155422
A step-down transformer increases the input current $4 \mathrm{~A}$ to $24 \mathrm{~A}$ at the secondary. If the number of turns in the primary coil is 330, the number of turns in the secondary coil is
1 60
2 50
3 65
4 45
5 55
Explanation:
E Given that, Primary current $\left(\mathrm{I}_{1}\right)=4 \mathrm{~A}$, Secondary current $\left(\mathrm{I}_{2}\right)=24 \mathrm{~A}$, Number of turns $\left(\mathrm{N}_{1}\right)=330$ turns From transformers equation - $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{330}{\mathrm{~N}_{2}}=\frac{24}{4}$ $\frac{330}{\mathrm{~N}_{2}}=6$ $\mathrm{~N}_{2}=55$
Kerala CEE- 2014
Alternating Current
155423
A transformer has an efficiency of $80 \%$. It is connected to a power input of $5 \mathrm{~kW}$ at $200 \mathrm{~V}$. If the secondary voltage is $250 \mathrm{~V}$, the primary and secondary currents are respectively
1 $25 \mathrm{~A}, 20 \mathrm{~A}$
2 $20 \mathrm{~A}, 16 \mathrm{~A}$
3 $25 \mathrm{~A}, 16 \mathrm{~A}$
4 $40 \mathrm{~A}, 25 \mathrm{~A}$
5 $40 \mathrm{~A}, 16 \mathrm{~A}$
Explanation:
C Given that, Efficiency of transformer $(\eta)=80 \%$ Power input $\left(\mathrm{P}_{\text {input }}\right)=5000 \mathrm{~W}$ Primary voltage $\left(\mathrm{V}_{\text {input }}\right)=200 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{\mathrm{s}}\right)=250 \mathrm{~V}$ We know that, $\eta=\frac{P_{\text {output }}}{P_{\text {input }}}$ $\frac{80}{100}=\frac{P_{\text {output }}}{5000}$ $P_{\text {output }}=4000 \mathrm{~W}$ And $\quad I_{p}=\frac{P_{\text {input }}}{V_{\text {input }}}$ $I_{p}=\frac{5000}{200}=25 \mathrm{~A} .$ $I_{\mathrm{s}}=\frac{P_{\text {ouput }}}{V_{\mathrm{s}}}=\frac{4000}{250}=16 \mathrm{~A}$
Kerala CEE 2007
Alternating Current
155424
A step-down transformer is used on a $1000 \mathrm{~V}$ line to deliver $20 \mathrm{~A}$ at $120 \mathrm{~V}$ at the secondary coil. If the efficiency of the transformer is $80 \%$, the current drawn from the line is:
1 $3 \mathrm{~A}$
2 $30 \mathrm{~A}$
3 $0.3 \mathrm{~A}$
4 $2.4 \mathrm{~A}$
5 $24 \mathrm{~A}$
Explanation:
A Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=1000 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=120 \mathrm{~V}$ Secondary current $\left(\mathrm{I}_{2}\right)=20 \mathrm{~A}$ Efficiency $=80 \%$ We know that, Efficiency of transformer $=\frac{\text { Power output }}{\text { Power Input }}=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{\mathrm{~V}_{1} \mathrm{I}_{1}}$. $\frac{80}{100}=\frac{120 \times 20}{1000 \times I_{1}}$ $I_{1}=3 \mathrm{~A}$
Kerala CEE 2005
Alternating Current
155425
The transformation ratio in the step-up transformer is
1 1
2 greater than one
3 less than one
4 the ratio greater or less than one depends on the other factors
Explanation:
B We know that, Transformer ratio is given by $K=\frac{N_{S}}{N_{P}}=\frac{E_{S}}{E_{P}}$ Where, $N_{P}=$ Number of turns of in primary coil $\mathrm{N}_{\mathrm{S}}=$ Number of turns of in secondary coil $E_{\mathrm{P}}=$ Emf's of primary coil $E_{S}=$ Emf's of secondary coil For a step-up transformer, $\mathrm{E}_{\mathrm{S}}>\mathrm{E}_{\mathrm{P}}$ $\therefore \quad \mathrm{K}>1$ For a step-down transformer, $\mathrm{E}_{\mathrm{S}} \lt \mathrm{E}_{\mathrm{P}}$ $\therefore \quad \mathrm{K} \lt 1$
155422
A step-down transformer increases the input current $4 \mathrm{~A}$ to $24 \mathrm{~A}$ at the secondary. If the number of turns in the primary coil is 330, the number of turns in the secondary coil is
1 60
2 50
3 65
4 45
5 55
Explanation:
E Given that, Primary current $\left(\mathrm{I}_{1}\right)=4 \mathrm{~A}$, Secondary current $\left(\mathrm{I}_{2}\right)=24 \mathrm{~A}$, Number of turns $\left(\mathrm{N}_{1}\right)=330$ turns From transformers equation - $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{330}{\mathrm{~N}_{2}}=\frac{24}{4}$ $\frac{330}{\mathrm{~N}_{2}}=6$ $\mathrm{~N}_{2}=55$
Kerala CEE- 2014
Alternating Current
155423
A transformer has an efficiency of $80 \%$. It is connected to a power input of $5 \mathrm{~kW}$ at $200 \mathrm{~V}$. If the secondary voltage is $250 \mathrm{~V}$, the primary and secondary currents are respectively
1 $25 \mathrm{~A}, 20 \mathrm{~A}$
2 $20 \mathrm{~A}, 16 \mathrm{~A}$
3 $25 \mathrm{~A}, 16 \mathrm{~A}$
4 $40 \mathrm{~A}, 25 \mathrm{~A}$
5 $40 \mathrm{~A}, 16 \mathrm{~A}$
Explanation:
C Given that, Efficiency of transformer $(\eta)=80 \%$ Power input $\left(\mathrm{P}_{\text {input }}\right)=5000 \mathrm{~W}$ Primary voltage $\left(\mathrm{V}_{\text {input }}\right)=200 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{\mathrm{s}}\right)=250 \mathrm{~V}$ We know that, $\eta=\frac{P_{\text {output }}}{P_{\text {input }}}$ $\frac{80}{100}=\frac{P_{\text {output }}}{5000}$ $P_{\text {output }}=4000 \mathrm{~W}$ And $\quad I_{p}=\frac{P_{\text {input }}}{V_{\text {input }}}$ $I_{p}=\frac{5000}{200}=25 \mathrm{~A} .$ $I_{\mathrm{s}}=\frac{P_{\text {ouput }}}{V_{\mathrm{s}}}=\frac{4000}{250}=16 \mathrm{~A}$
Kerala CEE 2007
Alternating Current
155424
A step-down transformer is used on a $1000 \mathrm{~V}$ line to deliver $20 \mathrm{~A}$ at $120 \mathrm{~V}$ at the secondary coil. If the efficiency of the transformer is $80 \%$, the current drawn from the line is:
1 $3 \mathrm{~A}$
2 $30 \mathrm{~A}$
3 $0.3 \mathrm{~A}$
4 $2.4 \mathrm{~A}$
5 $24 \mathrm{~A}$
Explanation:
A Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=1000 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=120 \mathrm{~V}$ Secondary current $\left(\mathrm{I}_{2}\right)=20 \mathrm{~A}$ Efficiency $=80 \%$ We know that, Efficiency of transformer $=\frac{\text { Power output }}{\text { Power Input }}=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{\mathrm{~V}_{1} \mathrm{I}_{1}}$. $\frac{80}{100}=\frac{120 \times 20}{1000 \times I_{1}}$ $I_{1}=3 \mathrm{~A}$
Kerala CEE 2005
Alternating Current
155425
The transformation ratio in the step-up transformer is
1 1
2 greater than one
3 less than one
4 the ratio greater or less than one depends on the other factors
Explanation:
B We know that, Transformer ratio is given by $K=\frac{N_{S}}{N_{P}}=\frac{E_{S}}{E_{P}}$ Where, $N_{P}=$ Number of turns of in primary coil $\mathrm{N}_{\mathrm{S}}=$ Number of turns of in secondary coil $E_{\mathrm{P}}=$ Emf's of primary coil $E_{S}=$ Emf's of secondary coil For a step-up transformer, $\mathrm{E}_{\mathrm{S}}>\mathrm{E}_{\mathrm{P}}$ $\therefore \quad \mathrm{K}>1$ For a step-down transformer, $\mathrm{E}_{\mathrm{S}} \lt \mathrm{E}_{\mathrm{P}}$ $\therefore \quad \mathrm{K} \lt 1$
155422
A step-down transformer increases the input current $4 \mathrm{~A}$ to $24 \mathrm{~A}$ at the secondary. If the number of turns in the primary coil is 330, the number of turns in the secondary coil is
1 60
2 50
3 65
4 45
5 55
Explanation:
E Given that, Primary current $\left(\mathrm{I}_{1}\right)=4 \mathrm{~A}$, Secondary current $\left(\mathrm{I}_{2}\right)=24 \mathrm{~A}$, Number of turns $\left(\mathrm{N}_{1}\right)=330$ turns From transformers equation - $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{330}{\mathrm{~N}_{2}}=\frac{24}{4}$ $\frac{330}{\mathrm{~N}_{2}}=6$ $\mathrm{~N}_{2}=55$
Kerala CEE- 2014
Alternating Current
155423
A transformer has an efficiency of $80 \%$. It is connected to a power input of $5 \mathrm{~kW}$ at $200 \mathrm{~V}$. If the secondary voltage is $250 \mathrm{~V}$, the primary and secondary currents are respectively
1 $25 \mathrm{~A}, 20 \mathrm{~A}$
2 $20 \mathrm{~A}, 16 \mathrm{~A}$
3 $25 \mathrm{~A}, 16 \mathrm{~A}$
4 $40 \mathrm{~A}, 25 \mathrm{~A}$
5 $40 \mathrm{~A}, 16 \mathrm{~A}$
Explanation:
C Given that, Efficiency of transformer $(\eta)=80 \%$ Power input $\left(\mathrm{P}_{\text {input }}\right)=5000 \mathrm{~W}$ Primary voltage $\left(\mathrm{V}_{\text {input }}\right)=200 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{\mathrm{s}}\right)=250 \mathrm{~V}$ We know that, $\eta=\frac{P_{\text {output }}}{P_{\text {input }}}$ $\frac{80}{100}=\frac{P_{\text {output }}}{5000}$ $P_{\text {output }}=4000 \mathrm{~W}$ And $\quad I_{p}=\frac{P_{\text {input }}}{V_{\text {input }}}$ $I_{p}=\frac{5000}{200}=25 \mathrm{~A} .$ $I_{\mathrm{s}}=\frac{P_{\text {ouput }}}{V_{\mathrm{s}}}=\frac{4000}{250}=16 \mathrm{~A}$
Kerala CEE 2007
Alternating Current
155424
A step-down transformer is used on a $1000 \mathrm{~V}$ line to deliver $20 \mathrm{~A}$ at $120 \mathrm{~V}$ at the secondary coil. If the efficiency of the transformer is $80 \%$, the current drawn from the line is:
1 $3 \mathrm{~A}$
2 $30 \mathrm{~A}$
3 $0.3 \mathrm{~A}$
4 $2.4 \mathrm{~A}$
5 $24 \mathrm{~A}$
Explanation:
A Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=1000 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=120 \mathrm{~V}$ Secondary current $\left(\mathrm{I}_{2}\right)=20 \mathrm{~A}$ Efficiency $=80 \%$ We know that, Efficiency of transformer $=\frac{\text { Power output }}{\text { Power Input }}=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{\mathrm{~V}_{1} \mathrm{I}_{1}}$. $\frac{80}{100}=\frac{120 \times 20}{1000 \times I_{1}}$ $I_{1}=3 \mathrm{~A}$
Kerala CEE 2005
Alternating Current
155425
The transformation ratio in the step-up transformer is
1 1
2 greater than one
3 less than one
4 the ratio greater or less than one depends on the other factors
Explanation:
B We know that, Transformer ratio is given by $K=\frac{N_{S}}{N_{P}}=\frac{E_{S}}{E_{P}}$ Where, $N_{P}=$ Number of turns of in primary coil $\mathrm{N}_{\mathrm{S}}=$ Number of turns of in secondary coil $E_{\mathrm{P}}=$ Emf's of primary coil $E_{S}=$ Emf's of secondary coil For a step-up transformer, $\mathrm{E}_{\mathrm{S}}>\mathrm{E}_{\mathrm{P}}$ $\therefore \quad \mathrm{K}>1$ For a step-down transformer, $\mathrm{E}_{\mathrm{S}} \lt \mathrm{E}_{\mathrm{P}}$ $\therefore \quad \mathrm{K} \lt 1$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155422
A step-down transformer increases the input current $4 \mathrm{~A}$ to $24 \mathrm{~A}$ at the secondary. If the number of turns in the primary coil is 330, the number of turns in the secondary coil is
1 60
2 50
3 65
4 45
5 55
Explanation:
E Given that, Primary current $\left(\mathrm{I}_{1}\right)=4 \mathrm{~A}$, Secondary current $\left(\mathrm{I}_{2}\right)=24 \mathrm{~A}$, Number of turns $\left(\mathrm{N}_{1}\right)=330$ turns From transformers equation - $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ $\frac{330}{\mathrm{~N}_{2}}=\frac{24}{4}$ $\frac{330}{\mathrm{~N}_{2}}=6$ $\mathrm{~N}_{2}=55$
Kerala CEE- 2014
Alternating Current
155423
A transformer has an efficiency of $80 \%$. It is connected to a power input of $5 \mathrm{~kW}$ at $200 \mathrm{~V}$. If the secondary voltage is $250 \mathrm{~V}$, the primary and secondary currents are respectively
1 $25 \mathrm{~A}, 20 \mathrm{~A}$
2 $20 \mathrm{~A}, 16 \mathrm{~A}$
3 $25 \mathrm{~A}, 16 \mathrm{~A}$
4 $40 \mathrm{~A}, 25 \mathrm{~A}$
5 $40 \mathrm{~A}, 16 \mathrm{~A}$
Explanation:
C Given that, Efficiency of transformer $(\eta)=80 \%$ Power input $\left(\mathrm{P}_{\text {input }}\right)=5000 \mathrm{~W}$ Primary voltage $\left(\mathrm{V}_{\text {input }}\right)=200 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{\mathrm{s}}\right)=250 \mathrm{~V}$ We know that, $\eta=\frac{P_{\text {output }}}{P_{\text {input }}}$ $\frac{80}{100}=\frac{P_{\text {output }}}{5000}$ $P_{\text {output }}=4000 \mathrm{~W}$ And $\quad I_{p}=\frac{P_{\text {input }}}{V_{\text {input }}}$ $I_{p}=\frac{5000}{200}=25 \mathrm{~A} .$ $I_{\mathrm{s}}=\frac{P_{\text {ouput }}}{V_{\mathrm{s}}}=\frac{4000}{250}=16 \mathrm{~A}$
Kerala CEE 2007
Alternating Current
155424
A step-down transformer is used on a $1000 \mathrm{~V}$ line to deliver $20 \mathrm{~A}$ at $120 \mathrm{~V}$ at the secondary coil. If the efficiency of the transformer is $80 \%$, the current drawn from the line is:
1 $3 \mathrm{~A}$
2 $30 \mathrm{~A}$
3 $0.3 \mathrm{~A}$
4 $2.4 \mathrm{~A}$
5 $24 \mathrm{~A}$
Explanation:
A Given that, Primary voltage $\left(\mathrm{V}_{1}\right)=1000 \mathrm{~V}$ Secondary voltage $\left(\mathrm{V}_{2}\right)=120 \mathrm{~V}$ Secondary current $\left(\mathrm{I}_{2}\right)=20 \mathrm{~A}$ Efficiency $=80 \%$ We know that, Efficiency of transformer $=\frac{\text { Power output }}{\text { Power Input }}=\frac{\mathrm{V}_{2} \mathrm{I}_{2}}{\mathrm{~V}_{1} \mathrm{I}_{1}}$. $\frac{80}{100}=\frac{120 \times 20}{1000 \times I_{1}}$ $I_{1}=3 \mathrm{~A}$
Kerala CEE 2005
Alternating Current
155425
The transformation ratio in the step-up transformer is
1 1
2 greater than one
3 less than one
4 the ratio greater or less than one depends on the other factors
Explanation:
B We know that, Transformer ratio is given by $K=\frac{N_{S}}{N_{P}}=\frac{E_{S}}{E_{P}}$ Where, $N_{P}=$ Number of turns of in primary coil $\mathrm{N}_{\mathrm{S}}=$ Number of turns of in secondary coil $E_{\mathrm{P}}=$ Emf's of primary coil $E_{S}=$ Emf's of secondary coil For a step-up transformer, $\mathrm{E}_{\mathrm{S}}>\mathrm{E}_{\mathrm{P}}$ $\therefore \quad \mathrm{K}>1$ For a step-down transformer, $\mathrm{E}_{\mathrm{S}} \lt \mathrm{E}_{\mathrm{P}}$ $\therefore \quad \mathrm{K} \lt 1$
