155416
A transformer with efficiency $80 \%$ works at 4 $\mathrm{kW}$ and $100 \mathrm{~V}$. If the secondary voltage is 200 $\mathrm{V}$, then the primary and secondary currents are respectively
1 $40 \mathrm{~A}, 16 \mathrm{~A}$
2 $16 \mathrm{~A}, 40 \mathrm{~A}$
3 $20 \mathrm{~A}, 40 \mathrm{~A}$
4 $40 \mathrm{~A}, 20 \mathrm{~A}$
Explanation:
A Given, $P_{\text {input }}=4 \mathrm{~kW}=4000 \mathrm{~W}$, Primary voltage $\left(V_1\right)=100 \mathrm{~V}$ Secondary voltage $\left(V_2\right)=200 \mathrm{~V}$, Efficiency of transformer $=80 \%$ We know that, $P_{\text {iuput }}=V_1 I_1$ $4000=100 \times I_1$ $\mathrm{I}_1=40 \mathrm{~A}$ And $\% \eta=\frac{P_{\text {oupyut }}}{P_{\text {imput }}}$ $\therefore \frac{80}{100} =\frac{\mathrm{V}_2 \mathrm{I}_2}{\mathrm{~V}_1 \mathrm{I}_1}$ $\frac{80}{100} =\frac{200 \times \mathrm{I}_2}{100 \times 40}$ $\text { Or } \mathrm{I}_2 =16 \mathrm{~A}$
CG PET -2016]**#```
Alternating Current
155417
A transformer has 200 turns in primary and 150 turns in secondary. If the operating voltage at the load connected to the secondary is measured to be $300 \mathrm{~V}$, the voltage supplied at the primary is
1 $230 \mathrm{~V}$
2 $400 \mathrm{~V}$
3 $480 \mathrm{~V}$
4 $660 \mathrm{~V}$
Explanation:
B Given that, Number of turns in primary coil $\left(\mathrm{N}_1\right)=200$ Number of turns in secondary coil $\left(\mathrm{N}_2\right)=150$ Voltage in secondary coil $\left(\mathrm{V}_2\right)=300 \mathrm{~V}$ We know that, $\therefore \quad \frac{\mathrm{N}_1}{\mathrm{~N}_2}=\frac{\mathrm{V}_1}{\mathrm{~V}_2}$ $\frac{200}{150}=\frac{\mathrm{V}_1}{300}$ $\mathrm{~V}_1=\frac{200 \times 300}{150}$ $\mathrm{~V}_1=400 \mathrm{~V}$
CG PET- 2015
Alternating Current
155420
The number of turns in primary and secondary coils of a transformer is 50 and 200 respectively. If the current in the primary coil is $4 \mathrm{~A}$, then the current in the secondary coil is
1 $1 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $4 \mathrm{~A}$
4 $5 \mathrm{~A}$
Explanation:
A Given that, Number of turn in primary coil $\left(\mathrm{N}_{1}\right)=50$ Number of turn in secondary coil $\left(\mathrm{N}_{2}\right)=200$ Current in primary coil $\left(\mathrm{I}_{1}\right)=4 \mathrm{~A}$ We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{2}=\frac{\mathrm{N}_{1} \mathrm{I}_{1}}{\mathrm{~N}_{2}}$ $\mathrm{I}_{2}=\frac{50 \times 4}{200}=1 \mathrm{~A}$
UPSEE - 2012
Alternating Current
155421
A generator at a utility company produces 100 A of current at $4000 \mathrm{~V}$. The voltage is stepped up to $240000 \mathrm{~V}$ by a transformer before it is sent on a high voltage transmission line. The current in transmission line is
1 $3.67 \mathrm{~A}$
2 $2.67 \mathrm{~A}$
3 $1.67 \mathrm{~A}$
4 $2.40 \mathrm{~A}$
Explanation:
C Given that, Primary current $\left(\mathrm{I}_{1}\right)=100 \mathrm{~A}$ Primary voltage $\left(\mathrm{V}_{1}\right)=4000 \mathrm{~V}$ Stepped-up voltage $\left(\mathrm{V}_{2}\right)=240000$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{2}=\frac{\mathrm{V}_{1} \mathrm{I}_{1}}{\mathrm{~V}_{2}}=\frac{4000 \times 100}{240000}=\frac{40}{24}$ $\mathrm{I}_{2}=1.666$ $\mathrm{I}_{2} \approx 1.67 \mathrm{~A}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155416
A transformer with efficiency $80 \%$ works at 4 $\mathrm{kW}$ and $100 \mathrm{~V}$. If the secondary voltage is 200 $\mathrm{V}$, then the primary and secondary currents are respectively
1 $40 \mathrm{~A}, 16 \mathrm{~A}$
2 $16 \mathrm{~A}, 40 \mathrm{~A}$
3 $20 \mathrm{~A}, 40 \mathrm{~A}$
4 $40 \mathrm{~A}, 20 \mathrm{~A}$
Explanation:
A Given, $P_{\text {input }}=4 \mathrm{~kW}=4000 \mathrm{~W}$, Primary voltage $\left(V_1\right)=100 \mathrm{~V}$ Secondary voltage $\left(V_2\right)=200 \mathrm{~V}$, Efficiency of transformer $=80 \%$ We know that, $P_{\text {iuput }}=V_1 I_1$ $4000=100 \times I_1$ $\mathrm{I}_1=40 \mathrm{~A}$ And $\% \eta=\frac{P_{\text {oupyut }}}{P_{\text {imput }}}$ $\therefore \frac{80}{100} =\frac{\mathrm{V}_2 \mathrm{I}_2}{\mathrm{~V}_1 \mathrm{I}_1}$ $\frac{80}{100} =\frac{200 \times \mathrm{I}_2}{100 \times 40}$ $\text { Or } \mathrm{I}_2 =16 \mathrm{~A}$
CG PET -2016]**#```
Alternating Current
155417
A transformer has 200 turns in primary and 150 turns in secondary. If the operating voltage at the load connected to the secondary is measured to be $300 \mathrm{~V}$, the voltage supplied at the primary is
1 $230 \mathrm{~V}$
2 $400 \mathrm{~V}$
3 $480 \mathrm{~V}$
4 $660 \mathrm{~V}$
Explanation:
B Given that, Number of turns in primary coil $\left(\mathrm{N}_1\right)=200$ Number of turns in secondary coil $\left(\mathrm{N}_2\right)=150$ Voltage in secondary coil $\left(\mathrm{V}_2\right)=300 \mathrm{~V}$ We know that, $\therefore \quad \frac{\mathrm{N}_1}{\mathrm{~N}_2}=\frac{\mathrm{V}_1}{\mathrm{~V}_2}$ $\frac{200}{150}=\frac{\mathrm{V}_1}{300}$ $\mathrm{~V}_1=\frac{200 \times 300}{150}$ $\mathrm{~V}_1=400 \mathrm{~V}$
CG PET- 2015
Alternating Current
155420
The number of turns in primary and secondary coils of a transformer is 50 and 200 respectively. If the current in the primary coil is $4 \mathrm{~A}$, then the current in the secondary coil is
1 $1 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $4 \mathrm{~A}$
4 $5 \mathrm{~A}$
Explanation:
A Given that, Number of turn in primary coil $\left(\mathrm{N}_{1}\right)=50$ Number of turn in secondary coil $\left(\mathrm{N}_{2}\right)=200$ Current in primary coil $\left(\mathrm{I}_{1}\right)=4 \mathrm{~A}$ We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{2}=\frac{\mathrm{N}_{1} \mathrm{I}_{1}}{\mathrm{~N}_{2}}$ $\mathrm{I}_{2}=\frac{50 \times 4}{200}=1 \mathrm{~A}$
UPSEE - 2012
Alternating Current
155421
A generator at a utility company produces 100 A of current at $4000 \mathrm{~V}$. The voltage is stepped up to $240000 \mathrm{~V}$ by a transformer before it is sent on a high voltage transmission line. The current in transmission line is
1 $3.67 \mathrm{~A}$
2 $2.67 \mathrm{~A}$
3 $1.67 \mathrm{~A}$
4 $2.40 \mathrm{~A}$
Explanation:
C Given that, Primary current $\left(\mathrm{I}_{1}\right)=100 \mathrm{~A}$ Primary voltage $\left(\mathrm{V}_{1}\right)=4000 \mathrm{~V}$ Stepped-up voltage $\left(\mathrm{V}_{2}\right)=240000$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{2}=\frac{\mathrm{V}_{1} \mathrm{I}_{1}}{\mathrm{~V}_{2}}=\frac{4000 \times 100}{240000}=\frac{40}{24}$ $\mathrm{I}_{2}=1.666$ $\mathrm{I}_{2} \approx 1.67 \mathrm{~A}$
155416
A transformer with efficiency $80 \%$ works at 4 $\mathrm{kW}$ and $100 \mathrm{~V}$. If the secondary voltage is 200 $\mathrm{V}$, then the primary and secondary currents are respectively
1 $40 \mathrm{~A}, 16 \mathrm{~A}$
2 $16 \mathrm{~A}, 40 \mathrm{~A}$
3 $20 \mathrm{~A}, 40 \mathrm{~A}$
4 $40 \mathrm{~A}, 20 \mathrm{~A}$
Explanation:
A Given, $P_{\text {input }}=4 \mathrm{~kW}=4000 \mathrm{~W}$, Primary voltage $\left(V_1\right)=100 \mathrm{~V}$ Secondary voltage $\left(V_2\right)=200 \mathrm{~V}$, Efficiency of transformer $=80 \%$ We know that, $P_{\text {iuput }}=V_1 I_1$ $4000=100 \times I_1$ $\mathrm{I}_1=40 \mathrm{~A}$ And $\% \eta=\frac{P_{\text {oupyut }}}{P_{\text {imput }}}$ $\therefore \frac{80}{100} =\frac{\mathrm{V}_2 \mathrm{I}_2}{\mathrm{~V}_1 \mathrm{I}_1}$ $\frac{80}{100} =\frac{200 \times \mathrm{I}_2}{100 \times 40}$ $\text { Or } \mathrm{I}_2 =16 \mathrm{~A}$
CG PET -2016]**#```
Alternating Current
155417
A transformer has 200 turns in primary and 150 turns in secondary. If the operating voltage at the load connected to the secondary is measured to be $300 \mathrm{~V}$, the voltage supplied at the primary is
1 $230 \mathrm{~V}$
2 $400 \mathrm{~V}$
3 $480 \mathrm{~V}$
4 $660 \mathrm{~V}$
Explanation:
B Given that, Number of turns in primary coil $\left(\mathrm{N}_1\right)=200$ Number of turns in secondary coil $\left(\mathrm{N}_2\right)=150$ Voltage in secondary coil $\left(\mathrm{V}_2\right)=300 \mathrm{~V}$ We know that, $\therefore \quad \frac{\mathrm{N}_1}{\mathrm{~N}_2}=\frac{\mathrm{V}_1}{\mathrm{~V}_2}$ $\frac{200}{150}=\frac{\mathrm{V}_1}{300}$ $\mathrm{~V}_1=\frac{200 \times 300}{150}$ $\mathrm{~V}_1=400 \mathrm{~V}$
CG PET- 2015
Alternating Current
155420
The number of turns in primary and secondary coils of a transformer is 50 and 200 respectively. If the current in the primary coil is $4 \mathrm{~A}$, then the current in the secondary coil is
1 $1 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $4 \mathrm{~A}$
4 $5 \mathrm{~A}$
Explanation:
A Given that, Number of turn in primary coil $\left(\mathrm{N}_{1}\right)=50$ Number of turn in secondary coil $\left(\mathrm{N}_{2}\right)=200$ Current in primary coil $\left(\mathrm{I}_{1}\right)=4 \mathrm{~A}$ We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{2}=\frac{\mathrm{N}_{1} \mathrm{I}_{1}}{\mathrm{~N}_{2}}$ $\mathrm{I}_{2}=\frac{50 \times 4}{200}=1 \mathrm{~A}$
UPSEE - 2012
Alternating Current
155421
A generator at a utility company produces 100 A of current at $4000 \mathrm{~V}$. The voltage is stepped up to $240000 \mathrm{~V}$ by a transformer before it is sent on a high voltage transmission line. The current in transmission line is
1 $3.67 \mathrm{~A}$
2 $2.67 \mathrm{~A}$
3 $1.67 \mathrm{~A}$
4 $2.40 \mathrm{~A}$
Explanation:
C Given that, Primary current $\left(\mathrm{I}_{1}\right)=100 \mathrm{~A}$ Primary voltage $\left(\mathrm{V}_{1}\right)=4000 \mathrm{~V}$ Stepped-up voltage $\left(\mathrm{V}_{2}\right)=240000$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{2}=\frac{\mathrm{V}_{1} \mathrm{I}_{1}}{\mathrm{~V}_{2}}=\frac{4000 \times 100}{240000}=\frac{40}{24}$ $\mathrm{I}_{2}=1.666$ $\mathrm{I}_{2} \approx 1.67 \mathrm{~A}$
155416
A transformer with efficiency $80 \%$ works at 4 $\mathrm{kW}$ and $100 \mathrm{~V}$. If the secondary voltage is 200 $\mathrm{V}$, then the primary and secondary currents are respectively
1 $40 \mathrm{~A}, 16 \mathrm{~A}$
2 $16 \mathrm{~A}, 40 \mathrm{~A}$
3 $20 \mathrm{~A}, 40 \mathrm{~A}$
4 $40 \mathrm{~A}, 20 \mathrm{~A}$
Explanation:
A Given, $P_{\text {input }}=4 \mathrm{~kW}=4000 \mathrm{~W}$, Primary voltage $\left(V_1\right)=100 \mathrm{~V}$ Secondary voltage $\left(V_2\right)=200 \mathrm{~V}$, Efficiency of transformer $=80 \%$ We know that, $P_{\text {iuput }}=V_1 I_1$ $4000=100 \times I_1$ $\mathrm{I}_1=40 \mathrm{~A}$ And $\% \eta=\frac{P_{\text {oupyut }}}{P_{\text {imput }}}$ $\therefore \frac{80}{100} =\frac{\mathrm{V}_2 \mathrm{I}_2}{\mathrm{~V}_1 \mathrm{I}_1}$ $\frac{80}{100} =\frac{200 \times \mathrm{I}_2}{100 \times 40}$ $\text { Or } \mathrm{I}_2 =16 \mathrm{~A}$
CG PET -2016]**#```
Alternating Current
155417
A transformer has 200 turns in primary and 150 turns in secondary. If the operating voltage at the load connected to the secondary is measured to be $300 \mathrm{~V}$, the voltage supplied at the primary is
1 $230 \mathrm{~V}$
2 $400 \mathrm{~V}$
3 $480 \mathrm{~V}$
4 $660 \mathrm{~V}$
Explanation:
B Given that, Number of turns in primary coil $\left(\mathrm{N}_1\right)=200$ Number of turns in secondary coil $\left(\mathrm{N}_2\right)=150$ Voltage in secondary coil $\left(\mathrm{V}_2\right)=300 \mathrm{~V}$ We know that, $\therefore \quad \frac{\mathrm{N}_1}{\mathrm{~N}_2}=\frac{\mathrm{V}_1}{\mathrm{~V}_2}$ $\frac{200}{150}=\frac{\mathrm{V}_1}{300}$ $\mathrm{~V}_1=\frac{200 \times 300}{150}$ $\mathrm{~V}_1=400 \mathrm{~V}$
CG PET- 2015
Alternating Current
155420
The number of turns in primary and secondary coils of a transformer is 50 and 200 respectively. If the current in the primary coil is $4 \mathrm{~A}$, then the current in the secondary coil is
1 $1 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $4 \mathrm{~A}$
4 $5 \mathrm{~A}$
Explanation:
A Given that, Number of turn in primary coil $\left(\mathrm{N}_{1}\right)=50$ Number of turn in secondary coil $\left(\mathrm{N}_{2}\right)=200$ Current in primary coil $\left(\mathrm{I}_{1}\right)=4 \mathrm{~A}$ We know that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{2}=\frac{\mathrm{N}_{1} \mathrm{I}_{1}}{\mathrm{~N}_{2}}$ $\mathrm{I}_{2}=\frac{50 \times 4}{200}=1 \mathrm{~A}$
UPSEE - 2012
Alternating Current
155421
A generator at a utility company produces 100 A of current at $4000 \mathrm{~V}$. The voltage is stepped up to $240000 \mathrm{~V}$ by a transformer before it is sent on a high voltage transmission line. The current in transmission line is
1 $3.67 \mathrm{~A}$
2 $2.67 \mathrm{~A}$
3 $1.67 \mathrm{~A}$
4 $2.40 \mathrm{~A}$
Explanation:
C Given that, Primary current $\left(\mathrm{I}_{1}\right)=100 \mathrm{~A}$ Primary voltage $\left(\mathrm{V}_{1}\right)=4000 \mathrm{~V}$ Stepped-up voltage $\left(\mathrm{V}_{2}\right)=240000$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\mathrm{I}_{2}=\frac{\mathrm{V}_{1} \mathrm{I}_{1}}{\mathrm{~V}_{2}}=\frac{4000 \times 100}{240000}=\frac{40}{24}$ $\mathrm{I}_{2}=1.666$ $\mathrm{I}_{2} \approx 1.67 \mathrm{~A}$