NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155394
A $220 \mathrm{~V}$ input is supplied to a transformer. The output circuit draws a current of $2.0 \mathrm{~A}$ at 440 $V$. If the efficiency of the transformer is $80 \%$, the current drawn by the primary windings of the transformer is
1 $3.6 \mathrm{~A}$
2 $2.8 \mathrm{~A}$
3 $2.5 \mathrm{~A}$
4 $5.0 \mathrm{~A}$
Explanation:
D Given, Input voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Output current $\left(\mathrm{I}_{2}\right)=2 \mathrm{~A}$ Output voltage $\left(\mathrm{V}_{2}\right)=440 \mathrm{~V}$ Efficiency $(\eta)=80 \%=0.8$ We know that, $\eta =\frac{P_{\text {output }}}{P_{\text {input }}}$ $0.8 =\frac{V_{2} I_{2}}{V_{1} I_{1}}$ $I_{1} =\frac{V_{2} I_{2}}{0.8 \times 220}=\frac{440 \times 2}{0.8 \times 220}$ $I_{1} =5 \mathrm{~A}$
Assam CEE-2018
Alternating Current
155395
A step-up transformer operates on a $230 \mathrm{~V}$ line and a load current of $2 \mathrm{~A}$. The ratio of primary and secondary windings is $1: 25$. Then, the current in the primary is :
1 $25 \mathrm{~A}$
2 $50 \mathrm{~A}$
3 $15 \mathrm{~A}$
4 $12.5 \mathrm{~A}$
Explanation:
B Given, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{1}{25}$ Load current $\left(\mathrm{I}_{2}\right)=2 \mathrm{~A}$ From the transformer ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{1}{25}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\mathrm{I}_{2} \times 25=2 \times 25$ $\mathrm{I}_{1}=50 \mathrm{~A}$
Karnataka CET-2018
Alternating Current
155396
An ideal transformer converts $220 \mathrm{~V}$ a.c. to 3.3 $\mathrm{kV}$ a.c. to transmit a power of $4.4 \mathrm{~kW}$. If primary coil has 600 turns, then alternating current in secondary coil is
155397
A current of $5 \mathrm{~A}$ is flowing at $220 \mathrm{~V}$ in the primary coil of a transformer. If the voltage produced in secondary coil is $2200 \mathrm{~V}$ and $50 \%$ of power is lost, then the current in the secondary coil is-
1 $0.25 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 $0.5 \mathrm{~A}$
4 $5 \mathrm{~A}$
Explanation:
A Given, Current in the primary coil $\left(\mathrm{I}_{\mathrm{P}}\right)=5$ A Potential difference across primary coil $\left(\mathrm{V}_{\mathrm{P}}\right)=220 \mathrm{~V}$ $\mathrm{V}_{\mathrm{S}}=2200 \mathrm{~V}$ Output power is half of input power - $\mathrm{P}_{\mathrm{S}} =0.5 \mathrm{P}_{\mathrm{P}}$ $\mathrm{V}_{\mathrm{S}} \mathrm{I}_{\mathrm{S}} =0.5 \mathrm{~V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}$ $2200 \times \mathrm{I}_{\mathrm{S}} =0.5 \times 220 \times 5$ $\mathrm{I}_{\mathrm{S}} =\frac{0.5 \times 220 \times 5}{2200}$ $\mathrm{I}_{\mathrm{S}} =0.25 \mathrm{~A}$
155394
A $220 \mathrm{~V}$ input is supplied to a transformer. The output circuit draws a current of $2.0 \mathrm{~A}$ at 440 $V$. If the efficiency of the transformer is $80 \%$, the current drawn by the primary windings of the transformer is
1 $3.6 \mathrm{~A}$
2 $2.8 \mathrm{~A}$
3 $2.5 \mathrm{~A}$
4 $5.0 \mathrm{~A}$
Explanation:
D Given, Input voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Output current $\left(\mathrm{I}_{2}\right)=2 \mathrm{~A}$ Output voltage $\left(\mathrm{V}_{2}\right)=440 \mathrm{~V}$ Efficiency $(\eta)=80 \%=0.8$ We know that, $\eta =\frac{P_{\text {output }}}{P_{\text {input }}}$ $0.8 =\frac{V_{2} I_{2}}{V_{1} I_{1}}$ $I_{1} =\frac{V_{2} I_{2}}{0.8 \times 220}=\frac{440 \times 2}{0.8 \times 220}$ $I_{1} =5 \mathrm{~A}$
Assam CEE-2018
Alternating Current
155395
A step-up transformer operates on a $230 \mathrm{~V}$ line and a load current of $2 \mathrm{~A}$. The ratio of primary and secondary windings is $1: 25$. Then, the current in the primary is :
1 $25 \mathrm{~A}$
2 $50 \mathrm{~A}$
3 $15 \mathrm{~A}$
4 $12.5 \mathrm{~A}$
Explanation:
B Given, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{1}{25}$ Load current $\left(\mathrm{I}_{2}\right)=2 \mathrm{~A}$ From the transformer ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{1}{25}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\mathrm{I}_{2} \times 25=2 \times 25$ $\mathrm{I}_{1}=50 \mathrm{~A}$
Karnataka CET-2018
Alternating Current
155396
An ideal transformer converts $220 \mathrm{~V}$ a.c. to 3.3 $\mathrm{kV}$ a.c. to transmit a power of $4.4 \mathrm{~kW}$. If primary coil has 600 turns, then alternating current in secondary coil is
155397
A current of $5 \mathrm{~A}$ is flowing at $220 \mathrm{~V}$ in the primary coil of a transformer. If the voltage produced in secondary coil is $2200 \mathrm{~V}$ and $50 \%$ of power is lost, then the current in the secondary coil is-
1 $0.25 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 $0.5 \mathrm{~A}$
4 $5 \mathrm{~A}$
Explanation:
A Given, Current in the primary coil $\left(\mathrm{I}_{\mathrm{P}}\right)=5$ A Potential difference across primary coil $\left(\mathrm{V}_{\mathrm{P}}\right)=220 \mathrm{~V}$ $\mathrm{V}_{\mathrm{S}}=2200 \mathrm{~V}$ Output power is half of input power - $\mathrm{P}_{\mathrm{S}} =0.5 \mathrm{P}_{\mathrm{P}}$ $\mathrm{V}_{\mathrm{S}} \mathrm{I}_{\mathrm{S}} =0.5 \mathrm{~V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}$ $2200 \times \mathrm{I}_{\mathrm{S}} =0.5 \times 220 \times 5$ $\mathrm{I}_{\mathrm{S}} =\frac{0.5 \times 220 \times 5}{2200}$ $\mathrm{I}_{\mathrm{S}} =0.25 \mathrm{~A}$
155394
A $220 \mathrm{~V}$ input is supplied to a transformer. The output circuit draws a current of $2.0 \mathrm{~A}$ at 440 $V$. If the efficiency of the transformer is $80 \%$, the current drawn by the primary windings of the transformer is
1 $3.6 \mathrm{~A}$
2 $2.8 \mathrm{~A}$
3 $2.5 \mathrm{~A}$
4 $5.0 \mathrm{~A}$
Explanation:
D Given, Input voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Output current $\left(\mathrm{I}_{2}\right)=2 \mathrm{~A}$ Output voltage $\left(\mathrm{V}_{2}\right)=440 \mathrm{~V}$ Efficiency $(\eta)=80 \%=0.8$ We know that, $\eta =\frac{P_{\text {output }}}{P_{\text {input }}}$ $0.8 =\frac{V_{2} I_{2}}{V_{1} I_{1}}$ $I_{1} =\frac{V_{2} I_{2}}{0.8 \times 220}=\frac{440 \times 2}{0.8 \times 220}$ $I_{1} =5 \mathrm{~A}$
Assam CEE-2018
Alternating Current
155395
A step-up transformer operates on a $230 \mathrm{~V}$ line and a load current of $2 \mathrm{~A}$. The ratio of primary and secondary windings is $1: 25$. Then, the current in the primary is :
1 $25 \mathrm{~A}$
2 $50 \mathrm{~A}$
3 $15 \mathrm{~A}$
4 $12.5 \mathrm{~A}$
Explanation:
B Given, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{1}{25}$ Load current $\left(\mathrm{I}_{2}\right)=2 \mathrm{~A}$ From the transformer ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{1}{25}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\mathrm{I}_{2} \times 25=2 \times 25$ $\mathrm{I}_{1}=50 \mathrm{~A}$
Karnataka CET-2018
Alternating Current
155396
An ideal transformer converts $220 \mathrm{~V}$ a.c. to 3.3 $\mathrm{kV}$ a.c. to transmit a power of $4.4 \mathrm{~kW}$. If primary coil has 600 turns, then alternating current in secondary coil is
155397
A current of $5 \mathrm{~A}$ is flowing at $220 \mathrm{~V}$ in the primary coil of a transformer. If the voltage produced in secondary coil is $2200 \mathrm{~V}$ and $50 \%$ of power is lost, then the current in the secondary coil is-
1 $0.25 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 $0.5 \mathrm{~A}$
4 $5 \mathrm{~A}$
Explanation:
A Given, Current in the primary coil $\left(\mathrm{I}_{\mathrm{P}}\right)=5$ A Potential difference across primary coil $\left(\mathrm{V}_{\mathrm{P}}\right)=220 \mathrm{~V}$ $\mathrm{V}_{\mathrm{S}}=2200 \mathrm{~V}$ Output power is half of input power - $\mathrm{P}_{\mathrm{S}} =0.5 \mathrm{P}_{\mathrm{P}}$ $\mathrm{V}_{\mathrm{S}} \mathrm{I}_{\mathrm{S}} =0.5 \mathrm{~V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}$ $2200 \times \mathrm{I}_{\mathrm{S}} =0.5 \times 220 \times 5$ $\mathrm{I}_{\mathrm{S}} =\frac{0.5 \times 220 \times 5}{2200}$ $\mathrm{I}_{\mathrm{S}} =0.25 \mathrm{~A}$
155394
A $220 \mathrm{~V}$ input is supplied to a transformer. The output circuit draws a current of $2.0 \mathrm{~A}$ at 440 $V$. If the efficiency of the transformer is $80 \%$, the current drawn by the primary windings of the transformer is
1 $3.6 \mathrm{~A}$
2 $2.8 \mathrm{~A}$
3 $2.5 \mathrm{~A}$
4 $5.0 \mathrm{~A}$
Explanation:
D Given, Input voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Output current $\left(\mathrm{I}_{2}\right)=2 \mathrm{~A}$ Output voltage $\left(\mathrm{V}_{2}\right)=440 \mathrm{~V}$ Efficiency $(\eta)=80 \%=0.8$ We know that, $\eta =\frac{P_{\text {output }}}{P_{\text {input }}}$ $0.8 =\frac{V_{2} I_{2}}{V_{1} I_{1}}$ $I_{1} =\frac{V_{2} I_{2}}{0.8 \times 220}=\frac{440 \times 2}{0.8 \times 220}$ $I_{1} =5 \mathrm{~A}$
Assam CEE-2018
Alternating Current
155395
A step-up transformer operates on a $230 \mathrm{~V}$ line and a load current of $2 \mathrm{~A}$. The ratio of primary and secondary windings is $1: 25$. Then, the current in the primary is :
1 $25 \mathrm{~A}$
2 $50 \mathrm{~A}$
3 $15 \mathrm{~A}$
4 $12.5 \mathrm{~A}$
Explanation:
B Given, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{1}{25}$ Load current $\left(\mathrm{I}_{2}\right)=2 \mathrm{~A}$ From the transformer ratio, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{1}{25}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{1}=\mathrm{I}_{2} \times 25=2 \times 25$ $\mathrm{I}_{1}=50 \mathrm{~A}$
Karnataka CET-2018
Alternating Current
155396
An ideal transformer converts $220 \mathrm{~V}$ a.c. to 3.3 $\mathrm{kV}$ a.c. to transmit a power of $4.4 \mathrm{~kW}$. If primary coil has 600 turns, then alternating current in secondary coil is
155397
A current of $5 \mathrm{~A}$ is flowing at $220 \mathrm{~V}$ in the primary coil of a transformer. If the voltage produced in secondary coil is $2200 \mathrm{~V}$ and $50 \%$ of power is lost, then the current in the secondary coil is-
1 $0.25 \mathrm{~A}$
2 $2.5 \mathrm{~A}$
3 $0.5 \mathrm{~A}$
4 $5 \mathrm{~A}$
Explanation:
A Given, Current in the primary coil $\left(\mathrm{I}_{\mathrm{P}}\right)=5$ A Potential difference across primary coil $\left(\mathrm{V}_{\mathrm{P}}\right)=220 \mathrm{~V}$ $\mathrm{V}_{\mathrm{S}}=2200 \mathrm{~V}$ Output power is half of input power - $\mathrm{P}_{\mathrm{S}} =0.5 \mathrm{P}_{\mathrm{P}}$ $\mathrm{V}_{\mathrm{S}} \mathrm{I}_{\mathrm{S}} =0.5 \mathrm{~V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}$ $2200 \times \mathrm{I}_{\mathrm{S}} =0.5 \times 220 \times 5$ $\mathrm{I}_{\mathrm{S}} =\frac{0.5 \times 220 \times 5}{2200}$ $\mathrm{I}_{\mathrm{S}} =0.25 \mathrm{~A}$
