Basic Quantities in AC
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PHXII07:ALTERNATING CURRENT

356080 Using an \(AC\) voltmeter the potential difference in the electrical line in a house is read to be \(234\;V\). If line frequency is known to be 50
cycles \(/s\), the equation for the line voltage is

1 \(V=165 \sin (100\, \pi t)\)
2 \(V=331 \sin (100\, \pi t)\)
3 \(V=220 \sin (100\, \pi t)\)
4 \(V=440 \sin (100\, \pi t)\)
PHXII07:ALTERNATING CURRENT

356081 The frequency of an alternating current is 50 \(Hz\), What is the minimum time taken by current to reach its peak value from \(rms\) value ?

1 \(0.2\,{\rm{s}}\)
2 \(5 \times {10^{ - 3}}\,{\rm{s}}\)
3 \(10 \times {10^{ - 3}}\,{\rm{s}}\)
4 \(2.5 \times {10^{ - 3}}\,{\rm{s}}\)
PHXII07:ALTERNATING CURRENT

356082 An \(ac\) is given by equation \(I = {I_1}\cos \omega t + {I_2}\sin \omega t.\) The \(rms\) value of current is given by

1 \(\frac{{{{\left( {{I_1} + {I_2}} \right)}^2}}}{{\sqrt 2 }}\)
2 \(\frac{{I_1^2 + I_2^2}}{2}\)
3 \(\frac{{{I_1} + {I_2}}}{2}\)
4 \(\sqrt {\frac{{I_1^2 + I_2^2}}{2}} \)
PHXII07:ALTERNATING CURRENT

356083 In an a.c circuit, the \(r m s\) voltage is \(100\sqrt 2 \,V\). The peak value of voltage and its mean value during a positive half cycle are

1 \(200\;V,127.4\;V\)
2 \(100\;V,127.4\;V\)
3 \(120\;V,150\;V\)
4 \(200\;V,150\;V\)
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PHXII07:ALTERNATING CURRENT

356080 Using an \(AC\) voltmeter the potential difference in the electrical line in a house is read to be \(234\;V\). If line frequency is known to be 50
cycles \(/s\), the equation for the line voltage is

1 \(V=165 \sin (100\, \pi t)\)
2 \(V=331 \sin (100\, \pi t)\)
3 \(V=220 \sin (100\, \pi t)\)
4 \(V=440 \sin (100\, \pi t)\)
PHXII07:ALTERNATING CURRENT

356081 The frequency of an alternating current is 50 \(Hz\), What is the minimum time taken by current to reach its peak value from \(rms\) value ?

1 \(0.2\,{\rm{s}}\)
2 \(5 \times {10^{ - 3}}\,{\rm{s}}\)
3 \(10 \times {10^{ - 3}}\,{\rm{s}}\)
4 \(2.5 \times {10^{ - 3}}\,{\rm{s}}\)
PHXII07:ALTERNATING CURRENT

356082 An \(ac\) is given by equation \(I = {I_1}\cos \omega t + {I_2}\sin \omega t.\) The \(rms\) value of current is given by

1 \(\frac{{{{\left( {{I_1} + {I_2}} \right)}^2}}}{{\sqrt 2 }}\)
2 \(\frac{{I_1^2 + I_2^2}}{2}\)
3 \(\frac{{{I_1} + {I_2}}}{2}\)
4 \(\sqrt {\frac{{I_1^2 + I_2^2}}{2}} \)
PHXII07:ALTERNATING CURRENT

356083 In an a.c circuit, the \(r m s\) voltage is \(100\sqrt 2 \,V\). The peak value of voltage and its mean value during a positive half cycle are

1 \(200\;V,127.4\;V\)
2 \(100\;V,127.4\;V\)
3 \(120\;V,150\;V\)
4 \(200\;V,150\;V\)
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PHXII07:ALTERNATING CURRENT

356080 Using an \(AC\) voltmeter the potential difference in the electrical line in a house is read to be \(234\;V\). If line frequency is known to be 50
cycles \(/s\), the equation for the line voltage is

1 \(V=165 \sin (100\, \pi t)\)
2 \(V=331 \sin (100\, \pi t)\)
3 \(V=220 \sin (100\, \pi t)\)
4 \(V=440 \sin (100\, \pi t)\)
PHXII07:ALTERNATING CURRENT

356081 The frequency of an alternating current is 50 \(Hz\), What is the minimum time taken by current to reach its peak value from \(rms\) value ?

1 \(0.2\,{\rm{s}}\)
2 \(5 \times {10^{ - 3}}\,{\rm{s}}\)
3 \(10 \times {10^{ - 3}}\,{\rm{s}}\)
4 \(2.5 \times {10^{ - 3}}\,{\rm{s}}\)
PHXII07:ALTERNATING CURRENT

356082 An \(ac\) is given by equation \(I = {I_1}\cos \omega t + {I_2}\sin \omega t.\) The \(rms\) value of current is given by

1 \(\frac{{{{\left( {{I_1} + {I_2}} \right)}^2}}}{{\sqrt 2 }}\)
2 \(\frac{{I_1^2 + I_2^2}}{2}\)
3 \(\frac{{{I_1} + {I_2}}}{2}\)
4 \(\sqrt {\frac{{I_1^2 + I_2^2}}{2}} \)
PHXII07:ALTERNATING CURRENT

356083 In an a.c circuit, the \(r m s\) voltage is \(100\sqrt 2 \,V\). The peak value of voltage and its mean value during a positive half cycle are

1 \(200\;V,127.4\;V\)
2 \(100\;V,127.4\;V\)
3 \(120\;V,150\;V\)
4 \(200\;V,150\;V\)
For More Updates join with Us
PHXII07:ALTERNATING CURRENT

356080 Using an \(AC\) voltmeter the potential difference in the electrical line in a house is read to be \(234\;V\). If line frequency is known to be 50
cycles \(/s\), the equation for the line voltage is

1 \(V=165 \sin (100\, \pi t)\)
2 \(V=331 \sin (100\, \pi t)\)
3 \(V=220 \sin (100\, \pi t)\)
4 \(V=440 \sin (100\, \pi t)\)
PHXII07:ALTERNATING CURRENT

356081 The frequency of an alternating current is 50 \(Hz\), What is the minimum time taken by current to reach its peak value from \(rms\) value ?

1 \(0.2\,{\rm{s}}\)
2 \(5 \times {10^{ - 3}}\,{\rm{s}}\)
3 \(10 \times {10^{ - 3}}\,{\rm{s}}\)
4 \(2.5 \times {10^{ - 3}}\,{\rm{s}}\)
PHXII07:ALTERNATING CURRENT

356082 An \(ac\) is given by equation \(I = {I_1}\cos \omega t + {I_2}\sin \omega t.\) The \(rms\) value of current is given by

1 \(\frac{{{{\left( {{I_1} + {I_2}} \right)}^2}}}{{\sqrt 2 }}\)
2 \(\frac{{I_1^2 + I_2^2}}{2}\)
3 \(\frac{{{I_1} + {I_2}}}{2}\)
4 \(\sqrt {\frac{{I_1^2 + I_2^2}}{2}} \)
PHXII07:ALTERNATING CURRENT

356083 In an a.c circuit, the \(r m s\) voltage is \(100\sqrt 2 \,V\). The peak value of voltage and its mean value during a positive half cycle are

1 \(200\;V,127.4\;V\)
2 \(100\;V,127.4\;V\)
3 \(120\;V,150\;V\)
4 \(200\;V,150\;V\)
NEET DIGITAL LIBRARY