356196
In an \(A.C.\) circuit, \(V\) and \(I\) are given by\({V}=50 \sin [1000 {t}]\) volt and\(I=500 \sin \left[1000 t+\dfrac{\pi}{3}\right] {mA}\). What would the power dissipated in the circuit ?
356197
In an a.c. circuit, voltage and current are given by \(V=100 \sin (100 t) V\) and \(I=100 \sin \left(100 t+\dfrac{\pi}{3}\right) m A\) respectively. The average power dissipated in one cycle is
1 \(10\,W\)
2 \(5\,W\)
3 \(25\,W\)
4 \(2.5\,W\)
Explanation:
Given, \(V=100 \sin (100 t) V\) \(I=100 \sin (100 t+\pi / 3) m A\) Phase difference between \(V\) and \(I, \phi=\dfrac{\pi}{3}\) \({V_0} = 100\;V,{I_0} = 100\;mA,\) Average power dissipated in one cycle is \(P_{a v}=I_{r m s} \times V_{r m s} \cos \phi\) \(P_{a v}=\dfrac{100}{\sqrt{2}} \times \dfrac{100}{\sqrt{2}}\left(\cos \dfrac{\pi}{3}\right) \times 10^{-3}\) \( = \frac{{10}}{2} \times \frac{1}{2} = 2.5\;W\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356198
Assertion : When capacitive reactance is smaller than the inductive reactance in LCR circuit, e.m.f. leads the current. Reason : Power is zero when power factor is 1.
1 Both assertion and reason are correct and reason isthe correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(\tan \phi = \frac{{{X_L} - {X_C}}}{R} = \frac{{\omega L - \frac{1}{{\omega C}}}}{R}\) When \({X_L} > {X_C}\) then \(\tan \phi \) is positive, i.e., \(\phi \) ispositive (between 0 and \(\frac{\pi }{2}\) ). Hence e.m.f. leads the current. So option (3) is correct.
PHXII07:ALTERNATING CURRENT
356199
The instantaneous value of alternating current and voltages in a circuit are given as \(I = \frac{1}{{\sqrt 2 }}\sin (100\pi t)amp\) \(e = \frac{1}{{\sqrt 2 }}\sin (100\pi t + \pi /3)volt\) The average power in watts consumed in the circuit is:
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PHXII07:ALTERNATING CURRENT
356196
In an \(A.C.\) circuit, \(V\) and \(I\) are given by\({V}=50 \sin [1000 {t}]\) volt and\(I=500 \sin \left[1000 t+\dfrac{\pi}{3}\right] {mA}\). What would the power dissipated in the circuit ?
356197
In an a.c. circuit, voltage and current are given by \(V=100 \sin (100 t) V\) and \(I=100 \sin \left(100 t+\dfrac{\pi}{3}\right) m A\) respectively. The average power dissipated in one cycle is
1 \(10\,W\)
2 \(5\,W\)
3 \(25\,W\)
4 \(2.5\,W\)
Explanation:
Given, \(V=100 \sin (100 t) V\) \(I=100 \sin (100 t+\pi / 3) m A\) Phase difference between \(V\) and \(I, \phi=\dfrac{\pi}{3}\) \({V_0} = 100\;V,{I_0} = 100\;mA,\) Average power dissipated in one cycle is \(P_{a v}=I_{r m s} \times V_{r m s} \cos \phi\) \(P_{a v}=\dfrac{100}{\sqrt{2}} \times \dfrac{100}{\sqrt{2}}\left(\cos \dfrac{\pi}{3}\right) \times 10^{-3}\) \( = \frac{{10}}{2} \times \frac{1}{2} = 2.5\;W\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356198
Assertion : When capacitive reactance is smaller than the inductive reactance in LCR circuit, e.m.f. leads the current. Reason : Power is zero when power factor is 1.
1 Both assertion and reason are correct and reason isthe correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(\tan \phi = \frac{{{X_L} - {X_C}}}{R} = \frac{{\omega L - \frac{1}{{\omega C}}}}{R}\) When \({X_L} > {X_C}\) then \(\tan \phi \) is positive, i.e., \(\phi \) ispositive (between 0 and \(\frac{\pi }{2}\) ). Hence e.m.f. leads the current. So option (3) is correct.
PHXII07:ALTERNATING CURRENT
356199
The instantaneous value of alternating current and voltages in a circuit are given as \(I = \frac{1}{{\sqrt 2 }}\sin (100\pi t)amp\) \(e = \frac{1}{{\sqrt 2 }}\sin (100\pi t + \pi /3)volt\) The average power in watts consumed in the circuit is:
356196
In an \(A.C.\) circuit, \(V\) and \(I\) are given by\({V}=50 \sin [1000 {t}]\) volt and\(I=500 \sin \left[1000 t+\dfrac{\pi}{3}\right] {mA}\). What would the power dissipated in the circuit ?
356197
In an a.c. circuit, voltage and current are given by \(V=100 \sin (100 t) V\) and \(I=100 \sin \left(100 t+\dfrac{\pi}{3}\right) m A\) respectively. The average power dissipated in one cycle is
1 \(10\,W\)
2 \(5\,W\)
3 \(25\,W\)
4 \(2.5\,W\)
Explanation:
Given, \(V=100 \sin (100 t) V\) \(I=100 \sin (100 t+\pi / 3) m A\) Phase difference between \(V\) and \(I, \phi=\dfrac{\pi}{3}\) \({V_0} = 100\;V,{I_0} = 100\;mA,\) Average power dissipated in one cycle is \(P_{a v}=I_{r m s} \times V_{r m s} \cos \phi\) \(P_{a v}=\dfrac{100}{\sqrt{2}} \times \dfrac{100}{\sqrt{2}}\left(\cos \dfrac{\pi}{3}\right) \times 10^{-3}\) \( = \frac{{10}}{2} \times \frac{1}{2} = 2.5\;W\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356198
Assertion : When capacitive reactance is smaller than the inductive reactance in LCR circuit, e.m.f. leads the current. Reason : Power is zero when power factor is 1.
1 Both assertion and reason are correct and reason isthe correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(\tan \phi = \frac{{{X_L} - {X_C}}}{R} = \frac{{\omega L - \frac{1}{{\omega C}}}}{R}\) When \({X_L} > {X_C}\) then \(\tan \phi \) is positive, i.e., \(\phi \) ispositive (between 0 and \(\frac{\pi }{2}\) ). Hence e.m.f. leads the current. So option (3) is correct.
PHXII07:ALTERNATING CURRENT
356199
The instantaneous value of alternating current and voltages in a circuit are given as \(I = \frac{1}{{\sqrt 2 }}\sin (100\pi t)amp\) \(e = \frac{1}{{\sqrt 2 }}\sin (100\pi t + \pi /3)volt\) The average power in watts consumed in the circuit is:
356196
In an \(A.C.\) circuit, \(V\) and \(I\) are given by\({V}=50 \sin [1000 {t}]\) volt and\(I=500 \sin \left[1000 t+\dfrac{\pi}{3}\right] {mA}\). What would the power dissipated in the circuit ?
356197
In an a.c. circuit, voltage and current are given by \(V=100 \sin (100 t) V\) and \(I=100 \sin \left(100 t+\dfrac{\pi}{3}\right) m A\) respectively. The average power dissipated in one cycle is
1 \(10\,W\)
2 \(5\,W\)
3 \(25\,W\)
4 \(2.5\,W\)
Explanation:
Given, \(V=100 \sin (100 t) V\) \(I=100 \sin (100 t+\pi / 3) m A\) Phase difference between \(V\) and \(I, \phi=\dfrac{\pi}{3}\) \({V_0} = 100\;V,{I_0} = 100\;mA,\) Average power dissipated in one cycle is \(P_{a v}=I_{r m s} \times V_{r m s} \cos \phi\) \(P_{a v}=\dfrac{100}{\sqrt{2}} \times \dfrac{100}{\sqrt{2}}\left(\cos \dfrac{\pi}{3}\right) \times 10^{-3}\) \( = \frac{{10}}{2} \times \frac{1}{2} = 2.5\;W\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356198
Assertion : When capacitive reactance is smaller than the inductive reactance in LCR circuit, e.m.f. leads the current. Reason : Power is zero when power factor is 1.
1 Both assertion and reason are correct and reason isthe correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(\tan \phi = \frac{{{X_L} - {X_C}}}{R} = \frac{{\omega L - \frac{1}{{\omega C}}}}{R}\) When \({X_L} > {X_C}\) then \(\tan \phi \) is positive, i.e., \(\phi \) ispositive (between 0 and \(\frac{\pi }{2}\) ). Hence e.m.f. leads the current. So option (3) is correct.
PHXII07:ALTERNATING CURRENT
356199
The instantaneous value of alternating current and voltages in a circuit are given as \(I = \frac{1}{{\sqrt 2 }}\sin (100\pi t)amp\) \(e = \frac{1}{{\sqrt 2 }}\sin (100\pi t + \pi /3)volt\) The average power in watts consumed in the circuit is:
