356166
The average power dissipated in pure inductor is
1 \(V{I^2}\)
2 \({\rm{zero}}\)
3 \(\frac{1}{2}VI\)
4 \(\frac{{V{I^2}}}{4}\)
Explanation:
The average power dissipated in a pure inductor is zero, as the phase angle between voltage and current is always \(\pi /2\)
PHXII07:ALTERNATING CURRENT
356167
An \(a c\) supply gives \({(30\;V)_{r,m.s}}\) which passed through a \(100\,\Omega \) resistance. The power dissipated in it is:
1 \(30\;W\)
2 \(90\;W\)
3 \(40\,W\)
4 \(9\;W\)
Explanation:
\(P = \left( {\frac{{V_{rms}^2}}{R}} \right) = \frac{{{{(30)}^2}}}{{100}} = \frac{{900}}{{100}} = 9\;W\) So, correct option is (4).
PHXII07:ALTERNATING CURRENT
356168
Power factor value is one for
1 Pure inductor
2 Pure resistor
3 Either an inductor or a capacitor
4 Pure capacitor
Explanation:
Conceptual Question
PHXII07:ALTERNATING CURRENT
356169
A group of electric lamps having a total power rating of 1000 watt is supplied by an \(AC\) voltage \(E = 200\sin \left( {3\,10t + 60^\circ } \right).\) Then the rms value of the circuit current is[ Here \(i = {i_o}\sin 310t\)]
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII07:ALTERNATING CURRENT
356166
The average power dissipated in pure inductor is
1 \(V{I^2}\)
2 \({\rm{zero}}\)
3 \(\frac{1}{2}VI\)
4 \(\frac{{V{I^2}}}{4}\)
Explanation:
The average power dissipated in a pure inductor is zero, as the phase angle between voltage and current is always \(\pi /2\)
PHXII07:ALTERNATING CURRENT
356167
An \(a c\) supply gives \({(30\;V)_{r,m.s}}\) which passed through a \(100\,\Omega \) resistance. The power dissipated in it is:
1 \(30\;W\)
2 \(90\;W\)
3 \(40\,W\)
4 \(9\;W\)
Explanation:
\(P = \left( {\frac{{V_{rms}^2}}{R}} \right) = \frac{{{{(30)}^2}}}{{100}} = \frac{{900}}{{100}} = 9\;W\) So, correct option is (4).
PHXII07:ALTERNATING CURRENT
356168
Power factor value is one for
1 Pure inductor
2 Pure resistor
3 Either an inductor or a capacitor
4 Pure capacitor
Explanation:
Conceptual Question
PHXII07:ALTERNATING CURRENT
356169
A group of electric lamps having a total power rating of 1000 watt is supplied by an \(AC\) voltage \(E = 200\sin \left( {3\,10t + 60^\circ } \right).\) Then the rms value of the circuit current is[ Here \(i = {i_o}\sin 310t\)]
356166
The average power dissipated in pure inductor is
1 \(V{I^2}\)
2 \({\rm{zero}}\)
3 \(\frac{1}{2}VI\)
4 \(\frac{{V{I^2}}}{4}\)
Explanation:
The average power dissipated in a pure inductor is zero, as the phase angle between voltage and current is always \(\pi /2\)
PHXII07:ALTERNATING CURRENT
356167
An \(a c\) supply gives \({(30\;V)_{r,m.s}}\) which passed through a \(100\,\Omega \) resistance. The power dissipated in it is:
1 \(30\;W\)
2 \(90\;W\)
3 \(40\,W\)
4 \(9\;W\)
Explanation:
\(P = \left( {\frac{{V_{rms}^2}}{R}} \right) = \frac{{{{(30)}^2}}}{{100}} = \frac{{900}}{{100}} = 9\;W\) So, correct option is (4).
PHXII07:ALTERNATING CURRENT
356168
Power factor value is one for
1 Pure inductor
2 Pure resistor
3 Either an inductor or a capacitor
4 Pure capacitor
Explanation:
Conceptual Question
PHXII07:ALTERNATING CURRENT
356169
A group of electric lamps having a total power rating of 1000 watt is supplied by an \(AC\) voltage \(E = 200\sin \left( {3\,10t + 60^\circ } \right).\) Then the rms value of the circuit current is[ Here \(i = {i_o}\sin 310t\)]
356166
The average power dissipated in pure inductor is
1 \(V{I^2}\)
2 \({\rm{zero}}\)
3 \(\frac{1}{2}VI\)
4 \(\frac{{V{I^2}}}{4}\)
Explanation:
The average power dissipated in a pure inductor is zero, as the phase angle between voltage and current is always \(\pi /2\)
PHXII07:ALTERNATING CURRENT
356167
An \(a c\) supply gives \({(30\;V)_{r,m.s}}\) which passed through a \(100\,\Omega \) resistance. The power dissipated in it is:
1 \(30\;W\)
2 \(90\;W\)
3 \(40\,W\)
4 \(9\;W\)
Explanation:
\(P = \left( {\frac{{V_{rms}^2}}{R}} \right) = \frac{{{{(30)}^2}}}{{100}} = \frac{{900}}{{100}} = 9\;W\) So, correct option is (4).
PHXII07:ALTERNATING CURRENT
356168
Power factor value is one for
1 Pure inductor
2 Pure resistor
3 Either an inductor or a capacitor
4 Pure capacitor
Explanation:
Conceptual Question
PHXII07:ALTERNATING CURRENT
356169
A group of electric lamps having a total power rating of 1000 watt is supplied by an \(AC\) voltage \(E = 200\sin \left( {3\,10t + 60^\circ } \right).\) Then the rms value of the circuit current is[ Here \(i = {i_o}\sin 310t\)]
