NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII07:ALTERNATING CURRENT
356338
A transformer of 100% efficiency has 200 turns in the primary and 40000 turns in secondary. It is connected to a 220 \(V\) main supply and secondary feeds to a \(100\,k\Omega \) resistance. The potential difference per turn is:
1 \(1.1\,V\)
2 \(25\,V\)
3 \(18\,V\)
4 \(11\,V\)
Explanation:
From transformer ratio, \(\frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}} \Rightarrow {V_s} = \frac{{{V_p} \times {N_s}}}{{{N_p}}}\) \({V_s} = \frac{{220 \times 40000}}{{200}} = 44000V\) Potential difference per turn is \(\frac{{{V_s}}}{{{N_s}}} = \frac{{44000}}{{40000}} = 1.1V\)
PHXII07:ALTERNATING CURRENT
356339
How much current is drawn by the primary coil of a transformer which steps down \(220\,volt\) to \(22\,volt\) to operate a device with an impedance of \(220\,\Omega \)?
356340
A power transformer is used to step up an alternating emf of \(220\,volt\) to \(11\,kV\) to transmit \(4.4\,kW\) of power. If the primary coil has 1000 turns, what is the current in the secondary coil?
356341
A transformer with turns ratio \(\dfrac{N_{1}}{N_{2}}=\dfrac{50}{1}\) is connected to a \(120\,V\,AC\) supply. If primary and secondary circuit resistance are \(1.5\,k\,\Omega \) and \(1\,\Omega \), then find out power of output.
1 \(5.76\;W\)
2 \(11.4\;W\)
3 \(2.89\;W\)
4 \(7.56\;W\)
Explanation:
Given, turn ratio of a transformer, \(\dfrac{N_{1}}{N_{2}}=\dfrac{50}{1}\) \(\Rightarrow N_{1}=50 N_{2}\) Since, \(\dfrac{N_{1}}{N_{2}}=\dfrac{V_{1}}{V_{2}}\) \(50 = \frac{{120}}{{{V_2}}}\left[ {{V_1} = 120\;V(} \right.\)given\(\left.)\right]\) \( \Rightarrow {V_2} = \frac{{12}}{5}\;V\) Output power at secondary coil, \(P_{s}=\dfrac{V_{2}^{2}}{R_{2}} \quad\left[R_{2}=1 \Omega(\right.\) given \(\left.)\right]\) \(=\dfrac{\left(\dfrac{12}{5}\right)^{2}}{1}=\dfrac{144}{25}\) \( = 5.76\;W\)
356338
A transformer of 100% efficiency has 200 turns in the primary and 40000 turns in secondary. It is connected to a 220 \(V\) main supply and secondary feeds to a \(100\,k\Omega \) resistance. The potential difference per turn is:
1 \(1.1\,V\)
2 \(25\,V\)
3 \(18\,V\)
4 \(11\,V\)
Explanation:
From transformer ratio, \(\frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}} \Rightarrow {V_s} = \frac{{{V_p} \times {N_s}}}{{{N_p}}}\) \({V_s} = \frac{{220 \times 40000}}{{200}} = 44000V\) Potential difference per turn is \(\frac{{{V_s}}}{{{N_s}}} = \frac{{44000}}{{40000}} = 1.1V\)
PHXII07:ALTERNATING CURRENT
356339
How much current is drawn by the primary coil of a transformer which steps down \(220\,volt\) to \(22\,volt\) to operate a device with an impedance of \(220\,\Omega \)?
356340
A power transformer is used to step up an alternating emf of \(220\,volt\) to \(11\,kV\) to transmit \(4.4\,kW\) of power. If the primary coil has 1000 turns, what is the current in the secondary coil?
356341
A transformer with turns ratio \(\dfrac{N_{1}}{N_{2}}=\dfrac{50}{1}\) is connected to a \(120\,V\,AC\) supply. If primary and secondary circuit resistance are \(1.5\,k\,\Omega \) and \(1\,\Omega \), then find out power of output.
1 \(5.76\;W\)
2 \(11.4\;W\)
3 \(2.89\;W\)
4 \(7.56\;W\)
Explanation:
Given, turn ratio of a transformer, \(\dfrac{N_{1}}{N_{2}}=\dfrac{50}{1}\) \(\Rightarrow N_{1}=50 N_{2}\) Since, \(\dfrac{N_{1}}{N_{2}}=\dfrac{V_{1}}{V_{2}}\) \(50 = \frac{{120}}{{{V_2}}}\left[ {{V_1} = 120\;V(} \right.\)given\(\left.)\right]\) \( \Rightarrow {V_2} = \frac{{12}}{5}\;V\) Output power at secondary coil, \(P_{s}=\dfrac{V_{2}^{2}}{R_{2}} \quad\left[R_{2}=1 \Omega(\right.\) given \(\left.)\right]\) \(=\dfrac{\left(\dfrac{12}{5}\right)^{2}}{1}=\dfrac{144}{25}\) \( = 5.76\;W\)
356338
A transformer of 100% efficiency has 200 turns in the primary and 40000 turns in secondary. It is connected to a 220 \(V\) main supply and secondary feeds to a \(100\,k\Omega \) resistance. The potential difference per turn is:
1 \(1.1\,V\)
2 \(25\,V\)
3 \(18\,V\)
4 \(11\,V\)
Explanation:
From transformer ratio, \(\frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}} \Rightarrow {V_s} = \frac{{{V_p} \times {N_s}}}{{{N_p}}}\) \({V_s} = \frac{{220 \times 40000}}{{200}} = 44000V\) Potential difference per turn is \(\frac{{{V_s}}}{{{N_s}}} = \frac{{44000}}{{40000}} = 1.1V\)
PHXII07:ALTERNATING CURRENT
356339
How much current is drawn by the primary coil of a transformer which steps down \(220\,volt\) to \(22\,volt\) to operate a device with an impedance of \(220\,\Omega \)?
356340
A power transformer is used to step up an alternating emf of \(220\,volt\) to \(11\,kV\) to transmit \(4.4\,kW\) of power. If the primary coil has 1000 turns, what is the current in the secondary coil?
356341
A transformer with turns ratio \(\dfrac{N_{1}}{N_{2}}=\dfrac{50}{1}\) is connected to a \(120\,V\,AC\) supply. If primary and secondary circuit resistance are \(1.5\,k\,\Omega \) and \(1\,\Omega \), then find out power of output.
1 \(5.76\;W\)
2 \(11.4\;W\)
3 \(2.89\;W\)
4 \(7.56\;W\)
Explanation:
Given, turn ratio of a transformer, \(\dfrac{N_{1}}{N_{2}}=\dfrac{50}{1}\) \(\Rightarrow N_{1}=50 N_{2}\) Since, \(\dfrac{N_{1}}{N_{2}}=\dfrac{V_{1}}{V_{2}}\) \(50 = \frac{{120}}{{{V_2}}}\left[ {{V_1} = 120\;V(} \right.\)given\(\left.)\right]\) \( \Rightarrow {V_2} = \frac{{12}}{5}\;V\) Output power at secondary coil, \(P_{s}=\dfrac{V_{2}^{2}}{R_{2}} \quad\left[R_{2}=1 \Omega(\right.\) given \(\left.)\right]\) \(=\dfrac{\left(\dfrac{12}{5}\right)^{2}}{1}=\dfrac{144}{25}\) \( = 5.76\;W\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII07:ALTERNATING CURRENT
356338
A transformer of 100% efficiency has 200 turns in the primary and 40000 turns in secondary. It is connected to a 220 \(V\) main supply and secondary feeds to a \(100\,k\Omega \) resistance. The potential difference per turn is:
1 \(1.1\,V\)
2 \(25\,V\)
3 \(18\,V\)
4 \(11\,V\)
Explanation:
From transformer ratio, \(\frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}} \Rightarrow {V_s} = \frac{{{V_p} \times {N_s}}}{{{N_p}}}\) \({V_s} = \frac{{220 \times 40000}}{{200}} = 44000V\) Potential difference per turn is \(\frac{{{V_s}}}{{{N_s}}} = \frac{{44000}}{{40000}} = 1.1V\)
PHXII07:ALTERNATING CURRENT
356339
How much current is drawn by the primary coil of a transformer which steps down \(220\,volt\) to \(22\,volt\) to operate a device with an impedance of \(220\,\Omega \)?
356340
A power transformer is used to step up an alternating emf of \(220\,volt\) to \(11\,kV\) to transmit \(4.4\,kW\) of power. If the primary coil has 1000 turns, what is the current in the secondary coil?
356341
A transformer with turns ratio \(\dfrac{N_{1}}{N_{2}}=\dfrac{50}{1}\) is connected to a \(120\,V\,AC\) supply. If primary and secondary circuit resistance are \(1.5\,k\,\Omega \) and \(1\,\Omega \), then find out power of output.
1 \(5.76\;W\)
2 \(11.4\;W\)
3 \(2.89\;W\)
4 \(7.56\;W\)
Explanation:
Given, turn ratio of a transformer, \(\dfrac{N_{1}}{N_{2}}=\dfrac{50}{1}\) \(\Rightarrow N_{1}=50 N_{2}\) Since, \(\dfrac{N_{1}}{N_{2}}=\dfrac{V_{1}}{V_{2}}\) \(50 = \frac{{120}}{{{V_2}}}\left[ {{V_1} = 120\;V(} \right.\)given\(\left.)\right]\) \( \Rightarrow {V_2} = \frac{{12}}{5}\;V\) Output power at secondary coil, \(P_{s}=\dfrac{V_{2}^{2}}{R_{2}} \quad\left[R_{2}=1 \Omega(\right.\) given \(\left.)\right]\) \(=\dfrac{\left(\dfrac{12}{5}\right)^{2}}{1}=\dfrac{144}{25}\) \( = 5.76\;W\)