155398
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, $\mathrm{N}_{\mathrm{P}}=50, \mathrm{~N}_{\mathrm{S}}=200, \mathrm{I}_{\mathrm{P}}=4 \mathrm{~A}, \mathrm{I}_{\mathrm{S}}=$ ? We know that, in a transformer $\frac{\mathrm{N}_{\mathrm{P}}}{\mathrm{N}_{\mathrm{S}}}=\frac{\mathrm{I}_{\mathrm{S}}}{\mathrm{I}_{\mathrm{P}}}$ $\frac{50}{200}=\frac{\mathrm{I}_{\mathrm{S}}}{4}$ $\mathrm{I}_{\mathrm{S}}=1 \mathrm{~A}$
EAMCET-2018
Alternating Current
155399
A transformer has 100 turns in the primary coil and carries $8 \mathrm{~A}$ current. If input power is $1 \mathrm{~kW}$, the number of turns in secondary coil to have $500 \mathrm{~V}$ of output will be
1 100
2 200
3 400
4 300
Explanation:
C Given that, Primary number of turns $=100$ Primary current $=8 \mathrm{~A}$ Input power $=1 \mathrm{~kW}$ $\text { Power = V.I }$ $1000=500 . \mathrm{I}$ $\mathrm{I}=2 \mathrm{~A}$ Now, from the transformers equation - $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}$ $\frac{8}{2}=\frac{\mathrm{N}_{2}}{100}$ $\mathrm{~N}_{2}=400$
JCECE-2018
Alternating Current
155400
A transformer is used to glow a $140 \mathrm{~W}-24 \mathrm{~V}$ bulb at $240 \mathrm{~V}$ AC. If current in the primary coil is $0.7 \mathrm{~A}$, then the efficiency of the transformer is
1 $16.7 \%$
2 $50 \%$
3 $83.3 \%$
4 $25 \%$
Explanation:
C Given that, Input voltage $=240 \mathrm{~V}$ Primary coil current $=0.7 \mathrm{~A}$ Output power $=140 \mathrm{~W}$ Power input $=\mathrm{V} \times \mathrm{I}$ $\mathrm{P}=240 \times 0.7$ $\mathrm{P}=168.0 \mathrm{~W}$ $\therefore$ Efficiency of transformer, $\eta \%=\frac{P_{\text {output }}}{P_{\text {input }}} \times 100=\frac{140}{168} \times 100$ $\eta=83.33 \%$
CG PET -2018
Alternating Current
155401
A distribution transformer with an efficiency of $\mathbf{9 0 \%}$ supplies to a colony of 10 homes. All the 10 homes have electrical oven running at the same time, that draw 20 A current from $220 \mathrm{~V}$ lines. The power dissipated as heat in the transformer is
1 $12.2 \mathrm{~kW}$
2 $4.9 \mathrm{~kW}$
3 $8.4 \mathrm{~kW}$
4 $9.9 \mathrm{~kW}$
Explanation:
B Given that, Efficiency of transformer $(\eta)=90 \%$ $\mathrm{V}=220$ volt, $\mathrm{I}=20 \mathrm{~A}$ We know that, $\mathrm{P}_{1}=10 \mathrm{VI}=10 \times 220 \times 20=44 \times 10^{3} \mathrm{~W}$ So, power delivered by transformer $\mathrm{P}_{2}=\frac{100}{90} \times \mathrm{P}_{1}=\frac{10}{9} \times 44 \times 10^{3}$ $\mathrm{P}_{2}=48.9 \times 10^{3} \mathrm{~W}$ $\therefore$ Power dissipated as heat $\mathrm{P}_{\mathrm{H}}=\mathrm{P}_{2}-\mathrm{P}_{1}$ $\mathrm{P}_{\mathrm{H}}=(48.9-44) \times 10^{3}$ $\mathrm{P}_{\mathrm{H}}=4.9 \times 10^{3}$ $\mathrm{P}_{\mathrm{H}}=4.9 \mathrm{~kW}$
155398
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, $\mathrm{N}_{\mathrm{P}}=50, \mathrm{~N}_{\mathrm{S}}=200, \mathrm{I}_{\mathrm{P}}=4 \mathrm{~A}, \mathrm{I}_{\mathrm{S}}=$ ? We know that, in a transformer $\frac{\mathrm{N}_{\mathrm{P}}}{\mathrm{N}_{\mathrm{S}}}=\frac{\mathrm{I}_{\mathrm{S}}}{\mathrm{I}_{\mathrm{P}}}$ $\frac{50}{200}=\frac{\mathrm{I}_{\mathrm{S}}}{4}$ $\mathrm{I}_{\mathrm{S}}=1 \mathrm{~A}$
EAMCET-2018
Alternating Current
155399
A transformer has 100 turns in the primary coil and carries $8 \mathrm{~A}$ current. If input power is $1 \mathrm{~kW}$, the number of turns in secondary coil to have $500 \mathrm{~V}$ of output will be
1 100
2 200
3 400
4 300
Explanation:
C Given that, Primary number of turns $=100$ Primary current $=8 \mathrm{~A}$ Input power $=1 \mathrm{~kW}$ $\text { Power = V.I }$ $1000=500 . \mathrm{I}$ $\mathrm{I}=2 \mathrm{~A}$ Now, from the transformers equation - $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}$ $\frac{8}{2}=\frac{\mathrm{N}_{2}}{100}$ $\mathrm{~N}_{2}=400$
JCECE-2018
Alternating Current
155400
A transformer is used to glow a $140 \mathrm{~W}-24 \mathrm{~V}$ bulb at $240 \mathrm{~V}$ AC. If current in the primary coil is $0.7 \mathrm{~A}$, then the efficiency of the transformer is
1 $16.7 \%$
2 $50 \%$
3 $83.3 \%$
4 $25 \%$
Explanation:
C Given that, Input voltage $=240 \mathrm{~V}$ Primary coil current $=0.7 \mathrm{~A}$ Output power $=140 \mathrm{~W}$ Power input $=\mathrm{V} \times \mathrm{I}$ $\mathrm{P}=240 \times 0.7$ $\mathrm{P}=168.0 \mathrm{~W}$ $\therefore$ Efficiency of transformer, $\eta \%=\frac{P_{\text {output }}}{P_{\text {input }}} \times 100=\frac{140}{168} \times 100$ $\eta=83.33 \%$
CG PET -2018
Alternating Current
155401
A distribution transformer with an efficiency of $\mathbf{9 0 \%}$ supplies to a colony of 10 homes. All the 10 homes have electrical oven running at the same time, that draw 20 A current from $220 \mathrm{~V}$ lines. The power dissipated as heat in the transformer is
1 $12.2 \mathrm{~kW}$
2 $4.9 \mathrm{~kW}$
3 $8.4 \mathrm{~kW}$
4 $9.9 \mathrm{~kW}$
Explanation:
B Given that, Efficiency of transformer $(\eta)=90 \%$ $\mathrm{V}=220$ volt, $\mathrm{I}=20 \mathrm{~A}$ We know that, $\mathrm{P}_{1}=10 \mathrm{VI}=10 \times 220 \times 20=44 \times 10^{3} \mathrm{~W}$ So, power delivered by transformer $\mathrm{P}_{2}=\frac{100}{90} \times \mathrm{P}_{1}=\frac{10}{9} \times 44 \times 10^{3}$ $\mathrm{P}_{2}=48.9 \times 10^{3} \mathrm{~W}$ $\therefore$ Power dissipated as heat $\mathrm{P}_{\mathrm{H}}=\mathrm{P}_{2}-\mathrm{P}_{1}$ $\mathrm{P}_{\mathrm{H}}=(48.9-44) \times 10^{3}$ $\mathrm{P}_{\mathrm{H}}=4.9 \times 10^{3}$ $\mathrm{P}_{\mathrm{H}}=4.9 \mathrm{~kW}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155398
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, $\mathrm{N}_{\mathrm{P}}=50, \mathrm{~N}_{\mathrm{S}}=200, \mathrm{I}_{\mathrm{P}}=4 \mathrm{~A}, \mathrm{I}_{\mathrm{S}}=$ ? We know that, in a transformer $\frac{\mathrm{N}_{\mathrm{P}}}{\mathrm{N}_{\mathrm{S}}}=\frac{\mathrm{I}_{\mathrm{S}}}{\mathrm{I}_{\mathrm{P}}}$ $\frac{50}{200}=\frac{\mathrm{I}_{\mathrm{S}}}{4}$ $\mathrm{I}_{\mathrm{S}}=1 \mathrm{~A}$
EAMCET-2018
Alternating Current
155399
A transformer has 100 turns in the primary coil and carries $8 \mathrm{~A}$ current. If input power is $1 \mathrm{~kW}$, the number of turns in secondary coil to have $500 \mathrm{~V}$ of output will be
1 100
2 200
3 400
4 300
Explanation:
C Given that, Primary number of turns $=100$ Primary current $=8 \mathrm{~A}$ Input power $=1 \mathrm{~kW}$ $\text { Power = V.I }$ $1000=500 . \mathrm{I}$ $\mathrm{I}=2 \mathrm{~A}$ Now, from the transformers equation - $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}$ $\frac{8}{2}=\frac{\mathrm{N}_{2}}{100}$ $\mathrm{~N}_{2}=400$
JCECE-2018
Alternating Current
155400
A transformer is used to glow a $140 \mathrm{~W}-24 \mathrm{~V}$ bulb at $240 \mathrm{~V}$ AC. If current in the primary coil is $0.7 \mathrm{~A}$, then the efficiency of the transformer is
1 $16.7 \%$
2 $50 \%$
3 $83.3 \%$
4 $25 \%$
Explanation:
C Given that, Input voltage $=240 \mathrm{~V}$ Primary coil current $=0.7 \mathrm{~A}$ Output power $=140 \mathrm{~W}$ Power input $=\mathrm{V} \times \mathrm{I}$ $\mathrm{P}=240 \times 0.7$ $\mathrm{P}=168.0 \mathrm{~W}$ $\therefore$ Efficiency of transformer, $\eta \%=\frac{P_{\text {output }}}{P_{\text {input }}} \times 100=\frac{140}{168} \times 100$ $\eta=83.33 \%$
CG PET -2018
Alternating Current
155401
A distribution transformer with an efficiency of $\mathbf{9 0 \%}$ supplies to a colony of 10 homes. All the 10 homes have electrical oven running at the same time, that draw 20 A current from $220 \mathrm{~V}$ lines. The power dissipated as heat in the transformer is
1 $12.2 \mathrm{~kW}$
2 $4.9 \mathrm{~kW}$
3 $8.4 \mathrm{~kW}$
4 $9.9 \mathrm{~kW}$
Explanation:
B Given that, Efficiency of transformer $(\eta)=90 \%$ $\mathrm{V}=220$ volt, $\mathrm{I}=20 \mathrm{~A}$ We know that, $\mathrm{P}_{1}=10 \mathrm{VI}=10 \times 220 \times 20=44 \times 10^{3} \mathrm{~W}$ So, power delivered by transformer $\mathrm{P}_{2}=\frac{100}{90} \times \mathrm{P}_{1}=\frac{10}{9} \times 44 \times 10^{3}$ $\mathrm{P}_{2}=48.9 \times 10^{3} \mathrm{~W}$ $\therefore$ Power dissipated as heat $\mathrm{P}_{\mathrm{H}}=\mathrm{P}_{2}-\mathrm{P}_{1}$ $\mathrm{P}_{\mathrm{H}}=(48.9-44) \times 10^{3}$ $\mathrm{P}_{\mathrm{H}}=4.9 \times 10^{3}$ $\mathrm{P}_{\mathrm{H}}=4.9 \mathrm{~kW}$
155398
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, $\mathrm{N}_{\mathrm{P}}=50, \mathrm{~N}_{\mathrm{S}}=200, \mathrm{I}_{\mathrm{P}}=4 \mathrm{~A}, \mathrm{I}_{\mathrm{S}}=$ ? We know that, in a transformer $\frac{\mathrm{N}_{\mathrm{P}}}{\mathrm{N}_{\mathrm{S}}}=\frac{\mathrm{I}_{\mathrm{S}}}{\mathrm{I}_{\mathrm{P}}}$ $\frac{50}{200}=\frac{\mathrm{I}_{\mathrm{S}}}{4}$ $\mathrm{I}_{\mathrm{S}}=1 \mathrm{~A}$
EAMCET-2018
Alternating Current
155399
A transformer has 100 turns in the primary coil and carries $8 \mathrm{~A}$ current. If input power is $1 \mathrm{~kW}$, the number of turns in secondary coil to have $500 \mathrm{~V}$ of output will be
1 100
2 200
3 400
4 300
Explanation:
C Given that, Primary number of turns $=100$ Primary current $=8 \mathrm{~A}$ Input power $=1 \mathrm{~kW}$ $\text { Power = V.I }$ $1000=500 . \mathrm{I}$ $\mathrm{I}=2 \mathrm{~A}$ Now, from the transformers equation - $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}$ $\frac{8}{2}=\frac{\mathrm{N}_{2}}{100}$ $\mathrm{~N}_{2}=400$
JCECE-2018
Alternating Current
155400
A transformer is used to glow a $140 \mathrm{~W}-24 \mathrm{~V}$ bulb at $240 \mathrm{~V}$ AC. If current in the primary coil is $0.7 \mathrm{~A}$, then the efficiency of the transformer is
1 $16.7 \%$
2 $50 \%$
3 $83.3 \%$
4 $25 \%$
Explanation:
C Given that, Input voltage $=240 \mathrm{~V}$ Primary coil current $=0.7 \mathrm{~A}$ Output power $=140 \mathrm{~W}$ Power input $=\mathrm{V} \times \mathrm{I}$ $\mathrm{P}=240 \times 0.7$ $\mathrm{P}=168.0 \mathrm{~W}$ $\therefore$ Efficiency of transformer, $\eta \%=\frac{P_{\text {output }}}{P_{\text {input }}} \times 100=\frac{140}{168} \times 100$ $\eta=83.33 \%$
CG PET -2018
Alternating Current
155401
A distribution transformer with an efficiency of $\mathbf{9 0 \%}$ supplies to a colony of 10 homes. All the 10 homes have electrical oven running at the same time, that draw 20 A current from $220 \mathrm{~V}$ lines. The power dissipated as heat in the transformer is
1 $12.2 \mathrm{~kW}$
2 $4.9 \mathrm{~kW}$
3 $8.4 \mathrm{~kW}$
4 $9.9 \mathrm{~kW}$
Explanation:
B Given that, Efficiency of transformer $(\eta)=90 \%$ $\mathrm{V}=220$ volt, $\mathrm{I}=20 \mathrm{~A}$ We know that, $\mathrm{P}_{1}=10 \mathrm{VI}=10 \times 220 \times 20=44 \times 10^{3} \mathrm{~W}$ So, power delivered by transformer $\mathrm{P}_{2}=\frac{100}{90} \times \mathrm{P}_{1}=\frac{10}{9} \times 44 \times 10^{3}$ $\mathrm{P}_{2}=48.9 \times 10^{3} \mathrm{~W}$ $\therefore$ Power dissipated as heat $\mathrm{P}_{\mathrm{H}}=\mathrm{P}_{2}-\mathrm{P}_{1}$ $\mathrm{P}_{\mathrm{H}}=(48.9-44) \times 10^{3}$ $\mathrm{P}_{\mathrm{H}}=4.9 \times 10^{3}$ $\mathrm{P}_{\mathrm{H}}=4.9 \mathrm{~kW}$