356321
Primary side of a transformer is connected to \(230\;\,V,50\;\,Hz\) supply. Turns ratio of primary to secondary winding is \(10: 1\). Load resistance connected to secondary side is \(46\,\Omega .\) The power consumed in it is:
1 \(10.0\;\,W\)
2 \(11.5\;\,W\)
3 \(12.5\;\,W\)
4 \(12.0\;\,W\)
Explanation:
Induced voltage \(\propto\) Number of turns in winding. \(\therefore \dfrac{V_{2}}{V_{1}}=\dfrac{N_{2}}{N_{1}}=\dfrac{1}{10}\) \({V_2} = \frac{{{V_1}}}{{10}} = \frac{{230}}{{10}} = 23\;\,V\) \(P_{L}=\dfrac{V_{2}^{2}}{R_{L}}=\dfrac{(23)^{2}}{46}\) \({P_L} = 11.5\;\,W\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356322
The output voltage of a transformer connected to \(220\, V\) line is \(1100\;V\) at \(2 A\) current. Its efficiency is \(100\% \). The current coming from the line is
356323
A step up transformer is connected on the primary side to a \(DC\) battery which can deliver a large current. If a bulb is connected in the secondary, then
1 The bulb will glow very bright
2 The bulb will get fused
3 The bulb will glow, but with less brightness
4 The bulb will not glow
Explanation:
Conceptual Question
PHXII07:ALTERNATING CURRENT
356324
A transformer rated at \(10\;kW\) is used to connect a \(5\;KV\) transmission line to \(240\;V\) circuit. The ratio of turns in the windings of a transformer?
1 5
2 20.8
3 104
4 40
Explanation:
\(5000\;V\) to be connected by \(240\;V\) so, This is a step down transformer. \(\dfrac{N_{p}}{N_{s}}=\dfrac{5000}{240}=20.8\) \(\Rightarrow\) Primary has 20.8 times the turns in the secondary. So, correct option is (2).
356321
Primary side of a transformer is connected to \(230\;\,V,50\;\,Hz\) supply. Turns ratio of primary to secondary winding is \(10: 1\). Load resistance connected to secondary side is \(46\,\Omega .\) The power consumed in it is:
1 \(10.0\;\,W\)
2 \(11.5\;\,W\)
3 \(12.5\;\,W\)
4 \(12.0\;\,W\)
Explanation:
Induced voltage \(\propto\) Number of turns in winding. \(\therefore \dfrac{V_{2}}{V_{1}}=\dfrac{N_{2}}{N_{1}}=\dfrac{1}{10}\) \({V_2} = \frac{{{V_1}}}{{10}} = \frac{{230}}{{10}} = 23\;\,V\) \(P_{L}=\dfrac{V_{2}^{2}}{R_{L}}=\dfrac{(23)^{2}}{46}\) \({P_L} = 11.5\;\,W\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356322
The output voltage of a transformer connected to \(220\, V\) line is \(1100\;V\) at \(2 A\) current. Its efficiency is \(100\% \). The current coming from the line is
356323
A step up transformer is connected on the primary side to a \(DC\) battery which can deliver a large current. If a bulb is connected in the secondary, then
1 The bulb will glow very bright
2 The bulb will get fused
3 The bulb will glow, but with less brightness
4 The bulb will not glow
Explanation:
Conceptual Question
PHXII07:ALTERNATING CURRENT
356324
A transformer rated at \(10\;kW\) is used to connect a \(5\;KV\) transmission line to \(240\;V\) circuit. The ratio of turns in the windings of a transformer?
1 5
2 20.8
3 104
4 40
Explanation:
\(5000\;V\) to be connected by \(240\;V\) so, This is a step down transformer. \(\dfrac{N_{p}}{N_{s}}=\dfrac{5000}{240}=20.8\) \(\Rightarrow\) Primary has 20.8 times the turns in the secondary. So, correct option is (2).
356321
Primary side of a transformer is connected to \(230\;\,V,50\;\,Hz\) supply. Turns ratio of primary to secondary winding is \(10: 1\). Load resistance connected to secondary side is \(46\,\Omega .\) The power consumed in it is:
1 \(10.0\;\,W\)
2 \(11.5\;\,W\)
3 \(12.5\;\,W\)
4 \(12.0\;\,W\)
Explanation:
Induced voltage \(\propto\) Number of turns in winding. \(\therefore \dfrac{V_{2}}{V_{1}}=\dfrac{N_{2}}{N_{1}}=\dfrac{1}{10}\) \({V_2} = \frac{{{V_1}}}{{10}} = \frac{{230}}{{10}} = 23\;\,V\) \(P_{L}=\dfrac{V_{2}^{2}}{R_{L}}=\dfrac{(23)^{2}}{46}\) \({P_L} = 11.5\;\,W\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356322
The output voltage of a transformer connected to \(220\, V\) line is \(1100\;V\) at \(2 A\) current. Its efficiency is \(100\% \). The current coming from the line is
356323
A step up transformer is connected on the primary side to a \(DC\) battery which can deliver a large current. If a bulb is connected in the secondary, then
1 The bulb will glow very bright
2 The bulb will get fused
3 The bulb will glow, but with less brightness
4 The bulb will not glow
Explanation:
Conceptual Question
PHXII07:ALTERNATING CURRENT
356324
A transformer rated at \(10\;kW\) is used to connect a \(5\;KV\) transmission line to \(240\;V\) circuit. The ratio of turns in the windings of a transformer?
1 5
2 20.8
3 104
4 40
Explanation:
\(5000\;V\) to be connected by \(240\;V\) so, This is a step down transformer. \(\dfrac{N_{p}}{N_{s}}=\dfrac{5000}{240}=20.8\) \(\Rightarrow\) Primary has 20.8 times the turns in the secondary. So, correct option is (2).
356321
Primary side of a transformer is connected to \(230\;\,V,50\;\,Hz\) supply. Turns ratio of primary to secondary winding is \(10: 1\). Load resistance connected to secondary side is \(46\,\Omega .\) The power consumed in it is:
1 \(10.0\;\,W\)
2 \(11.5\;\,W\)
3 \(12.5\;\,W\)
4 \(12.0\;\,W\)
Explanation:
Induced voltage \(\propto\) Number of turns in winding. \(\therefore \dfrac{V_{2}}{V_{1}}=\dfrac{N_{2}}{N_{1}}=\dfrac{1}{10}\) \({V_2} = \frac{{{V_1}}}{{10}} = \frac{{230}}{{10}} = 23\;\,V\) \(P_{L}=\dfrac{V_{2}^{2}}{R_{L}}=\dfrac{(23)^{2}}{46}\) \({P_L} = 11.5\;\,W\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356322
The output voltage of a transformer connected to \(220\, V\) line is \(1100\;V\) at \(2 A\) current. Its efficiency is \(100\% \). The current coming from the line is
356323
A step up transformer is connected on the primary side to a \(DC\) battery which can deliver a large current. If a bulb is connected in the secondary, then
1 The bulb will glow very bright
2 The bulb will get fused
3 The bulb will glow, but with less brightness
4 The bulb will not glow
Explanation:
Conceptual Question
PHXII07:ALTERNATING CURRENT
356324
A transformer rated at \(10\;kW\) is used to connect a \(5\;KV\) transmission line to \(240\;V\) circuit. The ratio of turns in the windings of a transformer?
1 5
2 20.8
3 104
4 40
Explanation:
\(5000\;V\) to be connected by \(240\;V\) so, This is a step down transformer. \(\dfrac{N_{p}}{N_{s}}=\dfrac{5000}{240}=20.8\) \(\Rightarrow\) Primary has 20.8 times the turns in the secondary. So, correct option is (2).
