22 \mathrm{~A}$
Explanation:
B Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=1100 \mathrm{~V}$ Secondary current $\left(\mathrm{I}_{2}\right)=2 \mathrm{~A}$ Primary current $\left(\mathrm{I}_{1}\right)=$ ? Efficiency of transformer $=100 \%$ For $100 \%$ efficiency, $\mathrm{V}_{1} \mathrm{I}_{1}=\mathrm{V}_{2} \mathrm{I}_{2}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\therefore \quad \frac{220}{1100}=\frac{2}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\frac{2 \times 1100}{220}$ $\mathrm{I}_{1}=10 \mathrm{~A}$
