B Transformer used in only AC. If we used dc then, $\because \quad \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}$ $\mathrm{X}_{\mathrm{L}}=2 \pi \mathrm{f} \mathrm{L}$ Frequency in d.c. is zero Then, $X_{\mathrm{L}}=0$ It will offer zero impedance, it will drawn a very high current therefore winding may saturated.
MP PET-2012
Alternating Current
155461
A transformer having efficiency of $\mathbf{7 5 \%}$ is working on $220 \mathrm{~V}$ and $4.4 \mathrm{~kW}$ power supply. If the current in the secondary coil is $5 \mathrm{~A}$. What will be the voltage across secondary coil and the current in primary coil ?
C Given, Efficiency of transformer $(\eta)=75 \%=0.75$ Input voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Input power $\left(\mathrm{P}_{\text {input }}\right)=4.4 \mathrm{~kW}$ Current in secondary coil $\left(\mathrm{I}_{2}\right)=5 \mathrm{~A}$ We know that, $\eta=\frac{\mathrm{P}_{\text {output }}}{\mathrm{P}_{\text {input }}}$ $0.75=\frac{\mathrm{P}_{\mathrm{o} / \mathrm{p}}}{4.4 \times 10^{3}}$ $\mathrm{P}_{\mathrm{o} / \mathrm{p}}=0.75 \times 10^{3} \times 4.4$ $\mathrm{P}_{\mathrm{o} / \mathrm{p}}=3300 \mathrm{~W}$ $\mathrm{Or} \mathrm{V}_{2} \mathrm{I}_{2}=3300$ $\mathrm{~V}_{2}=\frac{3300}{5}$ $\mathrm{~V}_{2}=660 \mathrm{~V}$ $\mathrm{P}_{\mathrm{I} / \mathrm{p}}=\mathrm{V}_{1} \mathrm{I}_{1}$ $4.4 \times 1000=220 \times \mathrm{I}_{1}$ $\mathrm{I}_{1}=20 \mathrm{~A}$
JIPMER-2016
Alternating Current
155462
A transformer having efficiency of $90 \%$ is working on $200 \mathrm{~V}$ and $3 \mathrm{~kW}$ power supply. If the current in secondary coil is $6 \mathrm{~A}$, the voltage across the secondary coil and the current in the primary coil respectively are
1 $300 \mathrm{~V}, 15 \mathrm{~A}$
2 $450 \mathrm{~V}, 15 \mathrm{~A}$
3 $450 \mathrm{~V}, 13.5 \mathrm{~A}$
4 $600 \mathrm{~V}, 15 \mathrm{~A}$
Explanation:
B Given, Efficiency of transformer $(\eta)=90 \%$ Input voltage $\left(\mathrm{V}_{1}\right)=200 \mathrm{~V}$ Input power $=3 \mathrm{~kW}=3000 \mathrm{~W}$ Output current $\left(\mathrm{I}_{2}\right)=6 \mathrm{~A}$ Output voltage $\left(\mathrm{V}_{2}\right)=$ ? Input current $\left(\mathrm{I}_{1}\right)=$ ? Efficiency of transformer $(\eta)=\frac{\text { Output power }}{\text { Input power }}$ $\eta=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{3000}$ $\frac{90}{100}=\frac{\mathrm{V}_{2} \times 6}{3000}$ $\mathrm{V}_{2}=450 \mathrm{~V}$ Input power $=\mathrm{V}_{1} \times \mathrm{I}_{1}$ $3000=200 \times \mathrm{I}_{1}$ $\mathrm{I}_{1}=15 \mathrm{~A}$
AIPMT- 2014
Alternating Current
155463
A transformer has 1500 turns in the primary coil and 1125 turns in the secondary coil. If the voltage in the primary coil is $200 \mathrm{~V}$, then the voltage in the secondary coil is
1 $100 \mathrm{~V}$
2 $150 \mathrm{~V}$
3 $200 \mathrm{~V}$
4 $250 \mathrm{~V}$
Explanation:
B Given, Number of turns in primary coil $\left(\mathrm{N}_{1}\right)=1500$ Number of turns in secondary $\left(\mathrm{N}_{2}\right)=1125$ Voltage in primary coil $\left(\mathrm{V}_{1}\right)=200 \mathrm{~V}$ We know that, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\mathrm{V}_{2} =\frac{200 \times 1125}{1500}$ $\therefore \quad \mathrm{V}_{2} =150 \mathrm{~V}$
B Transformer used in only AC. If we used dc then, $\because \quad \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}$ $\mathrm{X}_{\mathrm{L}}=2 \pi \mathrm{f} \mathrm{L}$ Frequency in d.c. is zero Then, $X_{\mathrm{L}}=0$ It will offer zero impedance, it will drawn a very high current therefore winding may saturated.
MP PET-2012
Alternating Current
155461
A transformer having efficiency of $\mathbf{7 5 \%}$ is working on $220 \mathrm{~V}$ and $4.4 \mathrm{~kW}$ power supply. If the current in the secondary coil is $5 \mathrm{~A}$. What will be the voltage across secondary coil and the current in primary coil ?
C Given, Efficiency of transformer $(\eta)=75 \%=0.75$ Input voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Input power $\left(\mathrm{P}_{\text {input }}\right)=4.4 \mathrm{~kW}$ Current in secondary coil $\left(\mathrm{I}_{2}\right)=5 \mathrm{~A}$ We know that, $\eta=\frac{\mathrm{P}_{\text {output }}}{\mathrm{P}_{\text {input }}}$ $0.75=\frac{\mathrm{P}_{\mathrm{o} / \mathrm{p}}}{4.4 \times 10^{3}}$ $\mathrm{P}_{\mathrm{o} / \mathrm{p}}=0.75 \times 10^{3} \times 4.4$ $\mathrm{P}_{\mathrm{o} / \mathrm{p}}=3300 \mathrm{~W}$ $\mathrm{Or} \mathrm{V}_{2} \mathrm{I}_{2}=3300$ $\mathrm{~V}_{2}=\frac{3300}{5}$ $\mathrm{~V}_{2}=660 \mathrm{~V}$ $\mathrm{P}_{\mathrm{I} / \mathrm{p}}=\mathrm{V}_{1} \mathrm{I}_{1}$ $4.4 \times 1000=220 \times \mathrm{I}_{1}$ $\mathrm{I}_{1}=20 \mathrm{~A}$
JIPMER-2016
Alternating Current
155462
A transformer having efficiency of $90 \%$ is working on $200 \mathrm{~V}$ and $3 \mathrm{~kW}$ power supply. If the current in secondary coil is $6 \mathrm{~A}$, the voltage across the secondary coil and the current in the primary coil respectively are
1 $300 \mathrm{~V}, 15 \mathrm{~A}$
2 $450 \mathrm{~V}, 15 \mathrm{~A}$
3 $450 \mathrm{~V}, 13.5 \mathrm{~A}$
4 $600 \mathrm{~V}, 15 \mathrm{~A}$
Explanation:
B Given, Efficiency of transformer $(\eta)=90 \%$ Input voltage $\left(\mathrm{V}_{1}\right)=200 \mathrm{~V}$ Input power $=3 \mathrm{~kW}=3000 \mathrm{~W}$ Output current $\left(\mathrm{I}_{2}\right)=6 \mathrm{~A}$ Output voltage $\left(\mathrm{V}_{2}\right)=$ ? Input current $\left(\mathrm{I}_{1}\right)=$ ? Efficiency of transformer $(\eta)=\frac{\text { Output power }}{\text { Input power }}$ $\eta=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{3000}$ $\frac{90}{100}=\frac{\mathrm{V}_{2} \times 6}{3000}$ $\mathrm{V}_{2}=450 \mathrm{~V}$ Input power $=\mathrm{V}_{1} \times \mathrm{I}_{1}$ $3000=200 \times \mathrm{I}_{1}$ $\mathrm{I}_{1}=15 \mathrm{~A}$
AIPMT- 2014
Alternating Current
155463
A transformer has 1500 turns in the primary coil and 1125 turns in the secondary coil. If the voltage in the primary coil is $200 \mathrm{~V}$, then the voltage in the secondary coil is
1 $100 \mathrm{~V}$
2 $150 \mathrm{~V}$
3 $200 \mathrm{~V}$
4 $250 \mathrm{~V}$
Explanation:
B Given, Number of turns in primary coil $\left(\mathrm{N}_{1}\right)=1500$ Number of turns in secondary $\left(\mathrm{N}_{2}\right)=1125$ Voltage in primary coil $\left(\mathrm{V}_{1}\right)=200 \mathrm{~V}$ We know that, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\mathrm{V}_{2} =\frac{200 \times 1125}{1500}$ $\therefore \quad \mathrm{V}_{2} =150 \mathrm{~V}$
B Transformer used in only AC. If we used dc then, $\because \quad \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}$ $\mathrm{X}_{\mathrm{L}}=2 \pi \mathrm{f} \mathrm{L}$ Frequency in d.c. is zero Then, $X_{\mathrm{L}}=0$ It will offer zero impedance, it will drawn a very high current therefore winding may saturated.
MP PET-2012
Alternating Current
155461
A transformer having efficiency of $\mathbf{7 5 \%}$ is working on $220 \mathrm{~V}$ and $4.4 \mathrm{~kW}$ power supply. If the current in the secondary coil is $5 \mathrm{~A}$. What will be the voltage across secondary coil and the current in primary coil ?
C Given, Efficiency of transformer $(\eta)=75 \%=0.75$ Input voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Input power $\left(\mathrm{P}_{\text {input }}\right)=4.4 \mathrm{~kW}$ Current in secondary coil $\left(\mathrm{I}_{2}\right)=5 \mathrm{~A}$ We know that, $\eta=\frac{\mathrm{P}_{\text {output }}}{\mathrm{P}_{\text {input }}}$ $0.75=\frac{\mathrm{P}_{\mathrm{o} / \mathrm{p}}}{4.4 \times 10^{3}}$ $\mathrm{P}_{\mathrm{o} / \mathrm{p}}=0.75 \times 10^{3} \times 4.4$ $\mathrm{P}_{\mathrm{o} / \mathrm{p}}=3300 \mathrm{~W}$ $\mathrm{Or} \mathrm{V}_{2} \mathrm{I}_{2}=3300$ $\mathrm{~V}_{2}=\frac{3300}{5}$ $\mathrm{~V}_{2}=660 \mathrm{~V}$ $\mathrm{P}_{\mathrm{I} / \mathrm{p}}=\mathrm{V}_{1} \mathrm{I}_{1}$ $4.4 \times 1000=220 \times \mathrm{I}_{1}$ $\mathrm{I}_{1}=20 \mathrm{~A}$
JIPMER-2016
Alternating Current
155462
A transformer having efficiency of $90 \%$ is working on $200 \mathrm{~V}$ and $3 \mathrm{~kW}$ power supply. If the current in secondary coil is $6 \mathrm{~A}$, the voltage across the secondary coil and the current in the primary coil respectively are
1 $300 \mathrm{~V}, 15 \mathrm{~A}$
2 $450 \mathrm{~V}, 15 \mathrm{~A}$
3 $450 \mathrm{~V}, 13.5 \mathrm{~A}$
4 $600 \mathrm{~V}, 15 \mathrm{~A}$
Explanation:
B Given, Efficiency of transformer $(\eta)=90 \%$ Input voltage $\left(\mathrm{V}_{1}\right)=200 \mathrm{~V}$ Input power $=3 \mathrm{~kW}=3000 \mathrm{~W}$ Output current $\left(\mathrm{I}_{2}\right)=6 \mathrm{~A}$ Output voltage $\left(\mathrm{V}_{2}\right)=$ ? Input current $\left(\mathrm{I}_{1}\right)=$ ? Efficiency of transformer $(\eta)=\frac{\text { Output power }}{\text { Input power }}$ $\eta=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{3000}$ $\frac{90}{100}=\frac{\mathrm{V}_{2} \times 6}{3000}$ $\mathrm{V}_{2}=450 \mathrm{~V}$ Input power $=\mathrm{V}_{1} \times \mathrm{I}_{1}$ $3000=200 \times \mathrm{I}_{1}$ $\mathrm{I}_{1}=15 \mathrm{~A}$
AIPMT- 2014
Alternating Current
155463
A transformer has 1500 turns in the primary coil and 1125 turns in the secondary coil. If the voltage in the primary coil is $200 \mathrm{~V}$, then the voltage in the secondary coil is
1 $100 \mathrm{~V}$
2 $150 \mathrm{~V}$
3 $200 \mathrm{~V}$
4 $250 \mathrm{~V}$
Explanation:
B Given, Number of turns in primary coil $\left(\mathrm{N}_{1}\right)=1500$ Number of turns in secondary $\left(\mathrm{N}_{2}\right)=1125$ Voltage in primary coil $\left(\mathrm{V}_{1}\right)=200 \mathrm{~V}$ We know that, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\mathrm{V}_{2} =\frac{200 \times 1125}{1500}$ $\therefore \quad \mathrm{V}_{2} =150 \mathrm{~V}$
B Transformer used in only AC. If we used dc then, $\because \quad \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}$ $\mathrm{X}_{\mathrm{L}}=2 \pi \mathrm{f} \mathrm{L}$ Frequency in d.c. is zero Then, $X_{\mathrm{L}}=0$ It will offer zero impedance, it will drawn a very high current therefore winding may saturated.
MP PET-2012
Alternating Current
155461
A transformer having efficiency of $\mathbf{7 5 \%}$ is working on $220 \mathrm{~V}$ and $4.4 \mathrm{~kW}$ power supply. If the current in the secondary coil is $5 \mathrm{~A}$. What will be the voltage across secondary coil and the current in primary coil ?
C Given, Efficiency of transformer $(\eta)=75 \%=0.75$ Input voltage $\left(\mathrm{V}_{1}\right)=220 \mathrm{~V}$ Input power $\left(\mathrm{P}_{\text {input }}\right)=4.4 \mathrm{~kW}$ Current in secondary coil $\left(\mathrm{I}_{2}\right)=5 \mathrm{~A}$ We know that, $\eta=\frac{\mathrm{P}_{\text {output }}}{\mathrm{P}_{\text {input }}}$ $0.75=\frac{\mathrm{P}_{\mathrm{o} / \mathrm{p}}}{4.4 \times 10^{3}}$ $\mathrm{P}_{\mathrm{o} / \mathrm{p}}=0.75 \times 10^{3} \times 4.4$ $\mathrm{P}_{\mathrm{o} / \mathrm{p}}=3300 \mathrm{~W}$ $\mathrm{Or} \mathrm{V}_{2} \mathrm{I}_{2}=3300$ $\mathrm{~V}_{2}=\frac{3300}{5}$ $\mathrm{~V}_{2}=660 \mathrm{~V}$ $\mathrm{P}_{\mathrm{I} / \mathrm{p}}=\mathrm{V}_{1} \mathrm{I}_{1}$ $4.4 \times 1000=220 \times \mathrm{I}_{1}$ $\mathrm{I}_{1}=20 \mathrm{~A}$
JIPMER-2016
Alternating Current
155462
A transformer having efficiency of $90 \%$ is working on $200 \mathrm{~V}$ and $3 \mathrm{~kW}$ power supply. If the current in secondary coil is $6 \mathrm{~A}$, the voltage across the secondary coil and the current in the primary coil respectively are
1 $300 \mathrm{~V}, 15 \mathrm{~A}$
2 $450 \mathrm{~V}, 15 \mathrm{~A}$
3 $450 \mathrm{~V}, 13.5 \mathrm{~A}$
4 $600 \mathrm{~V}, 15 \mathrm{~A}$
Explanation:
B Given, Efficiency of transformer $(\eta)=90 \%$ Input voltage $\left(\mathrm{V}_{1}\right)=200 \mathrm{~V}$ Input power $=3 \mathrm{~kW}=3000 \mathrm{~W}$ Output current $\left(\mathrm{I}_{2}\right)=6 \mathrm{~A}$ Output voltage $\left(\mathrm{V}_{2}\right)=$ ? Input current $\left(\mathrm{I}_{1}\right)=$ ? Efficiency of transformer $(\eta)=\frac{\text { Output power }}{\text { Input power }}$ $\eta=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{3000}$ $\frac{90}{100}=\frac{\mathrm{V}_{2} \times 6}{3000}$ $\mathrm{V}_{2}=450 \mathrm{~V}$ Input power $=\mathrm{V}_{1} \times \mathrm{I}_{1}$ $3000=200 \times \mathrm{I}_{1}$ $\mathrm{I}_{1}=15 \mathrm{~A}$
AIPMT- 2014
Alternating Current
155463
A transformer has 1500 turns in the primary coil and 1125 turns in the secondary coil. If the voltage in the primary coil is $200 \mathrm{~V}$, then the voltage in the secondary coil is
1 $100 \mathrm{~V}$
2 $150 \mathrm{~V}$
3 $200 \mathrm{~V}$
4 $250 \mathrm{~V}$
Explanation:
B Given, Number of turns in primary coil $\left(\mathrm{N}_{1}\right)=1500$ Number of turns in secondary $\left(\mathrm{N}_{2}\right)=1125$ Voltage in primary coil $\left(\mathrm{V}_{1}\right)=200 \mathrm{~V}$ We know that, $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}$ $\mathrm{V}_{2} =\frac{200 \times 1125}{1500}$ $\therefore \quad \mathrm{V}_{2} =150 \mathrm{~V}$
