356329
The output of a step-down transformer is measured to be 24 \(V\) when connected to a 12 \(W\) light bulb. The value of the peak current is
1 \(\sqrt 2 \,A\)
2 \(\frac{1}{{\sqrt 2 \,}}A\)
3 \(2\sqrt 2 \,A\)
4 \(2\,A\)
Explanation:
Secondary voltage \({V_S} = 24V\) Power associated with secondary \({P_S} = 12\,W\) \({I_S} = \frac{{{P_S}}}{{{V_S}}} = \frac{{12}}{{24}} = \frac{1}{2}A\) Peak value of the current is \({I_0} = {I_S}\sqrt 2 = \frac{1}{{\sqrt 2 }}A\)
NCERT Exemplar
PHXII07:ALTERNATING CURRENT
356330
Statement A : The power is produced when a transformer steps up the voltage. Statement B : In an ideal transformer \(VI = {\mathop{\rm constant}\nolimits} \)
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Transformer cannot produce power, but it transfer from primary to secondary.Power \(P = VI\) is constant for an ideal transformer. So option (3) is correct.
PHXII07:ALTERNATING CURRENT
356331
The transformer ratio of a transformer is \(10: 1\). If the primary voltage is \(440\;V\), secondary emf is
1 \(44\;V\)
2 \(440\;V\)
3 \(4400\;V\)
4 \(44000\;V\)
Explanation:
\(\dfrac{N_{S}}{N_{p}}=\dfrac{V_{s}}{V_{p}} \Rightarrow V_{s}=(10)(440)\) \( \Rightarrow {V_s} = 4400\;V\) So, correct option is (3).
PHXII07:ALTERNATING CURRENT
356332
A transformer has 500 primary turns and 10 secondary turns. If the secondary has a resistive load of \(15 \Omega\), the currents in the primary and secondary respectively, are
We have, \(\frac{{{N_s}}}{{{N_p}}} = \frac{{{i_p}}}{{{i_s}}} \Rightarrow \frac{{10}}{{500}} = \frac{{{i_p}}}{{{i_s}}}\) \( \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{i_p}}}{{{i_s}}} = \frac{1}{{50}} \Rightarrow {i_s} = 50{i_p}\) This condition is satisfied only when current in primary is \(3.2 \times {10^{ - 3}}\;A\) and in secondary \(0.16\;A.\)
PHXII07:ALTERNATING CURRENT
356333
A step down transformer has 50 turns on secondary and 1000 turns on primary winding. If a transofrmer is connected to 220\(V\), 1\(A\) \(AC\). source, what is output current of the transformer?
1 \(100\,A\)
2 \(\frac{1}{{20}}\,A\)
3 \(2\,A\)
4 \(20\,A\)
Explanation:
\(\frac{{{I_P}}}{{{I_S}}} = \frac{{{N_S}}}{{{N_P}}},\) where subscripts ‘\(P\)’ and ‘\(S\)’ represenet primary and secondary. \(\therefore \;\;{I_S} = \left( {\frac{{{N_P}}}{{{N_S}}}} \right){I_P}\) Here,\({I_P} = 1A,{N_P} = 1000,{N_S} = 50\) \(\therefore \,{I_S} = \left( {\frac{{1000}}{{50}}} \right)(1\,A) = 20A\) Hence the output current of the transformer is \(20A\).
356329
The output of a step-down transformer is measured to be 24 \(V\) when connected to a 12 \(W\) light bulb. The value of the peak current is
1 \(\sqrt 2 \,A\)
2 \(\frac{1}{{\sqrt 2 \,}}A\)
3 \(2\sqrt 2 \,A\)
4 \(2\,A\)
Explanation:
Secondary voltage \({V_S} = 24V\) Power associated with secondary \({P_S} = 12\,W\) \({I_S} = \frac{{{P_S}}}{{{V_S}}} = \frac{{12}}{{24}} = \frac{1}{2}A\) Peak value of the current is \({I_0} = {I_S}\sqrt 2 = \frac{1}{{\sqrt 2 }}A\)
NCERT Exemplar
PHXII07:ALTERNATING CURRENT
356330
Statement A : The power is produced when a transformer steps up the voltage. Statement B : In an ideal transformer \(VI = {\mathop{\rm constant}\nolimits} \)
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Transformer cannot produce power, but it transfer from primary to secondary.Power \(P = VI\) is constant for an ideal transformer. So option (3) is correct.
PHXII07:ALTERNATING CURRENT
356331
The transformer ratio of a transformer is \(10: 1\). If the primary voltage is \(440\;V\), secondary emf is
1 \(44\;V\)
2 \(440\;V\)
3 \(4400\;V\)
4 \(44000\;V\)
Explanation:
\(\dfrac{N_{S}}{N_{p}}=\dfrac{V_{s}}{V_{p}} \Rightarrow V_{s}=(10)(440)\) \( \Rightarrow {V_s} = 4400\;V\) So, correct option is (3).
PHXII07:ALTERNATING CURRENT
356332
A transformer has 500 primary turns and 10 secondary turns. If the secondary has a resistive load of \(15 \Omega\), the currents in the primary and secondary respectively, are
We have, \(\frac{{{N_s}}}{{{N_p}}} = \frac{{{i_p}}}{{{i_s}}} \Rightarrow \frac{{10}}{{500}} = \frac{{{i_p}}}{{{i_s}}}\) \( \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{i_p}}}{{{i_s}}} = \frac{1}{{50}} \Rightarrow {i_s} = 50{i_p}\) This condition is satisfied only when current in primary is \(3.2 \times {10^{ - 3}}\;A\) and in secondary \(0.16\;A.\)
PHXII07:ALTERNATING CURRENT
356333
A step down transformer has 50 turns on secondary and 1000 turns on primary winding. If a transofrmer is connected to 220\(V\), 1\(A\) \(AC\). source, what is output current of the transformer?
1 \(100\,A\)
2 \(\frac{1}{{20}}\,A\)
3 \(2\,A\)
4 \(20\,A\)
Explanation:
\(\frac{{{I_P}}}{{{I_S}}} = \frac{{{N_S}}}{{{N_P}}},\) where subscripts ‘\(P\)’ and ‘\(S\)’ represenet primary and secondary. \(\therefore \;\;{I_S} = \left( {\frac{{{N_P}}}{{{N_S}}}} \right){I_P}\) Here,\({I_P} = 1A,{N_P} = 1000,{N_S} = 50\) \(\therefore \,{I_S} = \left( {\frac{{1000}}{{50}}} \right)(1\,A) = 20A\) Hence the output current of the transformer is \(20A\).
356329
The output of a step-down transformer is measured to be 24 \(V\) when connected to a 12 \(W\) light bulb. The value of the peak current is
1 \(\sqrt 2 \,A\)
2 \(\frac{1}{{\sqrt 2 \,}}A\)
3 \(2\sqrt 2 \,A\)
4 \(2\,A\)
Explanation:
Secondary voltage \({V_S} = 24V\) Power associated with secondary \({P_S} = 12\,W\) \({I_S} = \frac{{{P_S}}}{{{V_S}}} = \frac{{12}}{{24}} = \frac{1}{2}A\) Peak value of the current is \({I_0} = {I_S}\sqrt 2 = \frac{1}{{\sqrt 2 }}A\)
NCERT Exemplar
PHXII07:ALTERNATING CURRENT
356330
Statement A : The power is produced when a transformer steps up the voltage. Statement B : In an ideal transformer \(VI = {\mathop{\rm constant}\nolimits} \)
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Transformer cannot produce power, but it transfer from primary to secondary.Power \(P = VI\) is constant for an ideal transformer. So option (3) is correct.
PHXII07:ALTERNATING CURRENT
356331
The transformer ratio of a transformer is \(10: 1\). If the primary voltage is \(440\;V\), secondary emf is
1 \(44\;V\)
2 \(440\;V\)
3 \(4400\;V\)
4 \(44000\;V\)
Explanation:
\(\dfrac{N_{S}}{N_{p}}=\dfrac{V_{s}}{V_{p}} \Rightarrow V_{s}=(10)(440)\) \( \Rightarrow {V_s} = 4400\;V\) So, correct option is (3).
PHXII07:ALTERNATING CURRENT
356332
A transformer has 500 primary turns and 10 secondary turns. If the secondary has a resistive load of \(15 \Omega\), the currents in the primary and secondary respectively, are
We have, \(\frac{{{N_s}}}{{{N_p}}} = \frac{{{i_p}}}{{{i_s}}} \Rightarrow \frac{{10}}{{500}} = \frac{{{i_p}}}{{{i_s}}}\) \( \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{i_p}}}{{{i_s}}} = \frac{1}{{50}} \Rightarrow {i_s} = 50{i_p}\) This condition is satisfied only when current in primary is \(3.2 \times {10^{ - 3}}\;A\) and in secondary \(0.16\;A.\)
PHXII07:ALTERNATING CURRENT
356333
A step down transformer has 50 turns on secondary and 1000 turns on primary winding. If a transofrmer is connected to 220\(V\), 1\(A\) \(AC\). source, what is output current of the transformer?
1 \(100\,A\)
2 \(\frac{1}{{20}}\,A\)
3 \(2\,A\)
4 \(20\,A\)
Explanation:
\(\frac{{{I_P}}}{{{I_S}}} = \frac{{{N_S}}}{{{N_P}}},\) where subscripts ‘\(P\)’ and ‘\(S\)’ represenet primary and secondary. \(\therefore \;\;{I_S} = \left( {\frac{{{N_P}}}{{{N_S}}}} \right){I_P}\) Here,\({I_P} = 1A,{N_P} = 1000,{N_S} = 50\) \(\therefore \,{I_S} = \left( {\frac{{1000}}{{50}}} \right)(1\,A) = 20A\) Hence the output current of the transformer is \(20A\).
356329
The output of a step-down transformer is measured to be 24 \(V\) when connected to a 12 \(W\) light bulb. The value of the peak current is
1 \(\sqrt 2 \,A\)
2 \(\frac{1}{{\sqrt 2 \,}}A\)
3 \(2\sqrt 2 \,A\)
4 \(2\,A\)
Explanation:
Secondary voltage \({V_S} = 24V\) Power associated with secondary \({P_S} = 12\,W\) \({I_S} = \frac{{{P_S}}}{{{V_S}}} = \frac{{12}}{{24}} = \frac{1}{2}A\) Peak value of the current is \({I_0} = {I_S}\sqrt 2 = \frac{1}{{\sqrt 2 }}A\)
NCERT Exemplar
PHXII07:ALTERNATING CURRENT
356330
Statement A : The power is produced when a transformer steps up the voltage. Statement B : In an ideal transformer \(VI = {\mathop{\rm constant}\nolimits} \)
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Transformer cannot produce power, but it transfer from primary to secondary.Power \(P = VI\) is constant for an ideal transformer. So option (3) is correct.
PHXII07:ALTERNATING CURRENT
356331
The transformer ratio of a transformer is \(10: 1\). If the primary voltage is \(440\;V\), secondary emf is
1 \(44\;V\)
2 \(440\;V\)
3 \(4400\;V\)
4 \(44000\;V\)
Explanation:
\(\dfrac{N_{S}}{N_{p}}=\dfrac{V_{s}}{V_{p}} \Rightarrow V_{s}=(10)(440)\) \( \Rightarrow {V_s} = 4400\;V\) So, correct option is (3).
PHXII07:ALTERNATING CURRENT
356332
A transformer has 500 primary turns and 10 secondary turns. If the secondary has a resistive load of \(15 \Omega\), the currents in the primary and secondary respectively, are
We have, \(\frac{{{N_s}}}{{{N_p}}} = \frac{{{i_p}}}{{{i_s}}} \Rightarrow \frac{{10}}{{500}} = \frac{{{i_p}}}{{{i_s}}}\) \( \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{i_p}}}{{{i_s}}} = \frac{1}{{50}} \Rightarrow {i_s} = 50{i_p}\) This condition is satisfied only when current in primary is \(3.2 \times {10^{ - 3}}\;A\) and in secondary \(0.16\;A.\)
PHXII07:ALTERNATING CURRENT
356333
A step down transformer has 50 turns on secondary and 1000 turns on primary winding. If a transofrmer is connected to 220\(V\), 1\(A\) \(AC\). source, what is output current of the transformer?
1 \(100\,A\)
2 \(\frac{1}{{20}}\,A\)
3 \(2\,A\)
4 \(20\,A\)
Explanation:
\(\frac{{{I_P}}}{{{I_S}}} = \frac{{{N_S}}}{{{N_P}}},\) where subscripts ‘\(P\)’ and ‘\(S\)’ represenet primary and secondary. \(\therefore \;\;{I_S} = \left( {\frac{{{N_P}}}{{{N_S}}}} \right){I_P}\) Here,\({I_P} = 1A,{N_P} = 1000,{N_S} = 50\) \(\therefore \,{I_S} = \left( {\frac{{1000}}{{50}}} \right)(1\,A) = 20A\) Hence the output current of the transformer is \(20A\).
356329
The output of a step-down transformer is measured to be 24 \(V\) when connected to a 12 \(W\) light bulb. The value of the peak current is
1 \(\sqrt 2 \,A\)
2 \(\frac{1}{{\sqrt 2 \,}}A\)
3 \(2\sqrt 2 \,A\)
4 \(2\,A\)
Explanation:
Secondary voltage \({V_S} = 24V\) Power associated with secondary \({P_S} = 12\,W\) \({I_S} = \frac{{{P_S}}}{{{V_S}}} = \frac{{12}}{{24}} = \frac{1}{2}A\) Peak value of the current is \({I_0} = {I_S}\sqrt 2 = \frac{1}{{\sqrt 2 }}A\)
NCERT Exemplar
PHXII07:ALTERNATING CURRENT
356330
Statement A : The power is produced when a transformer steps up the voltage. Statement B : In an ideal transformer \(VI = {\mathop{\rm constant}\nolimits} \)
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Transformer cannot produce power, but it transfer from primary to secondary.Power \(P = VI\) is constant for an ideal transformer. So option (3) is correct.
PHXII07:ALTERNATING CURRENT
356331
The transformer ratio of a transformer is \(10: 1\). If the primary voltage is \(440\;V\), secondary emf is
1 \(44\;V\)
2 \(440\;V\)
3 \(4400\;V\)
4 \(44000\;V\)
Explanation:
\(\dfrac{N_{S}}{N_{p}}=\dfrac{V_{s}}{V_{p}} \Rightarrow V_{s}=(10)(440)\) \( \Rightarrow {V_s} = 4400\;V\) So, correct option is (3).
PHXII07:ALTERNATING CURRENT
356332
A transformer has 500 primary turns and 10 secondary turns. If the secondary has a resistive load of \(15 \Omega\), the currents in the primary and secondary respectively, are
We have, \(\frac{{{N_s}}}{{{N_p}}} = \frac{{{i_p}}}{{{i_s}}} \Rightarrow \frac{{10}}{{500}} = \frac{{{i_p}}}{{{i_s}}}\) \( \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{i_p}}}{{{i_s}}} = \frac{1}{{50}} \Rightarrow {i_s} = 50{i_p}\) This condition is satisfied only when current in primary is \(3.2 \times {10^{ - 3}}\;A\) and in secondary \(0.16\;A.\)
PHXII07:ALTERNATING CURRENT
356333
A step down transformer has 50 turns on secondary and 1000 turns on primary winding. If a transofrmer is connected to 220\(V\), 1\(A\) \(AC\). source, what is output current of the transformer?
1 \(100\,A\)
2 \(\frac{1}{{20}}\,A\)
3 \(2\,A\)
4 \(20\,A\)
Explanation:
\(\frac{{{I_P}}}{{{I_S}}} = \frac{{{N_S}}}{{{N_P}}},\) where subscripts ‘\(P\)’ and ‘\(S\)’ represenet primary and secondary. \(\therefore \;\;{I_S} = \left( {\frac{{{N_P}}}{{{N_S}}}} \right){I_P}\) Here,\({I_P} = 1A,{N_P} = 1000,{N_S} = 50\) \(\therefore \,{I_S} = \left( {\frac{{1000}}{{50}}} \right)(1\,A) = 20A\) Hence the output current of the transformer is \(20A\).
