NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII07:ALTERNATING CURRENT
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A transformer is used to light a 100 \(W\) and 110\(V\) lamp from a 220 \(V\) mains. If the main current is 0.5 \(A\), the efficiency of the transformer is approximately
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A current of 5 \(A\) is flowing at 220 \(V\) in the primary coil of a transformer. If the voltage produced in the secondary coil is 2200 \(V\) and 50% of power is lost, then the current is secondary will be:
356289
A transformer connected to \(220\,V\) line shows an output of \(2\,A\) at \(11000\,V\). The efficiency is \({100 \%}\). The current drawn from the line is
356290
For a transformer, the turns ratio is 3 and its efficiency is 0.75. The current flowing in the primary coil is 2\(A\) and the voltage applied to it is 100\(V\). Then the voltage and the current flowing in the secondary coil are respectively.
1 \(150{\rm{ }}V,{\rm{ }}1.5{\rm{ }}A\)
2 \(300{\rm{ }}V,{\rm{ 0}}.5{\rm{ }}A\)
3 \(300{\rm{ }}V,{\rm{ 1}}.5{\rm{ }}A\)
4 \(150{\rm{ }}V,{\rm{ 0}}.5{\rm{ }}A\)
Explanation:
Given,\(\frac{{{N_S}}}{{{N_P}}} = 3,\eta = 0.75m{I_P} = 2A,{V_P} = 100V\) For a transformer, turns ratio = voltage ratio \( \Rightarrow \;\;\frac{{{N_S}}}{{{N_P}}} = \frac{{{V_S}}}{{{V_P}}}\) \(\therefore \;\;{V_S} = \frac{{{N_S}}}{{{N_P}}} \times {V_P} = 3 \times 100\,V = 300\;V\) Efficiency of a transformer, \(\eta = \frac{{Output\,power}}{{Input\,power}} = \frac{{{V_S}{I_S}}}{{{V_P}{I_P}}}\) \(\therefore \;\;0.75 = \frac{{(300V)({I_S})}}{{(100V)(2A)}}\) \({I_s} = \frac{{1.5}}{3}A = 0.5A\)
356287
A transformer is used to light a 100 \(W\) and 110\(V\) lamp from a 220 \(V\) mains. If the main current is 0.5 \(A\), the efficiency of the transformer is approximately
356288
A current of 5 \(A\) is flowing at 220 \(V\) in the primary coil of a transformer. If the voltage produced in the secondary coil is 2200 \(V\) and 50% of power is lost, then the current is secondary will be:
356289
A transformer connected to \(220\,V\) line shows an output of \(2\,A\) at \(11000\,V\). The efficiency is \({100 \%}\). The current drawn from the line is
356290
For a transformer, the turns ratio is 3 and its efficiency is 0.75. The current flowing in the primary coil is 2\(A\) and the voltage applied to it is 100\(V\). Then the voltage and the current flowing in the secondary coil are respectively.
1 \(150{\rm{ }}V,{\rm{ }}1.5{\rm{ }}A\)
2 \(300{\rm{ }}V,{\rm{ 0}}.5{\rm{ }}A\)
3 \(300{\rm{ }}V,{\rm{ 1}}.5{\rm{ }}A\)
4 \(150{\rm{ }}V,{\rm{ 0}}.5{\rm{ }}A\)
Explanation:
Given,\(\frac{{{N_S}}}{{{N_P}}} = 3,\eta = 0.75m{I_P} = 2A,{V_P} = 100V\) For a transformer, turns ratio = voltage ratio \( \Rightarrow \;\;\frac{{{N_S}}}{{{N_P}}} = \frac{{{V_S}}}{{{V_P}}}\) \(\therefore \;\;{V_S} = \frac{{{N_S}}}{{{N_P}}} \times {V_P} = 3 \times 100\,V = 300\;V\) Efficiency of a transformer, \(\eta = \frac{{Output\,power}}{{Input\,power}} = \frac{{{V_S}{I_S}}}{{{V_P}{I_P}}}\) \(\therefore \;\;0.75 = \frac{{(300V)({I_S})}}{{(100V)(2A)}}\) \({I_s} = \frac{{1.5}}{3}A = 0.5A\)
356287
A transformer is used to light a 100 \(W\) and 110\(V\) lamp from a 220 \(V\) mains. If the main current is 0.5 \(A\), the efficiency of the transformer is approximately
356288
A current of 5 \(A\) is flowing at 220 \(V\) in the primary coil of a transformer. If the voltage produced in the secondary coil is 2200 \(V\) and 50% of power is lost, then the current is secondary will be:
356289
A transformer connected to \(220\,V\) line shows an output of \(2\,A\) at \(11000\,V\). The efficiency is \({100 \%}\). The current drawn from the line is
356290
For a transformer, the turns ratio is 3 and its efficiency is 0.75. The current flowing in the primary coil is 2\(A\) and the voltage applied to it is 100\(V\). Then the voltage and the current flowing in the secondary coil are respectively.
1 \(150{\rm{ }}V,{\rm{ }}1.5{\rm{ }}A\)
2 \(300{\rm{ }}V,{\rm{ 0}}.5{\rm{ }}A\)
3 \(300{\rm{ }}V,{\rm{ 1}}.5{\rm{ }}A\)
4 \(150{\rm{ }}V,{\rm{ 0}}.5{\rm{ }}A\)
Explanation:
Given,\(\frac{{{N_S}}}{{{N_P}}} = 3,\eta = 0.75m{I_P} = 2A,{V_P} = 100V\) For a transformer, turns ratio = voltage ratio \( \Rightarrow \;\;\frac{{{N_S}}}{{{N_P}}} = \frac{{{V_S}}}{{{V_P}}}\) \(\therefore \;\;{V_S} = \frac{{{N_S}}}{{{N_P}}} \times {V_P} = 3 \times 100\,V = 300\;V\) Efficiency of a transformer, \(\eta = \frac{{Output\,power}}{{Input\,power}} = \frac{{{V_S}{I_S}}}{{{V_P}{I_P}}}\) \(\therefore \;\;0.75 = \frac{{(300V)({I_S})}}{{(100V)(2A)}}\) \({I_s} = \frac{{1.5}}{3}A = 0.5A\)
356287
A transformer is used to light a 100 \(W\) and 110\(V\) lamp from a 220 \(V\) mains. If the main current is 0.5 \(A\), the efficiency of the transformer is approximately
356288
A current of 5 \(A\) is flowing at 220 \(V\) in the primary coil of a transformer. If the voltage produced in the secondary coil is 2200 \(V\) and 50% of power is lost, then the current is secondary will be:
356289
A transformer connected to \(220\,V\) line shows an output of \(2\,A\) at \(11000\,V\). The efficiency is \({100 \%}\). The current drawn from the line is
356290
For a transformer, the turns ratio is 3 and its efficiency is 0.75. The current flowing in the primary coil is 2\(A\) and the voltage applied to it is 100\(V\). Then the voltage and the current flowing in the secondary coil are respectively.
1 \(150{\rm{ }}V,{\rm{ }}1.5{\rm{ }}A\)
2 \(300{\rm{ }}V,{\rm{ 0}}.5{\rm{ }}A\)
3 \(300{\rm{ }}V,{\rm{ 1}}.5{\rm{ }}A\)
4 \(150{\rm{ }}V,{\rm{ 0}}.5{\rm{ }}A\)
Explanation:
Given,\(\frac{{{N_S}}}{{{N_P}}} = 3,\eta = 0.75m{I_P} = 2A,{V_P} = 100V\) For a transformer, turns ratio = voltage ratio \( \Rightarrow \;\;\frac{{{N_S}}}{{{N_P}}} = \frac{{{V_S}}}{{{V_P}}}\) \(\therefore \;\;{V_S} = \frac{{{N_S}}}{{{N_P}}} \times {V_P} = 3 \times 100\,V = 300\;V\) Efficiency of a transformer, \(\eta = \frac{{Output\,power}}{{Input\,power}} = \frac{{{V_S}{I_S}}}{{{V_P}{I_P}}}\) \(\therefore \;\;0.75 = \frac{{(300V)({I_S})}}{{(100V)(2A)}}\) \({I_s} = \frac{{1.5}}{3}A = 0.5A\)
